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  • Select the next N elements of an IEnumerable<T>

    - by sassafrass
    Say you've got some IEnumerable called S of length N. I would like to select all continuous subsequences of length n <= N from S. If S were, say, a string, this'd be pretty easy. There are (S.Length - n + 1) subsequences of length n. For example, "abcdefg" is length (7), so that means it has (5) substrings of length (3): "abc", "bcd", "cde", "def", "efg". But S could be any IEnumerable, so this route isn't open. How do I use extension methods to solve this?

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  • Point inside Oriented Bounding Box?

    - by Milo
    I have an OBB2D class based on SAT. This is my point in OBB method: public boolean pointInside(float x, float y) { float newy = (float) (Math.sin(angle) * (y - center.y) + Math.cos(angle) * (x - center.x)); float newx = (float) (Math.cos(angle) * (x - center.x) - Math.sin(angle) * (y - center.y)); return (newy > center.y - (getHeight() / 2)) && (newy < center.y + (getHeight() / 2)) && (newx > center.x - (getWidth() / 2)) && (newx < center.x + (getWidth() / 2)); } public boolean pointInside(Vector2D v) { return pointInside(v.x,v.y); } Here is the rest of the class; the parts that pertain: public class OBB2D { private Vector2D projVec = new Vector2D(); private static Vector2D projAVec = new Vector2D(); private static Vector2D projBVec = new Vector2D(); private static Vector2D tempNormal = new Vector2D(); private Vector2D deltaVec = new Vector2D(); private ArrayList<Vector2D> collisionPoints = new ArrayList<Vector2D>(); // 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(float centerx, float centery, float w, float h, float angle) { for(int i = 0; i < corner.length; ++i) { corner[i] = new Vector2D(); } for(int i = 0; i < axis.length; ++i) { axis[i] = new Vector2D(); } set(centerx,centery,w,h,angle); } public OBB2D(float left, float top, float width, float height) { for(int i = 0; i < corner.length; ++i) { corner[i] = new Vector2D(); } for(int i = 0; i < axis.length; ++i) { axis[i] = new Vector2D(); } set(left + (width / 2), top + (height / 2),width,height,0.0f); } public void set(float centerx,float centery,float w, float h,float angle) { float vxx = (float)Math.cos(angle); float vxy = (float)Math.sin(angle); float vyx = (float)-Math.sin(angle); float vyy = (float)Math.cos(angle); vxx *= w / 2; vxy *= (w / 2); vyx *= (h / 2); vyy *= (h / 2); corner[0].x = centerx - vxx - vyx; corner[0].y = centery - vxy - vyy; corner[1].x = centerx + vxx - vyx; corner[1].y = centery + vxy - vyy; corner[2].x = centerx + vxx + vyx; corner[2].y = centery + vxy + vyy; corner[3].x = centerx - vxx + vyx; corner[3].y = centery - vxy + vyy; this.center.x = centerx; this.center.y = centery; this.angle = angle; computeAxes(); extents.x = w / 2; extents.y = h / 2; computeBoundingRect(); } //Updates the axes after the corners move. Assumes the //corners actually form a rectangle. private void computeAxes() { axis[0].x = corner[1].x - corner[0].x; axis[0].y = corner[1].y - corner[0].y; axis[1].x = corner[3].x - corner[0].x; axis[1].y = corner[3].y - corner[0].y; // 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) { float l = axis[a].length(); float ll = l * l; axis[a].x = axis[a].x / ll; axis[a].y = axis[a].y / ll; origin[a] = corner[0].dot(axis[a]); } } public void computeBoundingRect() { boundingRect.left = JMath.min(JMath.min(corner[0].x, corner[3].x), JMath.min(corner[1].x, corner[2].x)); boundingRect.top = JMath.min(JMath.min(corner[0].y, corner[1].y),JMath.min(corner[2].y, corner[3].y)); boundingRect.right = JMath.max(JMath.max(corner[1].x, corner[2].x), JMath.max(corner[0].x, corner[3].x)); boundingRect.bottom = JMath.max(JMath.max(corner[2].y, corner[3].y),JMath.max(corner[0].y, corner[1].y)); } public void set(RectF rect) { set(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; } public void moveTo(float centerx, float centery) { float cx,cy; cx = center.x; cy = center.y; deltaVec.x = centerx - cx; deltaVec.y = centery - cy; for (int c = 0; c < 4; ++c) { corner[c].x += deltaVec.x; corner[c].y += deltaVec.y; } boundingRect.left += deltaVec.x; boundingRect.top += deltaVec.y; boundingRect.right += deltaVec.x; boundingRect.bottom += deltaVec.y; this.center.x = centerx; this.center.y = centery; computeAxes(); } // 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.x,center.y,getWidth(),getHeight(),angle); } public float getAngle() { return angle; } public void setSize(float w,float h) { set(center.x,center.y,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; } public static float distance(float ax, float ay,float bx, float by) { if (ax < bx) return bx - ay; else return ax - by; } public Vector2D project(float ax, float ay) { projVec.x = Float.MAX_VALUE; projVec.y = Float.MIN_VALUE; for (int i = 0; i < corner.length; ++i) { float dot = Vector2D.dot(corner[i].x,corner[i].y,ax,ay); projVec.x = JMath.min(dot, projVec.x); projVec.y = JMath.max(dot, projVec.y); } return projVec; } public Vector2D getCorner(int c) { return corner[c]; } public int getNumCorners() { return corner.length; } public boolean pointInside(float x, float y) { float newy = (float) (Math.sin(angle) * (y - center.y) + Math.cos(angle) * (x - center.x)); float newx = (float) (Math.cos(angle) * (x - center.x) - Math.sin(angle) * (y - center.y)); return (newy > center.y - (getHeight() / 2)) && (newy < center.y + (getHeight() / 2)) && (newx > center.x - (getWidth() / 2)) && (newx < center.x + (getWidth() / 2)); } public boolean pointInside(Vector2D v) { return pointInside(v.x,v.y); } public ArrayList<Vector2D> getCollsionPoints(OBB2D b) { collisionPoints.clear(); for(int i = 0; i < corner.length; ++i) { if(b.pointInside(corner[i])) { collisionPoints.add(corner[i]); } } for(int i = 0; i < b.corner.length; ++i) { if(pointInside(b.corner[i])) { collisionPoints.add(b.corner[i]); } } return collisionPoints; } }; What could be wrong? When I getCollisionPoints for 2 OBBs I know are penetrating, it returns no points. Thanks

