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  • C problem, left of '->' must point to class/struct/union/generic type ??

    - by Patrick
    Hello! Trying to understand why this doesn't work. I keep getting the following errors: left of '-nextNode' must point to class/struct/union/generic type (Also all the lines with a - in the function new_math_struct) Header file #ifndef MSTRUCT_H #define MSTRUCT_H #define PLUS 0 #define MINUS 1 #define DIVIDE 2 #define MULTIPLY 3 #define NUMBER 4 typedef struct math_struct { int type_of_value; int value; int sum; int is_used; struct math_struct* nextNode; } ; typedef struct math_struct* math_struct_ptr; #endif C file int get_input(math_struct_ptr* startNode) { /* character, input by the user */ char input_ch; char* input_ptr; math_struct_ptr* ptr; math_struct_ptr* previousNode; input_ptr = &input_ch; previousNode = startNode; /* as long as input is not ok */ while (1) { input_ch = get_input_character(); if (input_ch == ',') // Carrage return return 1; else if (input_ch == '.') // Illegal character return 0; if (input_ch == '+') ptr = new_math_struct(PLUS, 0); else if (input_ch == '-') ptr = new_math_struct(MINUS, 0); else if (input_ch == '/') ptr = new_math_struct(DIVIDE, 0); else if (input_ch == '*') ptr = new_math_struct(MULTIPLY, 0); else ptr = new_math_struct(NUMBER, atoi(input_ptr)); if (startNode == NULL) { startNode = previousNode = ptr; } else { previousNode->nextNode = ptr; previousNode = ptr; } } return 0; } math_struct_ptr* new_math_struct(int symbol, int value) { math_struct_ptr* ptr; ptr = (math_struct_ptr*)malloc(sizeof(math_struct_ptr)); ptr->type_of_value = symbol; ptr->value = value; ptr->sum = 0; ptr->is_used = 0; return ptr; } char get_input_character() { /* character, input by the user */ char input_ch; /* get the character */ scanf("%c", &input_ch); if (input_ch == '+' || input_ch == '-' || input_ch == '*' || input_ch == '/' || input_ch == ')') return input_ch; // A special character else if (input_ch == '\n') return ','; // A carrage return else if (input_ch < '0' || input_ch > '9') return '.'; // Not a number else return input_ch; // Number } The header for the C file just contains a reference to the struct header and the definitions of the functions. Language C.

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  • Returning the size of available virtual memory at run-time in C++

    - by Greenhouse Gases
    In C++ is there a predefined library function that will return the size of RAM currently available on a computer a program is being run on, at run-time? For instance, if an object is 4bytes, then can we divide the available virtual memory by 4 bytes to give an estimate of how many more objects could be stored by the program safely? I have used the sizeof() function to return the size of objects within my program. Thanks

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  • Exceptions handling in SQL?

    - by Vineet
    Is there any way to handle exceptions in sql(ORACLE 9i)? Since I was trying to divide values of a column that contains both numbers and literals ,I need to fetch out only numbers from it ,as if it divisible by any number then its number else if contains literals it would not get divided it will generate error. how to handle those errors? Please suggest!!

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  • Basic string and value functions in Objective-C for iPhone

    - by David Maitland
    I have very little programming knowledge; only a fair bit in Visual Basic. How do I take a value from a text field, then do some simple math such as divide the value by two, then display it back to the user in the same field? In Visual Basic you could just do txtBoxOne.text = txtBoxOne.text / 2 I understand this question is more than one question and is very basic stuff, but I need to get my head out of Visual Basic and into where I should be :)

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  • Safe division function

    - by bugspy.net
    I would like to define some kind of safe division (and modulo) function, one that would return some predefined value when attempting to divide by zero. I don't want to throw exceptions, just to return some "reasonable" value (1? 0?) and continue the program flow. Obviously there is no correct return value, but I wonder if there is some standard or known approach to this

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  • Move penetrating OBB out of another OBB to resolve collision

