C ++函数模板重载

Question

This example is from C++ templates book by Josuttis :

 #include <iostream>
 #include <cstring>
 #include <string>

 // maximum of two values of any type (call-by-reference)
 template <typename T>
 inline T const& max (T const& a, T const& b)
 {
     return  a < b  ?  b : a;
 }

 // maximum of two C-strings (call-by-value)
 inline char const* max (char const* a, char const* b)
 { 
    return  std::strcmp(a,b) < 0  ?  b : a;
 }

 // maximum of three values of any type (call-by-reference)
 template <typename T>
 inline T const& max (T const& a, T const& b, T const& c)
 {
     return max (max(a,b), c);  // error, if max(a,b) uses call-by-value
 }

 int main ()
 {
   ::max(7, 42, 68);     // OK

    const char* s1 = "frederic";
    const char* s2 = "anica";
    const char* s3 = "lucas";
    ::max(s1, s2, s3);    // ERROR

}

He says that the reason for error in ::max(s1, s2, s3) is that for C-strings max(max(a,b),c) calls max(a,b) that creates a new temporary local value that may be returned by the function by reference.

I am not getting how a new local value is getting created ?

Answer 1

For C-strings, this code creates a local value, ie, a local variable that stores an address (pointer of type char const *):

std::strcmp(a,b) < 0 ? b : a;

Hence returning a reference to this (using the template functions) leads to an error. In this case the reference of type char const * const & to a local is returned by the template function max after the C-string max has returned a copy. The template functions have to return pointers by value instead of reference.

The template functions need to be overloaded for pointer types.