Change of behavior when passing int array

Question

Possible Duplicate:
Passing an array as an argument in C++
Sizeof an array in the C programming language?

Can you please explain the output of the following code:

#include<iostream>
using namespace std;

void foo(int array[])
{
    int size = sizeof(array) / sizeof(array[0]);    
    cout<<size<<endl;
}
int main()
{
    int array[] = {1,2,3};

    int size = sizeof(array) / sizeof(array[0]);
    cout<<size<<endl;
    foo(array);
    return 0;
}

The corresponding output is:

3
2

Both the code inside foo() and inside main() looks similar to me so as to produce the same output, but it does not, can you please explain why?

ARhgh, bad learning material strikes again. No array is being passed here. — R. Martinho Fernandes, Jul 13 '12 at 08:12
Summary: Whenever you pass an array as a parameter you need to pass the length as well. Even better would be to just use `std::vector`. — Jon, Jul 13 '12 at 08:12

score 4 · Answer 1 · answered Jul 13 '12 at 08:15

4

void foo(int array[])

in C or C++ you cannot pass arrays by value, so the above declaration is interpretted as:

void foo(int * array)

Consider passing the array by reference:

template <size_t N>
void foo( int(&array)[N] )
{   
    cout << N << endl;
}

answered Jul 13 '12 at 08:15

Andrzej

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That's not really an answer, since the question was "why are the results of the output different?" – SingerOfTheFall Jul 13 '12 at 08:18

SingerOfTheFall · Accepted Answer · 2012-07-13T08:27:57.080

3

You are not passing the array into the function, you are passing the pointer to the array.

That means, in the function foo(), the result of sizeof(array) is the size of a pointer to an array of char, not the size of the array itself, so the function is effectively doing this:

cout << sizeof(int *) / sizeof(array[0]);

In this case size of int * is 8, size of int is 4, so you are getting an output of 2 from the function.

edited Jul 13 '12 at 08:27

answered Jul 13 '12 at 08:15

SingerOfTheFall

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The array elements are `int`, not `char`. Doesn't change the point, but you should still correct it. (`sizeof(char*) != sizeof(int*)` is allowed.) – Daniel Fischer Jul 13 '12 at 08:25
@DanielFischer, right, thanks! For some reason I thought it was 'char array[]' instead of 'int array[]' in OP's post – SingerOfTheFall Jul 13 '12 at 08:29

Change of behavior when passing int array

2 Answers2