Thread: Quick question with pointers, arrays, and malloc. malloc space for array of pointers

It's all wrong because you can't have a zero-sized array. But let's pretend it's an array of one.

1. It's undefined behavior to access beyond the end of an array. You might get a segfault, you might not. That's the nature of undefined behavior.

2. Not sure what you want. Currently it's a memory leak. You can just do pointers[0] = &a and now the first (and only) element of your array points to the local variable “a”.

3. See 1.

When you say that you want to “manage an array of pointers that is flexible”, perhaps you mean something like this:

Code:

void **pointers;

pointers = malloc(5 * sizeof *pointers);
pointers[0] = some_memory_location;
pointers[1] = some_memory_location;
...
pointers[4] = some_memory_location;

Like any “dynamic array”, you create a pointer to the type you want to store. Since you want to store void*, you create a pointer to a pointer to void. Then allocate space.

Thanks!

One other question if you wouldn't mind indulging me.

Code:

Code:

int main(int argc, char * argv[]) {
	int a = 5;
	int b = 6;
	
	void ** pointers = malloc(2 * sizeof(void *));
	pointers[0] = &a;
	pointers[1] = &b;
	
	int test[2];
	test[0] = 4;
	test[1] = 8;
	
	printf("%lu\n", sizeof(test) / sizeof(int) );
	printf("%lu\n", sizeof(pointers));
	printf("%lu\n", sizeof(pointers) / sizeof(void *));
	printf("%lu\n", sizeof(*pointers) / sizeof(void *));
        
	return 0;
}

Is it possible to get the number of items for dynamic arrays like this? I can do it with the var "test" above, but am not sure if you can even do it with the "pointers" var.

Thanks again