Why function does not know the array size?

Question

If I write

int main()
{
    int a[100] = {1,2,3,4,};
    cout<

        

Answer 1

When passing your array as a parameter to a function taking a pointer, the array decays as a pointer.

void bar(int *a)
{
    std::cout << sizeof(a) << std::endl; // outputs "4" (on a 32 bit compiler)
}

void foo()
{
    int a[100] ;
    std::cout << sizeof(a) << std::endl; // outputs "400" (on a 32 bit compiler)
    bar(a);
}

So perhaps the solution is to provide a correct function taking a reference to an array as a parameter :

template 
void bar(int (&a)[T_Size])
{
    std::cout << T_Size << std::endl;    // outputs "100" (on ALL compilers)
    std::cout << sizeof(a) << std::endl; // outputs "400" (on a 32 bit compiler)
}

void foo()
{
    int a[100] ;
    std::cout << sizeof(a) << std::endl; // outputs "400" (on a 32 bit compiler)
    bar(a);
}

Of course, the function must be templated.