    - by Milo
    I'm working on collision resolution for my game. I just need a good way to get an object out of another object if it gets stuck. In this case a car. Here is a typical scenario. The red car is in the green object. How do I correctly get it out so the car can slide along the edge of the object as it should. I tried: if(buildings.size() > 0) { Entity e = buildings.get(0); Vector2D vel = new Vector2D(); vel.x = vehicle.getVelocity().x; vel.y = vehicle.getVelocity().y; vel.normalize(); while(vehicle.getRect().overlaps(e.getRect())) { vehicle.setCenter(vehicle.getCenterX() - vel.x * 0.1f, vehicle.getCenterY() - vel.y * 0.1f); } colided = true; } But that does not work too well. Is there some sort of vector I could calculate to use as the vector to move the car away from the object? Thanks Here is my OBB2D class: public class OBB2D { // Corners of the box, where 0 is the lower left. private Vector2D corner[] = new Vector2D[4]; private Vector2D center = new Vector2D(); private Vector2D extents = new Vector2D(); private RectF boundingRect = new RectF(); private float angle; //Two edges of the box extended away from corner[0]. private Vector2D axis[] = new Vector2D[2]; private double origin[] = new double[2]; public OBB2D(Vector2D center, float w, float h, float angle) { set(center,w,h,angle); } public OBB2D(float left, float top, float width, float height) { set(new Vector2D(left + (width / 2), top + (height / 2)),width,height,0.0f); } public void set(Vector2D center,float w, float h,float angle) { Vector2D X = new Vector2D( (float)Math.cos(angle), (float)Math.sin(angle)); Vector2D Y = new Vector2D((float)-Math.sin(angle), (float)Math.cos(angle)); X = X.multiply( w / 2); Y = Y.multiply( h / 2); corner[0] = center.subtract(X).subtract(Y); corner[1] = center.add(X).subtract(Y); corner[2] = center.add(X).add(Y); corner[3] = center.subtract(X).add(Y); computeAxes(); extents.x = w / 2; extents.y = h / 2; computeDimensions(center,angle); } private void computeDimensions(Vector2D center,float angle) { this.center.x = center.x; this.center.y = center.y; this.angle = angle; boundingRect.left = Math.min(Math.min(corner[0].x, corner[3].x), Math.min(corner[1].x, corner[2].x)); boundingRect.top = Math.min(Math.min(corner[0].y, corner[1].y),Math.min(corner[2].y, corner[3].y)); boundingRect.right = Math.max(Math.max(corner[1].x, corner[2].x), Math.max(corner[0].x, corner[3].x)); boundingRect.bottom = Math.max(Math.max(corner[2].y, corner[3].y),Math.max(corner[0].y, corner[1].y)); } public void set(RectF rect) { set(new Vector2D(rect.centerX(),rect.centerY()),rect.width(),rect.height(),0.0f); } // Returns true if other overlaps one dimension of this. private boolean overlaps1Way(OBB2D other) { for (int a = 0; a < axis.length; ++a) { double t = other.corner[0].dot(axis[a]); // Find the extent of box 2 on axis a double tMin = t; double tMax = t; for (int c = 1; c < corner.length; ++c) { t = other.corner[c].dot(axis[a]); if (t < tMin) { tMin = t; } else if (t > tMax) { tMax = t; } } // We have to subtract off the origin // See if [tMin, tMax] intersects [0, 1] if ((tMin > 1 + origin[a]) || (tMax < origin[a])) { // There was no intersection along this dimension; // the boxes cannot possibly overlap. return false; } } // There was no dimension along which there is no intersection. // Therefore the boxes overlap. return true; } //Updates the axes after the corners move. Assumes the //corners actually form a rectangle. private void computeAxes() { axis[0] = corner[1].subtract(corner[0]); axis[1] = corner[3].subtract(corner[0]); // Make the length of each axis 1/edge length so we know any // dot product must be less than 1 to fall within the edge. for (int a = 0; a < axis.length; ++a) { axis[a] = axis[a].divide((axis[a].length() * axis[a].length())); origin[a] = corner[0].dot(axis[a]); } } public void moveTo(Vector2D center) { Vector2D centroid = (corner[0].add(corner[1]).add(corner[2]).add(corner[3])).divide(4.0f); Vector2D translation = center.subtract(centroid); for (int c = 0; c < 4; ++c) { corner[c] = corner[c].add(translation); } computeAxes(); computeDimensions(center,angle); } // Returns true if the intersection of the boxes is non-empty. public boolean overlaps(OBB2D other) { if(right() < other.left()) { return false; } if(bottom() < other.top()) { return false; } if(left() > other.right()) { return false; } if(top() > other.bottom()) { return false; } if(other.getAngle() == 0.0f && getAngle() == 0.0f) { return true; } return overlaps1Way(other) && other.overlaps1Way(this); } public Vector2D getCenter() { return center; } public float getWidth() { return extents.x * 2; } public float getHeight() { return extents.y * 2; } public void setAngle(float angle) { set(center,getWidth(),getHeight(),angle); } public float getAngle() { return angle; } public void setSize(float w,float h) { set(center,w,h,angle); } public float left() { return boundingRect.left; } public float right() { return boundingRect.right; } public float bottom() { return boundingRect.bottom; } public float top() { return boundingRect.top; } public RectF getBoundingRect() { return boundingRect; } public boolean overlaps(float left, float top, float right, float bottom) { if(right() < left) { return false; } if(bottom() < top) { return false; } if(left() > right) { return false; } if(top() > bottom) { return false; } return true; } };

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  • Finding the normal of OBB face with an OBB penetrating

