Size of 2 D array

Question

int array[][2] = {
{1,0}, 
{2,2},
{3,4},
{4,17}
};

int main()
{
    /* calculate array size */

    printf(" => number of positions to capture : %d", (int)(sizeof(array)/sizeof(array[0])));
    func(array);
    return 0;
}
void func(int actu[][2])
{
    /* calculate array size */
    printf(" => number of positions to capture : %d", (int)(sizeof(actu)/sizeof(actu[0])));
 }

Result:

 => number of positions to capture : 4 -- inside main
 => number of positions to capture : 0 -- inside func -- I believe I should get 4 here too

Size of a same arrray in calling and called function are giving different values. Please, help me to find the issue.

On of them is taking the size of the array, the other is taking the size of a pointer decayed from an array. — Shahbaz, Nov 08 '13 at 15:57

score 3 · Answer 1 · edited May 23 '17 at 12:29

3

Take a look here and here (note the first link is talking about C++, not C, so while the mechanism is the same, in C there are no such thing as references (&)). The array has decayed into a pointer, so sizeof(actu) is not equal to sizeof(array).

edited May 23 '17 at 12:29

Community

1
1

answered Nov 08 '13 at 15:57

anjruu

1,224
10
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1

Here too... second answer. http://stackoverflow.com/questions/37538/how-do-i-determine-the-size-of-my-array-in-c – Mike Andrews Nov 08 '13 at 15:58

score 2 · Accepted Answer · answered Nov 08 '13 at 15:59

It's because when arrays are passed as function argument, they are converted to a pointer to the first element.

So in main, sizeof(array)/sizeof(array[0]) gives the array's length 4, as you may expect.

In func, sizeof(actu) is the size of a pointer, typically 4 bytes in 32-bit machine or 8 bytes in 64-bit machine, while sizeof(actu[0] is still two int, which is 8 if int is 4 bytes. In your machine, the pointer is 4 bytes, so integer division 4/8 outputs 0.

Size of 2 D array

2 Answers2