[FIXED] why do I see different values of array size in main vs other function

Issue

I created an array of ten integers and print its size 1. in main 2. in a function named print

#include <iostream>
using namespace std;
void print(int *a)
{
    cout<<sizeof(a);
}
int main()
{
    int arr[10];
    cout<<sizeof(arr)<<endl;
    print(arr);
}

The output is :

40

8

I was expecting 40 in both cases (as size of 1 integer=4 times 10 ) but the second case shows size of a single pointer and not the whole array. What's happening here?

Solution

As you can see from the function declaration

void print(int *a)
{
    cout<<sizeof(a);
}

the function deals with a pointer. So this statement

cout<<sizeof(a);

outputs the size of the pointer that is equal either 4 or 8 bytes depending on the used system.

Pay attention to that even if you will declare the function like

void print(int a[])
{
    cout<<sizeof(a);
}

you will get the same result because the parameter is implicitly adjusted by the compiler to the type int *.

That is the previous and this function declarations are equivalent.

If you want that the function would deal with the original array instead of a pointer that points to the first element of the passed array as an argument then declare the function at least like.

void print( int ( &a )[10] )
{
    cout<<sizeof(a);
}

That is declare the parameter as a reference.

Or you can make the function a template function like

template <size_t N>
void print( int ( &a )[N] )
{
    cout<<sizeof(a);
}

As the passed array is not being changed in the function then the parameter should have the qualifier const

template <size_t N>
void print( const int ( &a )[N] )
{
    cout<<sizeof(a);
}

Here is a demonstrative program.

#include <iostream>

template <size_t N>
void print( const int ( &a )[N] )
{
    std::cout << sizeof( a ) << '\n';
}

int main() 
{
    int arr[10];

    std::cout << sizeof( arr ) << '\n';

    print( arr );

    return 0;
}

Its output is

40
40

Or you could define the template function with a type template parameter. But again the function parameter has a referenced type.

#include <iostream>

template <typename T>
void print( const T &a )
{
    std::cout << sizeof( a ) << '\n';
}

int main() 
{
    int arr[10];

    std::cout << sizeof( arr ) << '\n';

    print( arr );

    return 0;
}

The program output will be the same as shown above.

Answered By - Vlad from Moscow

Answer Checked By - Senaida (PHPFixing Volunteer)