OK some answers about malloc were already posted. The more interesting part is how free works (and in this direction, malloc too can be understood better). In many malloc/free implementations, free does normally not return the memory to the operating system (or at least only in rare cases). The reason is that you will get … Read more
The answer is right there in the man page (at least on Linux): RETURN VALUE The alloca() function returns a pointer to the beginning of the allocated space. If the allocation causes stack overflow, program behaviour is undefined. Which isn’t to say it should never be used. One of the OSS projects I work on … Read more
Unless you are forced to use C, you should never use malloc. Always use new. If you need a big chunk of data just do something like: char *pBuffer = new char[1024]; Be careful though this is not correct: //This is incorrect – may delete only one element, may corrupt the heap, or worse… delete … Read more
Just about every modern operating system will recover all the allocated memory space after a program exits. The only exception I can think of might be something like Palm OS where the program’s static storage and runtime memory are pretty much the same thing, so not freeing might cause the program to take up more … Read more
calloc() gives you a zero-initialized buffer, while malloc() leaves the memory uninitialized. For large allocations, most calloc implementations under mainstream OSes will get known-zeroed pages from the OS (e.g. via POSIX mmap(MAP_ANONYMOUS) or Windows VirtualAlloc) so it doesn’t need to write them in user-space. This is how normal malloc gets more pages from the OS … Read more
TL;DR int *sieve = (int *) malloc(sizeof(int) * length); has two problems. The cast and that you’re using the type instead of variable as argument for sizeof. Instead, do like this: int *sieve = malloc(sizeof *sieve * length); Long version No; you don’t cast the result, since: It is unnecessary, as void * is automatically … Read more