std::string addition to char*

Question

I'm trying to understand how addition of std::string to char* works. This code is compiling and working just as expected:

#include <string>
#include <cstdio>

void func (const char* str) {
  printf("%s\n", str);
}

int main () {
  char arr[] = {'a','b','c',0};
  char *str = arr;
  func((str + std::string("xyz")).c_str()); // THIS LINE
  return 0;
}

But I do not understand what constructors/methods is calling and in what order for this to work. This is addition of std::string to char* which gives another std::string, but char* is not a class and it does not have a addition operator.

Answer 1

You are using operator + for const char* on the left hand side and a temporary std::string on the right hand side. This is overload #4 here:

template< class CharT, class Traits, class Alloc >
basic_string<CharT,Traits,Alloc>
    operator+( const CharT* lhs,
               const basic_string<CharT,Traits,Alloc>& rhs );

Return value

A string containing characters from lhs followed by the characters from rhs

With std::string and char, the above template signature can be "interpreted" as

std::string operator+ (const char* lhs, const std::string& rhs);

The result is a new, temporary std::string object that owns the concatenated new buffer "abcxyz". It can bind to the function argument of type const char* and is valid as long as the function body executes.