重载char *和std :: string是否安全？

Question

I have just read about the overloading functions on a beginner book. Just out of curiosity I 'd like to ask whether it is safe to overload between char* and std::string.

I played with the below code and get some result. But I was not sure whether it is an undefined behavior.

void foo(std::string str) {
  cout << "This is the std::string version. " << endl;
}

void foo(char* str) {
  cout << "This is the char* version. " << endl;
}

int main(int argc, char *argv[]) {

  foo("Hello"); // result shows char* version is invoked

  std::string s = "Hello";
  foo(s); // result shows std::string version

  return 0;

}

Answer 1

Yes, it's safe, as long as you make it const char* , and actually often useful. String literals cannot be converted to char* since C++11 (and it was deprecated before that).

The const char* overload will be picked for a string literal because a string literal is a const char[N] (where N is the number of characters). Overloads have a kind of priority ordering over which one will be picked when multiple would work. It's considered a better match to perform array-to-pointer conversion than to construct a std::string .

Why can overloading std::string and const char* be useful? If you had, for example, one overload for std::string and one for an bool , the bool would get called when you passed a string literal. That's because the bool overload is still considered a better match than constructing a std::string . We can get around this by providing a const char* overload, which will beat the bool overload, and can just forward to the std::string overload.

Answer 2

Short Answer: Perfectly safe. Consider the following uses:

foo("bar");//uses c string 
foo(std::string("bar") );//uses std::string
char* bar = "bar";
foo(bar);//uses c string
std::string bar_string = "bar";
foo(bar_string);//uses std::string
foo(bar_string.c_str()); //uses c string

Word of warning, some compilers (namely those with c++11 enabled) require the const keyword in parameter specification in order to allow temporary strings to be used.

For instance, in order to get this: foo("bar"); You need this: void foo(const char* bar);