    - by Milo
    Below is an illustration: I have an OBB in an OBB (see below for OBB2D code if needed). What I need to determine is, what face it is in, and what direction do I point the normal? The goal is to get the OBB out of the OBB so the normal needs to face outward of the OBB. How could I go about: Finding what face the line is penetrating given the 4 corners of the OBB and the class below: if we define dx=x2-x1 and dy=y2-y1, then the normals are (-dy, dx) and (dy, -dx). Which normal points outward of the OBB? Thanks public class OBB2D { // Corners of the box, where 0 is the lower left. private Vector2D corner[] = new Vector2D[4]; private Vector2D center = new Vector2D(); private Vector2D extents = new Vector2D(); private RectF boundingRect = new RectF(); private float angle; //Two edges of the box extended away from corner[0]. private Vector2D axis[] = new Vector2D[2]; private double origin[] = new double[2]; public OBB2D(Vector2D center, float w, float h, float angle) { set(center,w,h,angle); } public OBB2D(float left, float top, float width, float height) { set(new Vector2D(left + (width / 2), top + (height / 2)),width,height,0.0f); } public void set(Vector2D center,float w, float h,float angle) { Vector2D X = new Vector2D( (float)Math.cos(angle), (float)Math.sin(angle)); Vector2D Y = new Vector2D((float)-Math.sin(angle), (float)Math.cos(angle)); X = X.multiply( w / 2); Y = Y.multiply( h / 2); corner[0] = center.subtract(X).subtract(Y); corner[1] = center.add(X).subtract(Y); corner[2] = center.add(X).add(Y); corner[3] = center.subtract(X).add(Y); computeAxes(); extents.x = w / 2; extents.y = h / 2; computeDimensions(center,angle); } private void computeDimensions(Vector2D center,float angle) { this.center.x = center.x; this.center.y = center.y; this.angle = angle; boundingRect.left = Math.min(Math.min(corner[0].x, corner[3].x), Math.min(corner[1].x, corner[2].x)); boundingRect.top = Math.min(Math.min(corner[0].y, corner[1].y),Math.min(corner[2].y, corner[3].y)); boundingRect.right = Math.max(Math.max(corner[1].x, corner[2].x), Math.max(corner[0].x, corner[3].x)); boundingRect.bottom = Math.max(Math.max(corner[2].y, corner[3].y),Math.max(corner[0].y, corner[1].y)); } public void set(RectF rect) { set(new Vector2D(rect.centerX(),rect.centerY()),rect.width(),rect.height(),0.0f); } // Returns true if other overlaps one dimension of this. private boolean overlaps1Way(OBB2D other) { for (int a = 0; a < axis.length; ++a) { double t = other.corner[0].dot(axis[a]); // Find the extent of box 2 on axis a double tMin = t; double tMax = t; for (int c = 1; c < corner.length; ++c) { t = other.corner[c].dot(axis[a]); if (t < tMin) { tMin = t; } else if (t > tMax) { tMax = t; } } // We have to subtract off the origin // See if [tMin, tMax] intersects [0, 1] if ((tMin > 1 + origin[a]) || (tMax < origin[a])) { // There was no intersection along this dimension; // the boxes cannot possibly overlap. return false; } } // There was no dimension along which there is no intersection. // Therefore the boxes overlap. return true; } //Updates the axes after the corners move. Assumes the //corners actually form a rectangle. private void computeAxes() { axis[0] = corner[1].subtract(corner[0]); axis[1] = corner[3].subtract(corner[0]); // Make the length of each axis 1/edge length so we know any // dot product must be less than 1 to fall within the edge. for (int a = 0; a < axis.length; ++a) { axis[a] = axis[a].divide((axis[a].length() * axis[a].length())); origin[a] = corner[0].dot(axis[a]); } } public void moveTo(Vector2D center) { Vector2D centroid = (corner[0].add(corner[1]).add(corner[2]).add(corner[3])).divide(4.0f); Vector2D translation = center.subtract(centroid); for (int c = 0; c < 4; ++c) { corner[c] = corner[c].add(translation); } computeAxes(); computeDimensions(center,angle); } // Returns true if the intersection of the boxes is non-empty. public boolean overlaps(OBB2D other) { if(right() < other.left()) { return false; } if(bottom() < other.top()) { return false; } if(left() > other.right()) { return false; } if(top() > other.bottom()) { return false; } if(other.getAngle() == 0.0f && getAngle() == 0.0f) { return true; } return overlaps1Way(other) && other.overlaps1Way(this); } public Vector2D getCenter() { return center; } public float getWidth() { return extents.x * 2; } public float getHeight() { return extents.y * 2; } public void setAngle(float angle) { set(center,getWidth(),getHeight(),angle); } public float getAngle() { return angle; } public void setSize(float w,float h) { set(center,w,h,angle); } public float left() { return boundingRect.left; } public float right() { return boundingRect.right; } public float bottom() { return boundingRect.bottom; } public float top() { return boundingRect.top; } public RectF getBoundingRect() { return boundingRect; } public boolean overlaps(float left, float top, float right, float bottom) { if(right() < left) { return false; } if(bottom() < top) { return false; } if(left() > right) { return false; } if(top() > bottom) { return false; } return true; } };

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  • How to follow object on CatmullRomSplines at constant speed (e.g. train and train carriage)?

