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  • std::map operator[] and automatically created new objects

    - by thomas-gies
    I'm a little bit scared about something like this: std::map<DWORD, DWORD> tmap; tmap[0]+=1; tmap[0]+=1; tmap[0]+=1; Since DWORD's are not automatically initialized, I'm always afraid of tmap[0] being a random number that is incremented. How does the map know hot to initialize a DWORD if the runtime does not know how to do it? Is it guaranteed, that the result is always tmap[0] == 3?

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  • Conditional operator in if-statement?

    - by Pindatjuh
    I've written the following if-statement in Java: if(methodName.equals("set" + this.name) || isBoolean() ? methodName.equals("is" + this.name) : methodName.equals("get" + this.name)) { ... } Is this a good practice to write such expressions in if, to separate state from condition? And can this expression be simplified?

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  • Ternary operator in if-statement?

    - by Pindatjuh
    I've written the following if-statement in Java: if(methodName.equals("set" + this.name) || isBoolean() ? methodName.equals("is" + this.name) : methodName.equals("get" + this.name)) { ... } Is this a good practice to write such expressions in if, to separate state from condition? And can this expression be simplified?

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  • post increment operator java

    - by srandpersonia
    I can't make heads or tails of the following code from "java puzzlers" by joshua bloch. public class Test22{ public static void main(String args[]){ int j=0; for(int i=0;i<100;i++){ j=j++; } System.out.println(j); //prints 0 int a=0,b=0; a=b++; System.out.println(a); System.out.println(b); //prints 1 } } I can't get the part where j prints 0. According to the author, j=j++ is similar to temp=j; j=j+1; j=temp; But a=b++ makes b 1. So it should've evaluated like this, a=b b=b+1 By following the same logic, shouldn't j=j++ be evaluated as, j=j j=j+1 Where does the temp come into picture here? Any explanations would be much appreciated. << I'm breaking my head over this. ;) Thanks in advance.

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  • The unary increment operator in pointer arithmetic

    - by RhymesWithDuck
    Hello, this is my first post. I have this function for reversing a string in C that I found. void reverse(char* c) { if (*c != 0) { reverse(c + 1); } printf("%c",*c); } It works fine but if I replace: reverse(c + 1); with: reverse(++c); the first character of the original string is truncated. My question is why would are the statements not equivalent in this instance? Thanks

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  • Is Operator Overloading supported in C

    - by caramel23
    Today when I was reading about LCC(windows) compiler I find out it has the implemention for operator overloading . I'm puzzled because after a bit of googling , it has been confirm that operator overloading ain't support in standard C , but I read some people's comment mentioning LCC is ANSI-compliant . So my real question is , is LCC really standard C or it's just like objective-c , a C variant with object-oriented feature ?

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  • Double Buffering for Game objects, what's a nice clean generic C++ way?

    - by Gary
    This is in C++. So, I'm starting from scratch writing a game engine for fun and learning from the ground up. One of the ideas I want to implement is to have game object state (a struct) be double-buffered. For instance, I can have subsystems updating the new game object data while a render thread is rendering from the old data by guaranteeing there is a consistent state stored within the game object (the data from last time). After rendering of old and updating of new is finished, I can swap buffers and do it again. Question is, what's a good forward-looking and generic OOP way to expose this to my classes while trying to hide implementation details as much as possible? Would like to know your thoughts and considerations. I was thinking operator overloading could be used, but how do I overload assign for a templated class's member within my buffer class? for instance, I think this is an example of what I want: doublebuffer<Vector3> data; data.x=5; //would write to the member x within the new buffer int a=data.x; //would read from the old buffer's x member data.x+=1; //I guess this shouldn't be allowed If this is possible, I could choose to enable or disable double-buffering structs without changing much code. This is what I was considering: template <class T> class doublebuffer{ T T1; T T2; T * current=T1; T * old=T2; public: doublebuffer(); ~doublebuffer(); void swap(); operator=()?... }; and a game object would be like this: struct MyObjectData{ int x; float afloat; } class MyObject: public Node { doublebuffer<MyObjectData> data; functions... } What I have right now is functions that return pointers to the old and new buffer, and I guess any classes that use them have to be aware of this. Is there a better way?

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  • How do you pronounce the '...' operator

    - by Uri
    Now, in c++ '...' became a first class operator. In speech, how do you pronounce it? So far I've heard: dot dot dot triple dot ellipsis related: Is it OK to replace ... with ellipsis in writing? e.g. "The ellipsis operator expands the pack" EDIT (clarification): We are all aware that '...' as a punctuation mark is indeed called ellipsis. But in the context of C++ we don't pronounce the names of the punctuation mark. For example, the '&' operator, depends on the context is pronounced as 'and', 'bitwise and', 'address of', 'logical and' (when && is used), or 'reference'. It is rarely pronounced as 'ampersand'. In speeches, I've a feeling that 'dot dot dot' is used more often. For example: http://channel9.msdn.com/Events/GoingNative/GoingNative-2012/Variadic-Templates-are-Funadic (an excellent presentation about variadic templates). On the other hand, 'dot dot dot' is awkward hard to pronouce ('d' and 't' are both pronounce with the tongue). Can we pronounce it 'unpack'?

