sizeof operator returns size of pointer instead of array

Question

I'm a little confused when I run the code below:

int Maxi(int A[])
{
    return sizeof(A)/sizeof(A[0]);
}

int main()
{
    int A[5] = {1,2,3,4,5};
    print("%d \n", Maxi(A));
    return 0;
}

And I just get the result 2. It seems that sizeof(A) turns out to be 8, which is the size of a pointer. I know this may not be a question worthy asking, but I must miss something and I just cannot figure it out right now.

But if I do this:

int main()
{
    int A[5] = {1,2,3,4,5};
    print("%d \n", sizeof(A)/sizeof(A[0]));
    return 0;
}

It's all right then, and I thought it is related to the passage of value between functions, but I'm confused.

Thanks!

Note that `int Maxi(int A[])` is just another syntax for `int Maxi(int *A)` — nos, Aug 12 '14 at 09:03
[see here](http://stackoverflow.com/questions/22677415/why-do-c-and-c-compilers-allow-array-lengths-in-function-signatures-when-they/) for explanation of `int A[]` in parameter list — M.M, Aug 12 '14 at 09:12

score 1 · Accepted Answer · edited May 23 '17 at 12:11

1

The A array decays to a pointer once you pass it as a parameter to a function. Hence it's size gets computed to 8 (I'm guessing you're on a 64-bit system). sizeof(A[0]) is 4 (an int) therefore that's your 2.

The reason it 'works' inside main is that the compiler is able to compute the size directly from your static initialization above.

edited May 23 '17 at 12:11

Community

1
1

answered Aug 12 '14 at 08:59

dragosht

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It's correct when doing the same in `main`. So how can I do it outside of `main`? – ray6080 Aug 12 '14 at 09:04
If you simply try to get the number of elements of your array into the function, how about `int Maxi(int A[], int nelements)` ? – dragosht Aug 12 '14 at 09:07
Yeah, I did it. I just want to figure out maybe other ways to do this elegantly. – ray6080 Aug 12 '14 at 09:09

score 1 · Answer 2 · answered Aug 12 '14 at 09:09

1

You can't pass an integer array to another function with array name.

int Maxi(int A[])

is equal to this

int Maxi(int *A)

syntax.

So whenever you are passing an array to another function you should pass number of elements in that array also.

int Maxi(int A[], int elements)
//....
// call should be
print("%d \n", Maxi(A, elements));

else it treats array as a integer pointer, and it will return the size of the pointer.

answered Aug 12 '14 at 09:09

Sathish

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So no way to compute the size of an array outside the scope of its definition? – ray6080 Aug 12 '14 at 09:11
@ray6080 ya exactly! if you want to pass an array means you need to pass its no of elements also – Sathish Aug 12 '14 at 09:12

score 0 · Answer 3 · answered Aug 12 '14 at 09:17

When you pass an array to a function, within the called function, Maxi in this case, the parameter declaration for Maxi as int A[] same as int *A

int Maxi(int A[])

is equivalent to:

int Maxi(int *A)

So, an array passed is seen as a pointer within the called function, and size of a pointer is of course not the size of the array within Maxi,so, the size has to passed explicitly as in:

int Maxi(int A[], int sz)
{
    return sz/sizeof(A[0]);
}

int main()
{
    int A[5] = {1,2,3,4,5};
    printf("%d \n", Maxi(A, sizeof(A)));
    return 0;
}

This will output 5, as desired

sizeof operator returns size of pointer instead of array

3 Answers3