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  • How do I draw a dotted or dashed line?

    - by Gagege
    I'm trying to draw a dashed or dotted line by placing individual segments(dashes) along a path and then separating them. The only algorithm I could come up with for this gave me a dash length that was variable based on the angle of the line. Like this: private function createDashedLine(fromX:Float, fromY:Float, toX:Float, toY:Float):Sprite { var line = new Sprite(); var currentX = fromX; var currentY = fromY; var addX = (toX - fromX) * 0.0075; var addY = (toY - fromY) * 0.0075; line.graphics.lineStyle(1, 0xFFFFFF); var count = 0; // while line is not complete while (!lineAtDestination(fromX, fromY, toX, toY, currentX, currentY)) { /// move line draw cursor to beginning of next dash line.graphics.moveTo(currentX, currentY); // if dash is even if (count % 2 == 0) { // draw the dash line.graphics.lineTo(currentX + addX, currentY + addY); } // add next dash's length to current cursor position currentX += addX; currentY += addY; count++; } return line; } This just happens to be written in Haxe, but the solution should be language neutral. What I would like is for the dash length to be the same no matter what angle the line is. As is, it's just adding 75 thousandths of the line length to the x and y, so if the line is and a 45 degree angle you get pretty much a solid line. If the line is at something shallow like 85 degrees then you get a nice looking dashed line. So, the dash length is variable, and I don't want that. How would I make a function that I can pass a "dash length" into and get that length of dash, no matter what the angle is? If you need to completely disregard my code, be my guest. I'm sure there's a better solution.

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  • Attempt at Merge Sort: Is this correct? [migrated]

    - by Beatrice
    I am trying to write a merge sort algo. I can't tell if this is actually a canonical merge sort. If I knew how to calculate the runtime I would give that a go. Does anyone have any pointers? Thanks. public static void main(String[] argsv) { int[] A = {2, 4, 5, 7, 1, 2, 3, 6}; int[] L, R; L = new int[A.length/2]; R = new int[A.length/2]; int i = 0, j = 0, k; for (k = 0; k < A.length; k++) { if (k < A.length/2) { L[i] = A[k]; i++; } else { R[j] = A[k]; j++; } } i = 0; j = 0; for (k = 0; k < A.length; k++) { System.out.println(i + " " + j + " " + k); if (i < L.length && j < R.length) { if (L[i] < R[j]) { A[k] = L[i]; i++; } else { A[k] = R[j]; j++; } } } }

