Question

For example, I have this structure and code:

typedef struct
{
long long *number;
} test;

test * test_structure;
test_structure = malloc(sizeof(test));
test_structure->number = malloc(sizeof(long long) * 100);
free(test_structure->number);
free(test_structure);

versus this structure and code:

typedef struct
{
long long number[100];
} test;

test * test_structure;
test_structure = malloc(sizeof(test));
free(test_structure);

Since i know that the number array will ALWAYS be 100, are either of these methods perfectly acceptable, or is one way better than the other, and why?

Answer 1

The second way is better for several reasons:

1. You only have to release one allocation, not two (reduced chance of memory leakage).
2. You have encoded the size of the array in the data structure, and this could help with code analysis. The first version could have any size, and no-one can say that any subscript is wrong. If you added a ‘size’ member that was appropriately initialized to the first structure, there would be less force to this argument, but not very much less force.
3. You can copy the entire second structure with a structure assignment; you can’t copy the first structure in the same way.
4. The assembler generated is slightly different (longer) for the first (it has to fetch the address from the member, then dereference it, instead of just dereferencing with the second).

With all that said, the two methods are substantially the same. You use (I was tempted to say ‘waste’ but that’s not quite fair) more space with the first, but it is unlikely to be a major problem (and if it was, you’d be likely to want to vary the size from 100, as that will save you more space than tinkering with pointer vs array).