C++ array size different result

Question

Possible Duplicate:
Sizeof an array in the C programming language?

    #include "stdafx.h"
    #include <string>
    #include <iostream>

    using namespace std;

    string a[] = {"some", "text"};

    void test(string a[])
    {
        int size_of_a = sizeof(a) /sizeof(a[0]);
        cout << size_of_a; 
    }
    int _tmain(int argc, _TCHAR* argv[])
    {
        test(a); //gives 0
        int size_of_a = sizeof(a) /sizeof(a[0]);
        cout << size_of_a; //gives 2
        return 0;
    }

as u can see in the comment test(a) gives 0 instead of 2 as i would expect. Could someone explain why and how could i correct it? thanks

It would be very difficult to get 0, as `sizeof(a)` would need to be zero. — Ed Heal, Sep 21 '12 at 17:07

score 1 · Accepted Answer · answered Sep 21 '12 at 17:04

1

When you pass an array to a function, it decays to a pointer to the first element of the array and so within your test(string a[]) function

sizeof(a);

actually returns the size of a pointer and not the size of your array.

answered Sep 21 '12 at 17:04

mathematician1975

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Thanks for the explonation. How should i correct it? Whats the best way to program this? – user1686999 Sep 21 '12 at 17:13
1

yay vectors! or C++11 std::array – im so confused Sep 21 '12 at 17:14
1

@user1686999 As you are using C++ use a vector - they make lives so so much easier. – mathematician1975 Sep 21 '12 at 17:14

Lol4t0 · Answer 2 · 2012-09-21T17:17:31.680

0

To prevent array decaing to pointer, you can pass reference to array to the function. But it causes types of array of function formal argument and actual argument must coincide (including their sizes). So you need to use template to make your function work with an array of any size:

template <int N>
void foo(const string (&a)[N])
{
    int size_of_a = sizeof(a) /sizeof(a[0]);
    cout << size_of_a; 
}

edited Sep 21 '12 at 17:17

answered Sep 21 '12 at 17:12

Lol4t0

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C++ array size different result

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