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  • Vagrant (Virtualbox) host-only multiple node networking issue

    - by Lorin Hochstein
    I'm trying to use a multi-VM vagrant environment as a testbed for deploying OpenStack, and I've run into a networking problem with trying to communicate from one VM, to a VM-inside-of-a-VM. I have two Vagrant nodes, a cloud controller node and a compute node. I'm using host-only networking. My Vagrantfile looks like this: Vagrant::Config.run do |config| config.vm.box = "precise64" config.vm.define :controller do |controller_config| controller_config.vm.network :hostonly, "192.168.206.130" # eth1 controller_config.vm.network :hostonly, "192.168.100.130" # eth2 controller_config.vm.host_name = "controller" end config.vm.define :compute1 do |compute1_config| compute1_config.vm.network :hostonly, "192.168.206.131" # eth1 compute1_config.vm.network :hostonly, "192.168.100.131" # eth2 compute1_config.vm.host_name = "compute1" compute1_config.vm.customize ["modifyvm", :id, "--memory", 1024] end end When I try to start up a (QEMU-based) VM, it boots successfully on compute1, and its virtual nic (vnet0) is connected via a bridge, br100: root@compute1:~# brctl show 100 bridge name bridge id STP enabled interfaces br100 8000.08002798c6ef no eth2 vnet0 When the QEMU VM makes a request to the DHCP server (dnsmasq) running on controller, I can see the request reaches the controller because of the output on the syslog on the controller: Aug 6 02:34:56 precise64 dnsmasq-dhcp[12042]: DHCPDISCOVER(br100) fa:16:3e:07:98:11 Aug 6 02:34:56 precise64 dnsmasq-dhcp[12042]: DHCPOFFER(br100) 192.168.100.2 fa:16:3e:07:98:11 However, the DHCPOFFER never makes it back to the VM running on compute1. If I watch the requests using tcpdump on the vboxnet3 interface on my host machine that runs Vagrant (Mac OS X), I can see both the requests and the replies $ sudo tcpdump -i vboxnet3 -n port 67 or port 68 tcpdump: WARNING: vboxnet3: That device doesn't support promiscuous mode (BIOCPROMISC: Operation not supported on socket) tcpdump: verbose output suppressed, use -v or -vv for full protocol decode listening on vboxnet3, link-type EN10MB (Ethernet), capture size 65535 bytes 22:51:20.694040 IP 0.0.0.0.68 > 255.255.255.255.67: BOOTP/DHCP, Request from fa:16:3e:07:98:11, length 280 22:51:20.694057 IP 0.0.0.0.68 > 255.255.255.255.67: BOOTP/DHCP, Request from fa:16:3e:07:98:11, length 280 22:51:20.696047 IP 192.168.100.1.67 > 192.168.100.2.68: BOOTP/DHCP, Reply, length 311 22:51:23.700845 IP 0.0.0.0.68 > 255.255.255.255.67: BOOTP/DHCP, Request from fa:16:3e:07:98:11, length 280 22:51:23.700876 IP 0.0.0.0.68 > 255.255.255.255.67: BOOTP/DHCP, Request from fa:16:3e:07:98:11, length 280 22:51:23.701591 IP 192.168.100.1.67 > 192.168.100.2.68: BOOTP/DHCP, Reply, length 311 22:51:26.705978 IP 0.0.0.0.68 > 255.255.255.255.67: BOOTP/DHCP, Request from fa:16:3e:07:98:11, length 280 22:51:26.705995 IP 0.0.0.0.68 > 255.255.255.255.67: BOOTP/DHCP, Request from fa:16:3e:07:98:11, length 280 22:51:26.706527 IP 192.168.100.1.67 > 192.168.100.2.68: BOOTP/DHCP, Reply, length 311 But, if I tcpdump on eth2 on compute, I only see the requests, not the replies: root@compute1:~# tcpdump -i eth2 -n port 67 or port 68 tcpdump: WARNING: eth2: no IPv4 address assigned tcpdump: verbose output suppressed, use -v or -vv for full protocol decode listening on eth2, link-type EN10MB (Ethernet), capture size 65535 bytes 02:51:20.240672 IP 0.0.0.0.68 > 255.255.255.255.67: BOOTP/DHCP, Request from fa:16:3e:07:98:11, length 280 02:51:23.249758 IP 0.0.0.0.68 > 255.255.255.255.67: BOOTP/DHCP, Request from fa:16:3e:07:98:11, length 280 02:51:26.258281 IP 0.0.0.0.68 > 255.255.255.255.67: BOOTP/DHCP, Request from fa:16:3e:07:98:11, length 280 At this point, I'm stuck. I'm not sure why the DHCP replies aren't making it to the compute node. Perhaps it has something to do with the configuration of the VirtualBox virtual switch/router? Note that eth2 interfaces on both nodes have been set to promiscuous mode.