    - by Simon
    I have a CatmullRomSpline, and using the very good example at https://github.com/libgdx/libgdx/wiki/Path-interface-%26-Splines I have my object moving at an even pace over the spline. Using a simple train and carriage example, I now want to have the carriage follow the train at the same speed as the train (not jolting along as it does with my code below). This leads into my main questions: How can I make the carriage have the same constant speed as the train and make it non jerky (it has something to do with the derivative I think, I don't understand how that part works)? Why do I need to divide by the line length to convert to metres per second, and is that correct? It wasn't done in the linked examples? I have used the example I linked to above, and modified for my specific example: private void process(CatmullRomSpline catmullRomSpline) { // Render path with precision of 1000 points renderPath(catmullRomSpline, 1000); float length = catmullRomSpline.approxLength(catmullRomSpline.spanCount * 1000); // Render the "train" Vector2 trainDerivative = new Vector2(); Vector2 trainLocation = new Vector2(); catmullRomSpline.derivativeAt(trainDerivative, current); // For some reason need to divide by length to convert from pixel speed to metres per second but I do not // really understand why I need it, it wasn't done in the examples??????? current += (Gdx.graphics.getDeltaTime() * speed / length) / trainDerivative.len(); catmullRomSpline.valueAt(trainLocation, current); renderCircleAtLocation(trainLocation); if (current >= 1) { current -= 1; } // Render the "carriage" Vector2 carriageLocation = new Vector2(); float carriagePercentageCovered = (((current * length) - 1f) / length); // I would like it to follow at 1 metre behind carriagePercentageCovered = Math.max(carriagePercentageCovered, 0); catmullRomSpline.valueAt(carriageLocation, carriagePercentageCovered); renderCircleAtLocation(carriageLocation); } private void renderPath(CatmullRomSpline catmullRomSpline, int k) { // catMulPoints would normally be cached when initialising, but for sake of example... Vector2[] catMulPoints = new Vector2[k]; for (int i = 0; i < k; ++i) { catMulPoints[i] = new Vector2(); catmullRomSpline.valueAt(catMulPoints[i], ((float) i) / ((float) k - 1)); } SHAPE_RENDERER.begin(ShapeRenderer.ShapeType.Line); SHAPE_RENDERER.setColor(Color.NAVY); for (int i = 0; i < k - 1; ++i) { SHAPE_RENDERER.line((Vector2) catMulPoints[i], (Vector2) catMulPoints[i + 1]); } SHAPE_RENDERER.end(); } private void renderCircleAtLocation(Vector2 location) { SHAPE_RENDERER.begin(ShapeRenderer.ShapeType.Filled); SHAPE_RENDERER.setColor(Color.YELLOW); SHAPE_RENDERER.circle(location.x, location.y, .5f); SHAPE_RENDERER.end(); } To create a decent sized CatmullRomSpline for testing this out: Vector2[] controlPoints = makeControlPointsArray(); CatmullRomSpline myCatmull = new CatmullRomSpline(controlPoints, false); .... private Vector2[] makeControlPointsArray() { Vector2[] pointsArray = new Vector2[78]; pointsArray[0] = new Vector2(1.681817f, 10.379999f); pointsArray[1] = new Vector2(2.045455f, 10.379999f); pointsArray[2] = new Vector2(2.663636f, 10.479999f); pointsArray[3] = new Vector2(3.027272f, 10.700000f); pointsArray[4] = new Vector2(3.663636f, 10.939999f); pointsArray[5] = new Vector2(4.245455f, 10.899999f); pointsArray[6] = new Vector2(4.736363f, 10.720000f); pointsArray[7] = new Vector2(4.754545f, 10.339999f); pointsArray[8] = new Vector2(4.518181f, 9.860000f); pointsArray[9] = new Vector2(3.790908f, 9.340000f); pointsArray[10] = new Vector2(3.172727f, 8.739999f); pointsArray[11] = new