When getting array as a parameter for a function, only one element is passed

I'm trying to pass an array in a function, but only one element gets through. For example, in this program,

Code:

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#include <iostream>
using namespace std;

void displayInts(int intergers[]);

int main()
{
        int intergers[4];

        intergers[0] = 1;
        intergers[1] = 2;
        intergers[2] = 3;
        intergers[3] = 4;

        displayInts(intergers);

        system("PAUSE");
}

void displayInts(int intergers[])
{
        int size = (sizeof(intergers) / sizeof(int));

        for(int i = 0; i < size; i++)
        {
                cout<<intergers[i]<<endl;
        }
}

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the output is:

Code:

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1
Press any key to continue...

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How can I fix this?

Quote:

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Originally Posted by bijan311

I'm trying to pass an array in a function, but only one element gets through.

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Actually, none of the elements get through. What does get passed is a pointer to the first element of the array. Using that pointer, you can access the various elements of the array. Hence, the sizeof approach only gives you the size of a pointer (which you then divide by the size of an int, which does not make a great deal of sense).

Quote:

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Originally Posted by bijan311

How can I fix this?

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You could use a container instead, or maybe std::array<int, 4>. The standard containers have a size() member function.

When an array is passed to a function it "decays" to a pointer, losing any size info. So you can't determine its size using sizeof. You need to pass a second parameter with the size.

BTW, integer only has one r.

try this:

Code:

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template<int size>
void displayInts(int (&intergers)[size])
{
    for(int i = 0; i < size; i++)
    {
        cout<<intergers[i]<<endl;
    }
}

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keep in mind that this will only work if your array is a true array, and not if it's a pointer to a block of memory allocated with new or by some other means.

Thanks to everyone, now my problem is solved.

Quote:

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BTW, integer only has one r.

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Well, I feel a little silly now...