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  • obtaining nimbuzz server certificate for nmdecrypt expert in NetMon

    - by lurscher
    I'm using Network Monitor 3.4 with the nmdecrypt expert. I'm opening a nimbuzz conversation node in the conversation window and i click Expert- nmDecrpt - run Expert that shows up a window where i have to add the server certificate. I am not sure how to retrieve the server certificate for nimbuzz XMPP chat service. Any idea how to do this? this question is a follow up question of this one. Edit for some background so it might be that this is encrypted with the server pubkey and i cannot retrieve the message, unless i debug the native binary and try to intercept the encryption code. I have a test client (using agsXMPP) that is able to connect with nimbuzz with no problems. the only thing that is not working is adding invisible mode. It seems this is some packet sent from the official client during login which i want to obtain. any suggestions to try to grab this info would be greatly appreciated. Maybe i should get myself (and learn) IDA pro? This is what i get inspecting the TLS frames on Network Monitor: Frame: Number = 81, Captured Frame Length = 769, MediaType = ETHERNET + Ethernet: Etype = Internet IP (IPv4),DestinationAddress:[...],SourceAddress:[....] + Ipv4: Src = ..., Dest = 192.168.2.101, Next Protocol = TCP, Packet ID = 9939, Total IP Length = 755 - Tcp: Flags=...AP..., SrcPort=5222, DstPort=3578, PayloadLen=715, Seq=4101074854 - 4101075569, Ack=1127356300, Win=4050 (scale factor 0x0) = 4050 SrcPort: 5222 DstPort: 3578 SequenceNumber: 4101074854 (0xF4716FA6) AcknowledgementNumber: 1127356300 (0x4332178C) + DataOffset: 80 (0x50) + Flags: ...AP... Window: 4050 (scale factor 0x0) = 4050 Checksum: 0x8841, Good UrgentPointer: 0 (0x0) TCPPayload: SourcePort = 5222, DestinationPort = 3578 TLSSSLData: Transport Layer Security (TLS) Payload Data - TLS: TLS Rec Layer-1 HandShake: Server Hello.; TLS Rec Layer-2 HandShake: Certificate.; TLS Rec Layer-3 HandShake: Server Hello Done. - TlsRecordLayer: TLS Rec Layer-1 HandShake: ContentType: HandShake: - Version: TLS 1.0 Major: 3 (0x3) Minor: 1 (0x1) Length: 42 (0x2A) - SSLHandshake: SSL HandShake ServerHello(0x02) HandShakeType: ServerHello(0x02) Length: 38 (0x26) - ServerHello: 0x1 + Version: TLS 1.0 + RandomBytes: SessionIDLength: 0 (0x0) TLSCipherSuite: TLS_RSA_WITH_AES_256_CBC_SHA { 0x00, 0x35 } CompressionMethod: 0 (0x0) - TlsRecordLayer: TLS Rec Layer-2 HandShake: ContentType: HandShake: - Version: TLS 1.0 Major: 3 (0x3) Minor: 1 (0x1) Length: 654 (0x28E) - SSLHandshake: SSL HandShake Certificate(0x0B) HandShakeType: Certificate(0x0B) Length: 650 (0x28A) - Cert: 0x1 CertLength: 647 (0x287) - Certificates: CertificateLength: 644 (0x284) - X509Cert: Issuer: nimbuzz.com,Nimbuzz,NL, Subject: nimbuzz.com,Nimbuzz,NL + SequenceHeader: - TbsCertificate: Issuer: nimbuzz.com,Nimbuzz,NL, Subject: nimbuzz.com,Nimbuzz,NL + SequenceHeader: + Tag0: + Version: (2) + SerialNumber: -1018418383 + Signature: Sha1WithRSAEncryption (1.2.840.113549.1.1.5) - Issuer: nimbuzz.com,Nimbuzz,NL - RdnSequence: nimbuzz.com,Nimbuzz,NL + SequenceOfHeader: 0x1 + Name: NL + Name: Nimbuzz + Name: nimbuzz.com + Validity: From: 02/22/10 20:22:32 UTC To: 02/20/20 20:22:32 UTC + Subject: nimbuzz.com,Nimbuzz,NL - SubjectPublicKeyInfo: RsaEncryption (1.2.840.113549.1.1.1) + SequenceHeader: + Algorithm: RsaEncryption (1.2.840.113549.1.1.1) - SubjectPublicKey: - AsnBitStringHeader: - AsnId: BitString type (Universal 3) - LowTag: Class: (00......) Universal (0) Type: (..0.....) Primitive TagValue: (...00011) 3 - AsnLen: Length = 141, LengthOfLength = 1 LengthType: LengthOfLength = 1 Length: 141 bytes BitString: + Tag3: + Extensions: - SignatureAlgorithm: Sha1WithRSAEncryption (1.2.840.113549.1.1.5) - SequenceHeader: - AsnId: Sequence and SequenceOf types (Universal 16) + LowTag: - AsnLen: Length = 13, LengthOfLength = 0 Length: 13 bytes, LengthOfLength = 0 + Algorithm: Sha1WithRSAEncryption (1.2.840.113549.1.1.5) - Parameters: Null Value - Sha1WithRSAEncryption: Null Value + AsnNullHeader: - Signature: - AsnBitStringHeader: - AsnId: BitString type (Universal 3) - LowTag: Class: (00......) Universal (0) Type: (..0.....) Primitive TagValue: (...00011) 3 - AsnLen: Length = 129, LengthOfLength = 1 LengthType: LengthOfLength = 1 Length: 129 bytes BitString: + TlsRecordLayer: TLS Rec Layer-3 HandShake:

