Type of Template Parameters (For function Templates) : C++ -

i reading tutorial:

http://www.learncpp.com/cpp-tutorial/144-expression-parameters-and-template-specialization/

and mentioned however, template type parameters not type of template parameters available. template classes **(not template functions)** can make use of kind of template parameter known expression parameter.

so wrote program:

#include <iostream>  using namespace std;      template<typename t,int n>     bool compare(t t,const char* c)     {     if (n != 1)     {      cout << "exit failure" << endl;      exit(exit_failure);      }      bool result=false;     cout << c << endl;     cout << t << endl;     cout << t.compare(c) << endl;     if(t.compare(c) == 0)     result = true;      return result;     }     int main() {     string name="michael";      if (compare<string,1>(name,"sam"))     cout << "it sam" << endl;     else     cout << "this not sam" << endl;     return 0;     } 

and got output:

$ ./expressionparameter sam michael -1 not sam 

clearly, here template parameter taking int n expression parameter. point mentioned in tutorial template classes (not template functions) can make use of kind of template parameter known expression parameter. seems incorrect.

further reading at

non-type template parameters

also suggest same.

so have understood that: no matter if function template or class template, template parameter template type parameter i.e. typename or expression parameter. constraint , expression parameter - must constant integral expression. compiler not differentiate if function or class.

is understanding correct?

yes, seem understand correctly.

one use case of write generic code still benefits of compile-time constant (such better compiler optimizations).

one example std::array, takes such template size_t parameter length. example std::enable_if, uses bool.