I have a large byte[] receiveBuffer from a socket connection that contains multiple packets inside it.
The following code (which compiles and executes properly, doing what I want) is a minimal example of an oddity I experienced while writing a class to store properties of various types that needed the
#include \"stdio.h\" void Square(int num, int *myPointer); int main(int argc, const char *argv[]) { int originalNum = 5;
Consider: #include <stdio.h>开发者_开发知识库 int f() { return 20; } int main() { void (*blah)() = f;
I made this little piece of code where I alloc and then free memory for a struct: struct item { int val;
Code executes, but nothing gets passed on image_data (line 44 the write_function) So I have no working array for the function (line 11), why is that so? I thought passing on the reference for image_d
Why do we need the *? char* test = "testing"; From what I understood, we only app开发者_如何学JAVAly * onto addresses.This is a char:
Whew! Long title...here\'s some pseudo-code to explain that verbiage: int main(){ int* ptr = function1(); //the data that ptr points to is correct here
It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical andcannot be reasonably answered in its current form. For help clari
I have the following C code: static void* heap; static unsigned int ptr; int main(void) { ... heap=(void*)malloc(10000*sizeof(char));
加载中,请稍侯......