Vector2(3.300000f, 8.340000f); pointsArray[12] = new Vector2(3.700000f, 8.159999f); pointsArray[13] = new Vector2(4.227272f, 8.520000f); pointsArray[14] = new Vector2(4.681818f, 8.819999f); pointsArray[15] = new Vector2(5.081817f, 9.200000f); pointsArray[16] = new Vector2(5.463636f, 9.460000f); pointsArray[17] = new Vector2(5.972727f, 9.300000f); pointsArray[18] = new Vector2(6.063636f, 8.780000f); pointsArray[19] = new Vector2(6.027272f, 8.259999f); pointsArray[20] = new Vector2(5.700000f, 7.739999f); pointsArray[21] = new Vector2(5.300000f, 7.440000f); pointsArray[22] = new Vector2(4.645454f, 7.179999f); pointsArray[23] = new Vector2(4.136363f, 6.940000f); pointsArray[24] = new Vector2(3.427272f, 6.720000f); pointsArray[25] = new Vector2(2.572727f, 6.559999f); pointsArray[26] = new Vector2(1.900000f, 7.100000f); pointsArray[27] = new Vector2(2.336362f, 7.440000f); pointsArray[28] = new Vector2(2.590908f, 7.940000f); pointsArray[29] = new Vector2(2.318181f, 8.500000f); pointsArray[30] = new Vector2(1.663636f, 8.599999f); pointsArray[31] = new Vector2(1.209090f, 8.299999f); pointsArray[32] = new Vector2(1.118181f, 7.700000f); pointsArray[33] = new Vector2(1.045455f, 6.880000f); pointsArray[34] = new Vector2(1.154545f, 6.100000f); pointsArray[35] = new Vector2(1.281817f, 5.580000f); pointsArray[36] = new Vector2(1.700000f, 5.320000f); pointsArray[37] = new Vector2(2.190908f, 5.199999f); pointsArray[38] = new Vector2(2.900000f, 5.100000f); pointsArray[39] = new Vector2(3.700000f, 5.100000f); pointsArray[40] = new Vector2(4.372727f, 5.220000f); pointsArray[41] = new Vector2(4.827272f, 5.220000f); pointsArray[42] = new Vector2(5.463636f, 5.160000f); pointsArray[43] = new Vector2(5.554545f, 4.700000f); pointsArray[44] = new Vector2(5.245453f, 4.340000f); pointsArray[45] = new Vector2(4.445455f, 4.280000f); pointsArray[46] = new Vector2(3.609091f, 4.260000f); pointsArray[47] = new Vector2(2.718181f, 4.160000f); pointsArray[48] = new Vector2(1.990908f, 4.140000f); pointsArray[49] = new Vector2(1.427272f, 3.980000f); pointsArray[50] = new Vector2(1.609090f, 3.580000f); pointsArray[51] = new Vector2(2.136363f, 3.440000f); pointsArray[52] = new Vector2(3.227272f, 3.280000f); pointsArray[53] = new Vector2(3.972727f, 3.340000f); pointsArray[54] = new Vector2(5.027272f, 3.360000f); pointsArray[55] = new Vector2(5.718181f, 3.460000f); pointsArray[56] = new Vector2(6.100000f, 4.240000f); pointsArray[57] = new Vector2(6.209091f, 4.500000f); pointsArray[58] = new Vector2(6.118181f, 5.320000f); pointsArray[59] = new Vector2(5.772727f, 5.920000f); pointsArray[60] = new Vector2(4.881817f, 6.140000f); pointsArray[61] = new Vector2(5.318181f, 6.580000f); pointsArray[62] = new Vector2(6.263636f, 7.020000f); pointsArray[63] = new Vector2(6.645453f, 7.420000f); pointsArray[64] = new Vector2(6.681817f, 8.179999f); pointsArray[65] = new Vector2(6.627272f, 9.080000f); pointsArray[66] = new Vector2(6.572727f, 9.699999f); pointsArray[67] = new Vector2(6.263636f, 10.820000f); pointsArray[68] = new Vector2(5.754546f, 11.479999f); pointsArray[69] = new Vector2(4.536363f, 11.599998f); pointsArray[70] = new Vector2(3.572727f, 11.700000f); pointsArray[71] = new Vector2(2.809090f, 11.660000f); pointsArray[72] = new Vector2(1.445455f, 11.559999f); pointsArray[73] = new Vector2(0.936363f, 11.280000f); pointsArray[74] = new Vector2(0.754545f, 10.879999f); pointsArray[75] = new Vector2(0.700000f, 9.939999f); pointsArray[76] = new Vector2(0.918181f, 9.620000f); pointsArray[77] = new Vector2(1.463636f, 9.600000f); return pointsArray; } Disclaimer: My math is very rusty, so please explain in lay mans terms....

