功能模板中的非类型参数

Question

I've a very simple template function, but there's confusion as to how to instantiate/call the function because of the non-type parameter. The definition of the template function is as goes:

template<typename Glorp, int size>
Glorp min(Glorp array[size])
{

Glorp minival = array[0];
for (int i = 0; i < size; i++)
if (array[i] < minival)
    minival = array[i];

return minival;
}

Now, in main() I have the following code:

void main()
{
const int size=5;
int array[size];
for (int i = 0; i < size; i++)
    cin >> array[i];

int p = min(array[size]);
cout << p;
}

This gets me the error message:

Error   1   error C2783: 'Glorp min(Glorp *)' : could not deduce template argument for 'size'   c:\users\tamara\documents\visual studio 2013\projects\nuevoprojecto\nuevoprojecto\main.cpp  23  1   NuevoProjecto

How DO I call this function from main() ? I can't find the answer for this, the only examples I saw were for non type parameters in template classes

Answer 1

I see two major problems in you code

1) the syntax for a template function receiving an array, deducing the type and the size, is the following

template <typename Glorp, int size>
Glorp min (Glorp (&array)[size])
 {
   // ...........^^^^^^^^
 }

2) you have to call it without [size]

int p = min(array[size]); // wrong
int p = min(array);       // correct

because passing array[size] you're trying to pass a single int from an un-allocated memory position (correct array values are from array[0] to array[size-1] ).

A minor problem: main() return a int , not a void .

Off topic suggestion: if you can use at least C++11, consider using std::array , instead of old C-style arrays, whenever possible.