c++ - Accessing a class member in an if statement using std::is_same -

i'm trying tackle following problem: if statement depending on whether argument of template specific object or not - , if is, call object's member function. let's want std::string

the snippet:

#include <iostream> #include <string>  template <typename t> void is_string(const t& arg) {     if (std::is_same<t, const std::string&>::value)         std::cout << arg.length() << std::endl;     else         std::cout << "the argument not string" << std::endl; }  int main() {     is_string(0);     return 0; } 

it doesn't compile, following error:

types.cpp: in instantiation of ‘void is_string(const t&) [with t = int]’: types.cpp:13:13:   required here types.cpp:7:13: error: request member ‘length’ in ‘arg’, of non-class type ‘const int’    std::cout << arg.length() << std::endl; 

i reckon i'm trying achieve might not possible in c++11, appreciate suggestions on how able such thing

in regular if statement, both branches must valid code. in case int.length() makes no sense.

in c++1z use constexpr if:

if constexpr(std::is_same<t, const std::string&>::value)     std::cout << arg.length() << std::endl; else     std::cout << "the argument not string" << std::endl; 

demo

in c++11 (or older) can employ overloading achieve similar result:

void foo(std::string const& str){     std::cout << str.length() << std::endl; }  template<typename t> void foo(t const&){     std::cout << "the argument not string" << std::endl; }  template <typename t> void is_string(const t& arg) {     foo(arg); } 

demo