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  • How to do RLE (run length encoding) in C# on a byte array?

    - by CuriousCoder
    I am trying to XOR two bitmap files (their byte arrays) to produce a byte array that can be used to change image A into image B or vice versa. I am sending this over the network so I would like to do some basic compression before this happens. Is there a way to do RLE (run length encoding) in C# (using a built-in, or fast reliable 3rd party library) on a byte array for this purpose? Notes: If you are going to suggest an alternative to my approach please keep in mind that the decompression and transformation on the remote machine has to be as quick and efficient as possible.

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  • How to speed up calculation of length of longest common substring?

    - by eSKay
    I have two very large strings and I am trying to find out their Longest Common Substring. One way is using suffix trees (supposed to have a very good complexity, though a complex implementation), and the another is the dynamic programming method (both are mentioned on the Wikipedia page linked above). Using dynamic programming The problem is that the dynamic programming method has a huge running time (complexity is O(n*m), where n and m are lengths of the two strings). What I want to know (before jumping to implement suffix trees): Is it possible to speed up the algorithm if I only want to know the length of the common substring (and not the common substring itself)?

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  • Python: How would i write this 'if' statement for a word of arbitrary length?

    - by ElCarlos
    This is what I currently have: wordlist = [fox, aced, definite, ace] for word in wordlist: a = len(word) if (ord(word[a-(a-1)] - ord(word[(a-a)])) == ord(word[a-(a-2)])-ord(word[a-(a-1)]: print "success", word else: print "fail", word What I'm trying to do is calculate the ASCII values between each of the letters in the word. And check to see if the ord of the letters are increasing by the same value. so for fox, it would check if the difference between the ord of 2nd and 1st letters are equal to the ord difference of the 3rd and 2nd letters. However, with my current 'if' statement, only the first 3 letters of a word are compared. How can I rewrite this statement to cover every letter in a word of length greater than 3? Sorry if I can't present this clearly, thanks for your time.

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  • Is there a default buffer length for 'sprintf' method?

    - by Isuru
    Hi, I used sprintf method to format data to a string which I want to write to a file, in C++ console application using VS 2008. The Input is a particular message, which has various variables and values (ex: Type 'int' and Value '10' / Type string and value "abc", etc.) When I send a two messages it works perfectly. But When I send more than two messages it gives a runtime error saying 0xC0000005: Access violation reading location 0xabababab. Why is this happening? Is it because the method 'sprintf' has a default buffer length? How can I overcome this problem?

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  • In Haskell, will calling length on a Lazy ByteString force the entire string into memory?

    - by me2
    I am reading a large data stream using lazy bytestrings, and want to know if at least X more bytes is available while parsing it. That is, I want to know if the bytestring is at least X bytes long. Will calling length on it result in the entire stream getting loaded, hence defeating the purpose of using the lazy bytestring? If yes, then the followup would be: How to tell if it has at least X bytes without loading the entire stream? EDIT: Originally I asked in the context of reading files but understand that there are better ways to determine filesize. Te ultimate solution I need however should not depend on the lazy bytestring source.

