Question

#include <iostream>

template <typename T>
inline
T accum (T const* beg, T const* end)
{
    T total = T();  // assume T() actually creates a zero value
    while (beg != end) {
        total += *beg;
        ++beg;
    }
    return total;
}

int main() 
{ 

// create array of 5 integer values 
int num[]={1,2,3,4,5}; 

// print average value 
std::cout << "the average value of the integer values is " 
          << accum(&num[0], &num[5]) / 5 
          << '\n'; 

// create array of character values 
char name[] = "templates"; 
int length = sizeof(name)-1; 

// (try to) print average character value 
std::cout << "the average value of the characters in \"" 
          << name << "\" is " 
          << accum(&name[0], &name[length]) / length 
          //<< accum<int>(&name[0], &name[length]) / length //but this give me error
          << '\n'; 
} 

I was reading c++ templates:the complete guide book and the author mentioned
that I can use template specialization
accum<int>(&name[0], &name[length]) / length
I try this in visual studio 2012 and got error
main.cpp(34): error C2664: ‘accum’ : cannot convert parameter 1 from ‘char *’ to ‘const int *’
My C++ is a bit rusty.
I’m just curious, if this “behavior” previously allowed but there’s changes in “latest” C++ standard making this illegal or this is an error on the book I’m reading.

Answer 1

The instantiation is int accum<int> (int const* beg, int const* end), you can’t pass char * arguments to this function.

The reason the uncommented line works is that it instantiates accum<char>.