size of array - why function gives size ONE only

the user-defined function size_of_array(numA) is meant to give the size of numA which is 5 but it gives 1... I wonder where went wrong? can anyone help pls...
insert

Code:

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#include <iostream>
using namespace std;
int size_of_array(int numA[]);

int main()
{
        int numA[]={0,1,2,3,4};
        int i=0;
        int total=0;

        for(i=0; i<size_of_array(numA); i++)
        {
                total+=numA[i];
        }
        float average=total/i;
        cout << "average is "<<average;
       
        return 0;
}
int size_of_array(int numA[])
{
                int num_of_elements=sizeof(numA)/sizeof(int);
                return num_of_elements;
}

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This has been answered elsewhere.

Obviously, sizeof(numA) is 5, and sizeof(int) is 4(depending on the compiler). You have made 5/4 and that results in 1.

Quote:

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Originally Posted by Dawnson

Obviously, sizeof(numA) is 5, and sizeof(int) is 4(depending on the compiler). You have made 5/4 and that results in 1.

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Wrong. sizeof(numA) is 4 here. Had numA been an array rather than a pointer, of 5 elements, sizeof(numA) would have been 20.
Yes, depending on the compiler.

Quote:

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Originally Posted by Dawnson

Obviously, sizeof(numA) is 5, and sizeof(int) is 4(depending on the compiler). You have made 5/4 and that results in 1.

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Absolutely wrong. If num[] would have contained 8 elements then according to you the answer should have been 2 which is incorrect. The correct reason is in the link given by laserlight. When you pass an array to a function it decomposes into pointer to the the first element and thus sizeof(num)==4 which is equal to sizeof(int).

You could do this instead.

Code:

---

#include <iostream>
using namespace std;

template<typename T, int N>
char (&array(T (&)[N]))[N];

int main()
{
        int numA[]={0,1,2,3,4};
        int i=0;
        int total=0;

        for(i=0; i<sizeof(array(numA)); i++)
        {
                total+=numA[i];
        }
        float average=total/i;
        cout << "average is "<<average;
       
        return 0;
}

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Or how about using some typedefs? I am having trouble discerning the types that goes into that function. It's pure luck I can see it's a function, too.

Code:

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template<typename T, int N>
void myfunction(T (& name)[N])
{
    //...
}

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But I also believe something like this works, where N is the number of elements. You could do a sizeof on name and get the real size in this case.
Or you could just use a vector and use the .size() method and multiple with the size of one element.

I think we need to wait til C++0x in order to really simplify that ugly function signature

Code:

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template<typename T, int N>
typedef T (&array_t)[N];

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** I assume that this will eventually be legal, although even Comeau with C++0x extensions doesn't recognise it right now, so perhaps I misunderstand what they actually mean by "adding support for template typedefs" :-)


Yes I would prefer an STL container too; Personally I only find myself using these to initialise containers using const arrays as lookup tables.

The reason I prefer the sizeof(char(&)[N]) approach is that the return type can be evaluated at compile time without ever actually calling the function, but I suppose a solution which actually calls a function would be OK aswell.

Code:

---

template<typename T, int N>
int array_size( T (&dummy_arg)[N] )
{
    return N;
}

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Or, if a range is needed for an STL container or algorithm then maybe returning a pointer would be better

Code:

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template<typename T, int N>
T* end_of( T (&array)[N] )
{
    return array+N;
}

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