Why am I losing bytes?

Question

I am simply trying to dynamically declare an array in C The code is as follows:

int maxSize = 3;
int *answer;
answer = malloc(maxSize * sizeof(int));
printf("this is max size: %d\n", maxSize);
printf("this is the mult result: %d\n", maxSize * sizeof(int));
printf("size of answer in bytes: %d\n", sizeof(answer));
printf("size of the answer array: %d\n", sizeof(answer) / sizeof(answer[0]));

Printing the result gives me:

this is max size: 3
this is the mult result: 12
size of answer in bytes: 8
size of the answer array: 2

I don't think it's an architecture thing (rather me being inexperienced), but I am running this on a Macbook Pro.

I do not understand why is malloc only allocating 8 bytes instead of 12 bytes for the integer array.

@George Stocker Because the size of and int in my machine is 4 bytes. Since `maxSize` is 3 bytes, I expect to allocate enough memory for an array of ints of length 3. — FSB, Mar 15 '16 at 17:51
@DigitalNinja Yes, I forgot to put that in the code example. — FSB, Mar 15 '16 at 17:51
@LeeDanielCrocker You can declare an array as a pointer in C — FSB, Mar 15 '16 at 18:25
Um, no. You can access an array through a pointer if you manually allocate the array and make the pointer point to it. — Lee Daniel Crocker, Mar 15 '16 at 18:28

score 1 · Accepted Answer · answered Mar 15 '16 at 17:51

1

Sizeof(answer) is returning the size of the variable answer in memory, which is the same as sizeOf(int *). Since your computer architecture is 64-bit, the size of a pointer is 8 bytes, which is your result. Your computer is indeed allocating 12 bytes, but you don't really have a way to validate that.

answered Mar 15 '16 at 17:51

matanso

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Thank you. This means the C compiler doesn't have knowledge of dynamically allocated memory and you can't use functions to keep track of the memory allocated, correct? – FSB Mar 15 '16 at 18:07
Yes, you're absolutely correct. – matanso Mar 15 '16 at 18:08

Why am I losing bytes?

1 Answers1