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  • How to limit NSTextField text length and keep it always upper case?

    - by carlosb
    Need to have an NSTextField with a text limit of 4 characters maximum and show always in upper case but can't figure out a good way of achieving that. I've tried to do it through a binding with a validation method but the validation only gets called when the control loses first responder and that's no good. Temporarly I made it work by observing the notification NSControlTextDidChangeNotification on the text field and having it call the method: - (void)textDidChange:(NSNotification*)notification { NSTextField* textField = [notification object]; NSString* value = [textField stringValue]; if ([value length] > 4) { [textField setStringValue:[[value uppercaseString] substringWithRange:NSMakeRange(0, 4)]]; } else { [textField setStringValue:[value uppercaseString]]; } } But this surely isn't the best way of doing it. Any better suggestion?

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  • maximum string length quota error consuming WCF webservice from Biztalk.

    - by TygerKrash
    I'm getting this error message "The Maximum string content length quota (8192) has been exceeded while reading XML data. This quotea may be increased by changing the MaxStringContentLength property on the XmlDictionaryReaderQuotas object used when creating the XML reader" In the one of my orchestrations that consumes a wcf webservice (stacktrace indicates the receive shape is where the issue is), It is likely that the response is very large. looking at some of the other questions with this error message the solution is to change a WCF bindings setting in the configuration file. However I can't find these configuration settings when I'm using biztalk. They don't seem to be generated anywhere, should I be trying to add them to BTSNTSVc.exe.config. Any suggestions welcome.

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  • [Sql-Server]what data type to use for password salt and hash values and what length?

    - by Pandiya Chendur
    I am generating salt and hash values from my passwords by using, string salt = CreateSalt(TxtPassword.Text.Length); string hash = CreatePasswordHash(TxtPassword.Text, salt); private static string CreateSalt(int size) { //Generate a cryptographic random number. RNGCryptoServiceProvider rng = new RNGCryptoServiceProvider(); byte[] buff = new byte[size]; rng.GetBytes(buff); // Return a Base64 string representation of the random number. return Convert.ToBase64String(buff); } private static string CreatePasswordHash(string pwd, string salt) { string saltAndPwd = String.Concat(pwd, salt); string hashedPwd = FormsAuthentication.HashPasswordForStoringInConfigFile( saltAndPwd, "sha1"); return hashedPwd; } What datatype you would suggest for storing these values in sql server? Any suggestion... Salt:9GsPWpFD Hash:E778AF0DC5F2953A00B35B35D80F6262CDBB8567

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  • Why doesn't Python's `re.split()` split on zero-length matches?

    - by Tim Pietzcker
    One particular quirk of the (otherwise quite powerful) re module in Python is that re.split() will never split a string on a zero-length match, for example if I want to split a string along word boundaries: >>> re.split(r"\s+|\b", "Split along words, preserve punctuation!") ['Split', 'along', 'words,', 'preserve', 'punctuation!'] instead of ['', 'Split', 'along', 'words', ',', 'preserve', 'punctuation', '!'] Why does it have this limitation? Is it by design? Do other regex flavors behave like this?

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  • How do I generate a random string of up to a certain length?

    - by slavy13
    I would like to generate a random string (or a series of random strings, repetitions allowed) of length between 1 and n characters from some (finite) alphabet. Each string should be equally likely (in other words, the strings should be uniformly distributed). The uniformity requirement means that an algorithm like this doesn't work: alphabet = "abcdefghijklmnopqrstuvwxyz" len = rand(1, n) s = "" for(i = 0; i < len; ++i) s = s + alphabet[rand(0, 25)] (pseudo code, rand(a, b) returns a integer between a and b, inclusively, each integer equally likely) It doesn't work because shorter lengths are as likely as longer ones, meaning it's more likely to generate a shorter string than a longer one, so the result is not uniform.

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  • how to set length and postion for repeat-x and scroll background-image?

    - by user550945
    Hi all, I have an to set background image. it is a red line under the input value. background-attachment:scroll; background-color:#FFFFFF; background-image:url("../images/errorLine.gif"); background-position:left bottom; background-repeat:repeat-x; the input value is like "123; 456; 789;". the red line cover all the string, its length is the same as the input width. Is there any way to make the red line only under the "456;"? Is there anyway to do it by CSS? Thanks a lot. Best regards, Ryanivanka

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