c++11 - C++ Template Equivalence of Prototypes -

the following compiles runs , executes expected:

#include <cstdlib> #include <cstring> #include <iostream> #include <type_traits>  class freaky { public:     template<         typename unsigned_type,         typename std::enable_if<(sizeof(unsigned_type)>=sizeof(int)),int>::type x = 0      >     static unsigned_type copything(int  x) ; };  template<     typename unsigned_type,     typename std::enable_if<(sizeof(unsigned_type)>=sizeof(int)),int>::type x >  unsigned_type freaky::copything(int x) {     unsigned_type r(0);     std::memcpy(&r,&x,sizeof(int));//please ignore. not point of question...     return r;    }  int main(int argc, char*argv[]) {     std::cout << "the answer ... " <<          freaky::copything<unsigned long>(10)<<std::endl;     return exit_success; } 

specimen output (actual output may depend on endian-ness , integer sizes):

the answer .... 10 

the following won't compile , complains prototype implementation of copything() doesn't match 1 declared in class.

#include <cstdlib> #include <cstring> #include <iostream> #include <type_traits>  class freaky { public:     template<         typename unsigned_type,         typename std::enable_if<(sizeof(unsigned_type)>=sizeof(int)),int>::type x = 0      >     static unsigned_type copything(int  x) ; };  template<     typename unsigned_type,     typename std::enable_if<(sizeof(int)<=sizeof(unsigned_type)),int>::type x >  unsigned_type freaky::copything(int x) {     unsigned_type r(0);     std::memcpy(&r,&x,sizeof(int));//please ignore. not point of question...     return r;    }  int main(int argc, char*argv[]) {     std::cout << "the answer ... " <<          freaky::copything<unsigned long>(10)<<std::endl;     return exit_success; } 

the difference between 2 sizeof(unsigned_type)>=sizeof(int) has been replaced sizeof(int)<=sizeof(unsigned_type) in implementation.

obviously 2 statements semantically equivalent. can find formal definition of how template prototypes determined equal?

it's level of lexical equivalence rather semantic equivalence.

i can't find part of standard explicitly specifies when redeclarations of class templates (or class template , specifier in out-of-line definition of members) same.

compilers follow rules redeclaration of function template, specified in c++11 14.5.6.1/5+6:

5 2 expressions involving template parameters considered equivalent if 2 function definitions containing expressions satisfy 1 definition rule (3.2), except [template parameter renaming]. 6 2 function template equivalent if declared in same scope, have same name, have identical template parameter lists, , have return types , parameter lists equivalent using rules described above compare expressions involving template parameters.

i can't find rule makes apply expressions in types of non-type template parameters redeclarations of class templates, though.