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  • Distinct operator in Linq

    - by Jalpesh P. Vadgama
    Linq operator provides great flexibility and easy way of coding. Let’s again take one more example of distinct operator. As name suggest it will find the distinct elements from IEnumerable. Let’s take an example of array in console application and then we will again print array to see is it working or not. Below is the code for that. In this application I have integer array which contains duplicate elements and then I will apply distinct operator to this and then I will print again result of distinct operators to actually see whether its working or not. using System; using System.Collections.Generic; using System.Linq; using System.Text; namespace Experiment { class Program { static void Main(string[] args) { int[] intArray = { 1, 2, 2, 3, 3, 3, 4, 4, 4, 4, 5, 5, 5, 5, 5 }; var uniqueIntegers = intArray.Distinct(); foreach (var uInteger in uniqueIntegers) { Console.WriteLine(uInteger); } Console.ReadKey(); } } } Below is output as expected.. That’s cool..Stay tuned for more.. Happy programming. Technorati Tags: Linq,Distinct

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  • Does NULL and nil are equal?

    - by monish
    Hi Guys, Actually my question here is does Null and nil are equal or not? I had an Example but I am confused when they are equal when they are not. NSNull *nullValue = [NSNull null]; NSArray *arrayWithNull = [NSArray arrayWithObject:nullValue]; NSLog(@"arrayWithNull: %@", arrayWithNull); id aValue = [arrayWithNull objectAtIndex:0]; if (aValue == nil) { NSLog(@"equals nil"); } else if (aValue == [NSNull null]) { NSLog(@"equals NSNull instance"); if ([aValue isEqual:nil]) { NSLog(@"isEqual:nil"); } } Here in the above case it shows that both Null and nil are not equal and it displays "equals NSNull instance" NSString *str=NULL; id str1=nil; if(str1 == str) { printf("\n IS EQUAL........"); } else { printf("\n NOT EQUAL........"); } And in the second case it shows both are equal and it displays "IS EQUAL". Anyone's help will be much appreciated. Thank you, Monish.

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  • Play music by Operator in asterisk?

    - by Rev
    Hi I want in call duration between operator and caller,play sound for operator(something like hold music). But in order to play this sound, operator must dial unique code and then sound will be play for caller, and caller only hear that sound file! After that (sound fully played), caller back to operator's queue or something like this. So is this possible to do or not? (if possible, post dial-plan for this too)

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  • Play music by Operator in asterisk?

    - by Rev
    Hi I want in call duration between operator and caller,play sound for operator(something like hold music). But in order to play this sound, operator must dial unique code and then sound will be play for caller, and caller only hear that sound file! After that (sound fully played), caller back to operator's queue or something like this. So is this possible to do or not? (if possible, post dial-plan for this too)

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  • how to refer to the current struct in an overloaded operator?

    - by genesys
    Hi! I have a struct for which i want to define a relative order by defining < , , <= and = operators. actually in my order there won't be any equality, so if one struct is not smaller than another, it's automatically larger. I defined the first operator like this: struct MyStruct{ ... ... bool operator < (const MyStruct &b) const {return (somefancycomputation);} }; now i'd like to define the other operators based on this operator, such that <= will return the same as < and the other two will simply return the oposite. so for example for the operator i'd like to write something like bool operator > (const MyStruct &b) const {return !(self<b);} but i don't know how to refere to this 'self' since i can refere only to the fields inside the current struct. whole is in C++ hope my question was understandable :) thank you for the help!

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  • Defining < for STL sort algorithm - operator overload, functor or standalone function?

    - by Andy
    I have a stl::list containing Widget class objects. They need to be sorted according to two members in the Widget class. For the sorting to work, I need to define a less-than comparator comparing two Widget objects. There seems to be a myriad of ways to do it. From what I can gather, one can either: a. Define a comparison operator overload in the class: bool Widget::operator< (const Widget &rhs) const b. Define a standalone function taking two Widgets: bool operator<(const Widget& lhs, const Widget& rhs); And then make the Widget class a friend of it: class Widget { // Various class definitions ... friend bool operator<(const Widget& lhs, const Widget& rhs); }; c. Define a functor and then include it as a parameter when calling the sort function: class Widget_Less : public binary_function<Widget, Widget, bool> { bool operator()(const Widget &lhs, const Widget& rhs) const; }; Does anybody know which method is better? In particular I am interested to know if I should do 1 or 2. I searched the book Effective STL by Scott Meyer but unfortunately it does not have anything to say about this. Thank you for your reply.