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  • Alright, I'm still stuck on this homework problem. C++

    - by Josh
    Okay, the past few days I have been trying to get some input on my programs. Well I decided to scrap them for the most part and try again. So once again, I'm in need of help. For the first program I'm trying to fix, it needs to show the sum of SEVEN numbers. Well, I'm trying to change is so that I don't need the mem[##] = ####. I just want the user to be able to input the numbers and the program run from there and go through my switch loop. And have some kind of display..saying like the sum is?.. Here's my code so far. #include <iostream> #include <iomanip> #include <ios> using namespace std; int main() { const int READ = 10; const int WRITE = 11; const int LOAD = 20; const int STORE = 21; const int ADD = 30; const int SUBTRACT = 31; const int DIVIDE = 32; const int MULTIPLY = 33; const int BRANCH = 40; const int BRANCHNEG = 41; const int BRANCHZERO = 42; const int HALT = 43; int mem[100] = {0}; //Making it 100, since simpletron contains a 100 word mem. int operation; //taking the rest of these variables straight out of the book seeing as how they were italisized. int operand; int accum = 0; // the special register is starting at 0 int counter; for ( counter=0; counter < 100; counter++) mem[counter] = 0; // This is for part a, it will take in positive variables in //a sent-controlled loop and compute + print their sum. Variables from example in text. mem[0] = 1009; mem[1] = 1109; mem[2] = 2010; mem[3] = 2111; mem[4] = 2011; mem[5] = 3100; mem[6] = 2113; mem[7] = 1113; mem[8] = 4300; counter = 0; //Makes the variable counter start at 0. while(true) { operand = mem[ counter ]%100; // Finds the op codes from the limit on the mem (100) operation = mem[ counter ]/100; //using a switch loop to set up the loops for the cases switch ( operation ){ case READ: //reads a variable into a word from loc. Enter in -1 to exit cout <<"\n Input a positive variable: "; cin >> mem[ operand ]; counter++; break; case WRITE: // takes a word from location cout << "\n\nThe content at location " << operand << " is " << mem[operand]; counter++; break; case LOAD:// loads accum = mem[ operand ];counter++; break; case STORE: //stores mem[ operand ] = accum;counter++; break; case ADD: //adds accum += mem[operand];counter++; break; case SUBTRACT: // subtracts accum-= mem[ operand ];counter++; break; case DIVIDE: //divides accum /=(mem[ operand ]);counter++; break; case MULTIPLY: // multiplies accum*= mem [ operand ];counter++; break; case BRANCH: // Branches to location counter = operand; break; case BRANCHNEG: //branches if acc. is < 0 if (accum < 0) counter = operand; else counter++; break; case BRANCHZERO: //branches if acc = 0 if (accum == 0) counter = operand; else counter++; break; case HALT: // Program ends break; } } return 0; } part B int main() { const int READ = 10; const int WRITE = 11; const int LOAD = 20; const int STORE = 21; const int ADD = 30; const int SUBTRACT = 31; const int DIVIDE = 32; const int MULTIPLY = 33; const int BRANCH = 40; const int BRANCHNEG = 41; const int BRANCHZERO = 41; const int HALT = 43; int mem[100] = {0}; int operation; int operand; int accum = 0; int pos = 0; int j; mem[22] = 7; // loop 7 times mem[25] = 1; // increment by 1 mem[00] = 4306; mem[01] = 2303; mem[02] = 3402; mem[03] = 6410; mem[04] = 3412; mem[05] = 2111; mem[06] = 2002; mem[07] = 2312; mem[08] = 4210; mem[09] = 2109; mem[10] = 4001; mem[11] = 2015; mem[12] = 3212; mem[13] = 2116; mem[14] = 1101; mem[15] = 1116; mem[16] = 4300; j = 0; while ( true ) { operand = memory[ j ]%100; // Finds the op codes from the limit on the memory (100) operation = memory[ j ]/100; //using a switch loop to set up the loops for the cases switch ( operation ){ case 1: //reads a variable into a word from loc. Enter in -1 to exit cout <<"\n enter #: "; cin >> memory[ operand ]; break; case 2: // takes a word from location cout << "\n\nThe content at location " << operand << "is " << memory[operand]; break; case 3:// loads accum = memory[ operand ]; break; case 4: //stores memory[ operand ] = accum; break; case 5: //adds accum += mem[operand];; break; case 6: // subtracts accum-= memory[ operand ]; break; case 7: //divides accum /=(memory[ operand ]); break; case 8: // multiplies accum*= memory [ operand ]; break; case 9: // Branches to location j = operand; break; case 10: //branches if acc. is < 0 break; case 11: //branches if acc = 0 if (accum == 0) j = operand; break; case 12: // Program ends exit(0); break; } j++; } return 0; }

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  • PDC and Tech-Ed Europe Slides and Code

    - by Stephen Walther
    I spent close to three weeks on the road giving talks at Tech-Ed Europe (Berlin), PDC (Los Angeles), and the Los Angeles Code Camp (Los Angeles). I got to talk about two topics that I am very passionate about: ASP.NET MVC and Ajax. Thanks everyone for coming to all my talks! At PDC, I announced all of the new features of our ASP.NET Ajax Library. In particular, I made five big announcements: ASP.NET Ajax Library Beta Released – You can download the beta from Ajax.CodePlex.com ASP.NET Ajax Library includes the AJAX Control Toolkit – You can use the Ajax Control Toolkit with ASP.NET MVC. ASP.NET Ajax Library being contributed to the CodePlex Foundation – ASP.NET Ajax is the founding project for the CodePlex Foundation (see CodePlex.org) ASP.NET Ajax Library is receiving full product support – Complain to Microsoft Customer Service at midnight on Christmas ASP.NET Ajax Library supports jQuery integration – Use (almost) all of the Ajax Control Toolkit controls in jQuery For more details on the Ajax announcements, see James Senior’s blog entry on the Ajax announcements at: http://jamessenior.com/post/News-on-the-ASPNET-Ajax-Library.aspx In my MVC talks, I discussed the new features being introduced with ASP.NET MVC 2. Here are three of my favorite new features: Client Validation – Client validation done the right way. Do your validation in your model and let the validation bubble up to JavaScript code automatically. Areas – Divide your ASP.NET MVC application into sub-applications. Great for managing both medium and large projects. RenderAction() – Finally, a way to add content to master pages and multiple pages without doing anything strange or twisted. There are demos of all of these features in the MVC downloads below. Here are the power point and code from all of the talks: PDC – Introducing the New ASP.NET Ajax Library PDC – ASP.NET MVC: The New Stuff Tech-Ed Europe - What's New in Microsoft ASP.NET Model-View-Controller Tech-Ed Europe - Microsoft ASP.NET AJAX: Taking AJAX to the Next Level

