c

问题 Normally, to indicate EOF to a program attached to standard input on a Linux terminal, I need to press Ctrl+D once if I just pressed Enter, or twice otherwise. I noticed that the patch command is different, though. With it, I need to press Ctrl+D twice if I just pressed Enter, or three times otherwise. (Doing cat | patch instead doesn't have this oddity. Also, If I press Ctrl+D before typing any real input at all, it doesn't have this oddity.) Digging into patch 's source code, I traced this

问题 This question already has answers here : Sockets - Using INADDR_ANY on client side (4 answers) Closed 3 years ago . In socket programming what does servaddr.sin_addr.s_addr=INADDR_ANY; in a client program do actually? What does INADDR_ANY mean? Also in another source found in the internet, I found servaddr.sin_addr.s_addr=inet_addr(argv[1]); What is this step doing? I think I have to enter the server ip as argument in that client wants to get connected to. Am I correct? 回答1: On a server,

问题 I want to insert a block in between two basic blocks in LLVM. So for example, if a basic block A was jumping to basic block B, I want to insert a basic block C in between them such that A jumps to C and C jumps to B. How can I do that? I do have the basic idea that I need to change the terminating instruction of Basic Block A, such that the target B is replaced by C, but how do I go on adding the new basic block C in between? 回答1: Yes, you need to change (or replace) the terminating

问题 I want to insert a block in between two basic blocks in LLVM. So for example, if a basic block A was jumping to basic block B, I want to insert a basic block C in between them such that A jumps to C and C jumps to B. How can I do that? I do have the basic idea that I need to change the terminating instruction of Basic Block A, such that the target B is replaced by C, but how do I go on adding the new basic block C in between? 回答1: Yes, you need to change (or replace) the terminating

问题 This is a sister question along with my first question Allow C# application built with .NET 2.0 to run on .NET 4.0/4.5. I simplified my situation a little bit there for understanding. Actually our case is a little special. Basically we wrote a C# DLL (Let's call E.dll ) for our MSI installer. Since our MSI installer is using older version of windows installer, it could not use C# DLL directly and it could only work with C type DLL, so we use an open source library called DLLExporter to

问题 I have code snippet that I can't understand how it works, because of one line that does a double dereference. The code looks like this: void afunction(int**x){ *x = malloc(2 * sizeof(int)); **x = 12; *(*x + 1) = 13; } int main(){ int *v = 10; afunction(&v); printf("%d %d\n", v[0], v[1]); return 1; } I understand that the first element of the pointer to pointer gets the value 12, but the line after that I just can't seem to understand. Does the second element in the first pointer get value 13?

问题 I'm writing Win32-API C code that needs to wait for new TCP connections and on the other side can be closed at any time by any other process/thread. Therefore, I need to somehow WaitForSingleObject on the stop event and wait for connections using WSAAccept simultaneously. I tried WaitForMultipleObjects on both socket and handle but new connection won't trigger the function (also WaitForSingleObject on the socket won't be triggered on a new connection). Any idea? 回答1: You need to use

问题 I've seen in many impletation of sine/cosine a so called extended modular precision arithmetic. But what it is for? For instance in the cephes implemetation, after reduction to the range [0,pi/4], they are doing this modular precision arithmetic to improve the precision. Hereunder the code: z = ((x - y * DP1) - y * DP2) - y * DP3; where DP1, DP2 and DP3 are some hardcoded coefficient. How to find those coefficient mathematically? I've understand the purpose of "modular extension arithmetic"

问题 I've seen in many impletation of sine/cosine a so called extended modular precision arithmetic. But what it is for? For instance in the cephes implemetation, after reduction to the range [0,pi/4], they are doing this modular precision arithmetic to improve the precision. Hereunder the code: z = ((x - y * DP1) - y * DP2) - y * DP3; where DP1, DP2 and DP3 are some hardcoded coefficient. How to find those coefficient mathematically? I've understand the purpose of "modular extension arithmetic"

问题 I had initial apprehensions about posting this question lest it be a duplicate.But even after googling with many keywords,I couldn't find any link on StackOverflow that explains static and dynamic binding for C.There are questions and answers for C++ though,but all involve classes and stuff that are clearly not for C.And the links outside StackExchange were quite dubious. I need to know the rigorous definition and contrast between these two bindings,exclusively in the context of C.I would