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

[bijan311]

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

#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;
	}
}

the output is:

1
Press any key to continue...

How can I fix this?

[laserlight]

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

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).

How can I fix this?

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

[oogabooga]

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.

[Elkvis]

try this:

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

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.

[bijan311]

Thanks to everyone, now my problem is solved.

BTW, integer only has one r.

Well, I feel a little silly now...