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  • Wireless AP Placement and Digramming

    - by Matt Simmons
    I'm trying to research the best placement of wireless APs in a given space, and I'm running into problems in gathering information. I found (what I thought was) a great source in this tech republic article: http://techrepublic.com.com/5206-6230-0.html?forumID=82&threadID=163120 While this diagram seems detailed and overall very informative, there were a lot of comments talking about how it was lacking in things like "wire racks, microwaves, concrete walls, motors..." etc. Maybe I'm rash, but I just sort of looked around my office (which is, albeit, somewhat smaller than the one diagrammed), and went "uhhh, there", and hooked up the AP. It seems to cover everywhere. I imagine if my office quadrupled in size, I'd logically divide it up and put four APs in, with a similar amount of thought devoted to each. So, suppose I had a much more complex office. What tools (both diagramming and surveying) do you use to plan your AP placement?

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  • Wireless AP Placement and Digramming

    - by Matt Simmons
    I'm trying to research the best placement of wireless APs in a given space, and I'm running into problems in gathering information. I found (what I thought was) a great source in this tech republic article: http://techrepublic.com.com/5206-6230-0.html?forumID=82&threadID=163120 While this diagram seems detailed and overall very informative, there were a lot of comments talking about how it was lacking in things like "wire racks, microwaves, concrete walls, motors..." etc. Maybe I'm rash, but I just sort of looked around my office (which is, albeit, somewhat smaller than the one diagrammed), and went "uhhh, there", and hooked up the AP. It seems to cover everywhere. I imagine if my office quadrupled in size, I'd logically divide it up and put four APs in, with a similar amount of thought devoted to each. So, suppose I had a much more complex office. What tools (both diagramming and surveying) do you use to plan your AP placement?

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  • QoS - split bandwidth across all IPs during high load

    - by Matthew Iselin
    We have a Linux-based router which is currently working fairly well, but our network only has a 1.5 mbps incoming connection. The network is small, but during high load periods some systems can end up dominating the bandwidth. For example, a client downloading a file can easily saturate the connection leaving everyone else with barely any access to the outside world. Naturally, I'd like to fix this. I believe a combination of iptables rules and tc is in order, but I have no idea how to go about distributing the bandwidth evenly across the clients. It would be nice if there was a way to divide the bandwidth only across clients that are actually utilising the connection as well, rather than hard limit each connection to (bandwidth / number of clients).

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  • SQL SERVER – DATEDIFF – Accuracy of Various Dateparts

    - by pinaldave
    I recently received the following question through email and I found it very interesting so I want to share it with you. “Hi Pinal, In SQL statement below the time difference between two given dates is 3 sec, but when checked in terms of Min it says 1 Min (whereas the actual min is 0.05Min) SELECT DATEDIFF(MI,'2011-10-14 02:18:58' , '2011-10-14 02:19:01') AS MIN_DIFF Is this is a BUG in SQL Server ?” Answer is NO. It is not a bug; it is a feature that works like that. Let us understand that in a bit more detail. When you instruct SQL Server to find the time difference in minutes, it just looks at the minute section only and completely ignores hour, second, millisecond, etc. So in terms of difference in minutes, it is indeed 1. The following will also clear how DATEDIFF works: SELECT DATEDIFF(YEAR,'2011-12-31 23:59:59' , '2012-01-01 00:00:00') AS YEAR_DIFF The difference between the above dates is just 1 second, but in terms of year difference it shows 1. If you want to have accuracy in seconds, you need to use a different approach. In the first example, the accurate method is to find the number of seconds first and then divide it by 60 to convert it to minutes. SELECT DATEDIFF(second,'2011-10-14 02:18:58' , '2011-10-14 02:19:01')/60.0 AS MIN_DIFF Even though the concept is very simple it is always a good idea to refresh it. Please share your related experience with me through your comments. Reference: Pinal Dave (http://blog.SQLAuthority.com) Filed under: Pinal Dave, PostADay, SQL, SQL Authority, SQL DateTime, SQL Query, SQL Scripts, SQL Server, SQL Tips and Tricks, T SQL, Technology

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  • Do Repeat Yourself in Unit Tests

    - by João Angelo
    Don’t get me wrong I’m a big supporter of the DRY (Don’t Repeat Yourself) Principle except however when it comes to unit tests. Why? Well, in my opinion a unit test should be a self-contained group of actions with the intent to test a very specific piece of code and should not depend on externals shared with other unit tests. In a typical unit test we can divide its code in two major groups: Preparation of preconditions for the code under test; Invocation of the code under test. It’s in the first group that you are tempted to refactor common code in several unit tests into helper methods that can then be called in each one of them. Another way to not duplicate code is to use the built-in infrastructure of some unit test frameworks such as SetUp/TearDown methods that automatically run before and after each unit test. I must admit that in the past I was guilty of both charges but what at first seemed a good idea since I was removing code duplication turnout to offer no added value and even complicate the process when a given test fails. We love unit tests because of their rapid feedback when something goes wrong. However, this feedback requires most of the times reading the code for the failed test. Given this, what do you prefer? To read a single method or wander through several methods like SetUp/TearDown and private common methods. I say it again, do repeat yourself in unit tests. It may feel wrong at first but I bet you won’t regret it later.

