Question

Why can I return from a function an array setup by malloc:

int *dog = (int*)malloc(n * sizeof(int));

but not an array setup by

 int cat[3] = {0,0,0};

The “cat[ ]” array is returned with a Warning.

Thanks all for your help

Answer 1

There are two key parts of memory in a running program: the stack, and the heap. The stack is also referred to as the call stack.

When you make a function call, information about the parameters, where to return, and all the variables defined in the scope of the function are pushed onto the stack. (It used to be the case that C variables could only be defined at the beginning of the function. Mostly because it made life easier for the compiler writers.)

When you return from a function, everything on the stack is popped off and is gone (and soon when you make some more function calls you’ll overwrite that memory, so you don’t want to be pointing at it!)

Anytime you allocate memory you are allocating if from the heap. That’s some other part of memory, maintained by the allocation manager. Once you “reserve” part of it, you are responsible for it, and if you want to stop pointing at it, you’re supposed to let the manager know. If you drop the pointer and can’t ask to have it released any more, that’s a leak.

You’re also supposed to only look at the part of memory you said you wanted. Overwriting not just the part you said you wanted, but past (or before) that part of memory is a classic technique for exploits: writing information into part of memory that is holding computer instructions instead of data. Knowledge of how the compiler and the runtime manage things helps experts figure out how to do this. Well designed operating systems prevent them from doing that.

heap:

int *dog = (int*)malloc(n*sizeof(int*));

stack:

int cat[3] = {0,0,0};