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  • [C++] Can all/any struct assignment operator be Overloaded? (and specifically struct tm = sql::Resu

    - by Luke Mcneice
    Hi all, Generally, i was wondering if there was any exceptions of types that cant have thier assignment operator overloaded. Specifically, I'm wanting to overload the assignment operator of a tm struct, (time.h) so i can assign a sql::ResultSet to it. I have already have the conversion logic: sscanf(sqlresult->getString("StoredAt").c_str(),"%d-%d-%d %d:%d:%d",&TempTimeStruct->tm_year,&TempTimeStruct->tm_mon,&TempTimeStruct->tm_mday,&TempTimeStruct->tm_hour,&TempTimeStruct->tm_min,&TempTimeStruct->tm_sec); //populating the struct I tried the overload with this: tm& tm::operator=(sql::ResultSet & results) { //CODE return *this; } however VS08 reports: error C2511: 'tm &tm::operator =(sql::ResultSet &)' : overloaded member function not found in 'tm'

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  • What is the purpose of Java's unary plus operator?

    - by Syntactic
    Java's unary plus operator appears to have come over from C, via C++. As near as I can tell, it has the following effects: promotes its operand to int, if it's not already an int or wider unboxes its operand, if it's a wrapper object complicates slightly the parsing of evil expressions containing large numbers of consecutive plus signs It seems to me that there are better (or, at least, clearer) ways to do all of these things. In this SO question, concerning the counterpart operator in C#, someone said that "It's there to be overloaded if you feel the need." But in Java, one cannot overload any operator. So does this operator exist in Java just because it existed in C++?

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  • In what situation should the built-in 'operator' module be used in python?

    - by apphacker
    I'm speaking of this module: http://docs.python.org/library/operator.html From the article: The operator module exports a set of functions implemented in C corresponding to the intrinsic operators of Python. For example, operator.add(x, y) is equivalent to the expression x+y. The function names are those used for special class methods; variants without leading and trailing __ are also provided for convenience. I'm not sure I understand the benefit or purpose of this module.

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  • C# Extend array type to overload operators

    - by Episodex
    I'd like to create my own class extending array of ints. Is that possible? What I need is array of ints that can be added by "+" operator to another array (each element added to each), and compared by "==", so it could (hopefully) be used as a key in dictionary. The thing is I don't want to implement whole IList interface to my new class, but only add those two operators to existing array class. I'm trying to do something like this: class MyArray : Array<int> But it's not working that way obviously ;). Sorry if I'm unclear but I'm searching solution for hours now... UPDATE: I tried something like this: class Zmienne : IEquatable<Zmienne> { public int[] x; public Zmienne(int ilosc) { x = new int[ilosc]; } public override bool Equals(object obj) { if (obj == null || GetType() != obj.GetType()) { return false; } return base.Equals((Zmienne)obj); } public bool Equals(Zmienne drugie) { if (x.Length != drugie.x.Length) return false; else { for (int i = 0; i < x.Length; i++) { if (x[i] != drugie.x[i]) return false; } } return true; } public override int GetHashCode() { int hash = x[0].GetHashCode(); for (int i = 1; i < x.Length; i++) hash = hash ^ x[i].GetHashCode(); return hash; } } Then use it like this: Zmienne tab1 = new Zmienne(2); Zmienne tab2 = new Zmienne(2); tab1.x[0] = 1; tab1.x[1] = 1; tab2.x[0] = 1; tab2.x[1] = 1; if (tab1 == tab2) Console.WriteLine("Works!"); And no effect. I'm not good with interfaces and overriding methods unfortunately :(. As for reason I'm trying to do it. I have some equations like: x1 + x2 = 0.45 x1 + x4 = 0.2 x2 + x4 = 0.11 There are a lot more of them, and I need to for example add first equation to second and search all others to find out if there is any that matches the combination of x'es resulting in that adding. Maybe I'm going in totally wrong direction?

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  • Why is Perl's smart-match operator considered broken?

    - by Sean McMillan
    I've seen a number of comments across the web Perl's smart-match operator is broken. I know it originally was part of Perl 6, then was implemented in Perl 5.10 off of an old version of the spec, and was then corrected in 5.10.1 to match the current Perl 6 spec. Is the problem fixed in 5.10.1+, or are there other problems with the smart-match operator that make it troublesome in practice? What are the problems? Is there a yet-more-updated version (Perl 6, perhaps) that fixes the problems?

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