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  • handling various frame layouts in android

    - by vaibhav
    i'm new to game development and am trying create a Contra or the old tmnt game (but a simple one) like game for android. for the game i decided to divide my main screen in three parts - upper for stats,mid for the game and lower for controls. my main.xml is <?xml version="1.0" encoding="utf-8"?> <LinearLayout xmlns:android="http://schemas.android.com/apk/res/android" android:layout_width="fill_parent" android:layout_height="fill_parent" android:orientation="vertical" > <FrameLayout android:id="@+id/upper_bar" android:layout_width="fill_parent" android:layout_height="fill_parent" android:layout_weight="1" > </FrameLayout> <FrameLayout android:id="@+id/fl" android:layout_width="fill_parent" android:layout_height="fill_parent" android:layout_weight="0.5" > </FrameLayout> <FrameLayout android:id="@+id/low_bar" android:layout_width="fill_parent" android:layout_height="fill_parent" android:layout_weight="0.85" > </FrameLayout> </LinearLayout> so i have created the gameview and gameloopthread classes for the mid surface(which is pretty standard). my problem is that how do i draw in the upper and lower frame layouts? should i make new classes for view and thread for each layout , should i do all this in the gameview class itself or is there any better way to implement this?

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  • MaxClients in apache. How to know the size of my proccess?

    - by Larry
    From http://httpd.apache.org/docs/2.2/misc/perf-tuning.html The single biggest hardware issue affecting webserver performance is RAM. A webserver should never ever have to swap, as swapping increases the latency of each request beyond a point that users consider "fast enough". This causes users to hit stop and reload, further increasing the load. You can, and should, control the MaxClients setting so that your server does not spawn so many children it starts swapping. This procedure for doing this is simple: determine the size of your average Apache process, by looking at your process list via a tool such as top, and divide this into your total available memory, leaving some room for other processes. The main issue is that I can't understand how to know the size, because, well i have the size of httpd on no more of 3888 But, if we need to determine the number for MaxClients, and I have 4GB of RAM, so I get: 972, so I should use like 900 in the MaxClients?

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  • Node.js MMO - process and/or map division

    - by Gipsy King
    I am in the phase of designing a mmo browser based game (certainly not massive, but all connected players are in the same universe), and I am struggling with finding a good solution to the problem of distributing players across processes. I'm using node.js with socket.io. I have read this helpful article, but I would like some advice since I am also concerned with different processes. Solution 1: Tie a process to a map location (like a map-cell), connect players to the process corresponding to their location. When a player performs an action, transmit it to all other players in this process. When a player moves away, he will eventually have to connect to another process (automatically). Pros: Easier to implement Cons: Must divide map into zones Player reconnection when moving into a different zone is probably annoying If one zone/process is always busy (has players in it), it doesn't really load-balance, unless I split the zone which may not be always viable There shouldn't be any visible borders Solution 1b: Same as 1, but connect processes of bordering cells, so that players on the other side of the border are visible and such. Maybe even let them interact. Solution 2: Spawn processes on demand, unrelated to a location. Have one special process to keep track of all connected player handles, their location, and the process they're connected to. Then when a player performs an action, the process finds all other nearby players (from the special player-process-location tracking node), and instructs their matching processes to relay the action. Pros: Easy load balancing: spawn more processes Avoids player reconnecting / borders between zones Cons: Harder to implement and test Additional steps of finding players, and relaying event/action to another process If the player-location-process tracking process fails, all other fail too I would like to hear if I'm missing something, or completely off track.

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  • How do partitions help in optimizing the harddrive?

    - by Fasih Khatib
    I was recently reading a guide on Tom's Hardware about how to optimize the harddrive. They listed creating partitions as one of them. They said keeping the various files seperate is a good idea as it reduces the read/write cycles required. Now my querry is: what size partitions do i make for my 500gb harddrive. Its completely blank. I will be installing WIN7 in it. My usual strategy is to divide it into two equal partitions. Is it the optimum size?

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  • Using VLANs that are routed together?

    - by dannymcc
    I have a quick question that's bugging me the more I read about VLANs. So far I understand that they are useful for dividing the network into sub-sections, but if you route them together does that not remove any security benefit? As an example, if I created a VLAN on my home network which was simply one computer, one server and one router.if I wanted to divide the network between computers and servers I could put the computer on VLAN 10 and the server on VLAN 20. Then the computer would no longer be able to communicate with the server - unless I added a static route to the router that connected the two together, basically telling VLAN 10 that VLAN 20 exists and how to communicate with it. The VLANs would then be connected in a similar way to a 'flat' network that has no VLANs. Therefore, surely, all security benefits are lost. Am I missing something?

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