C ++将字符串隐式转换为char *匹配错误的函数签名

Question

I am writing a program which is supposed to handle both c strings (char*) and c++ strings (std::string). I have isolated by concern to the example below.

#include <iostream>
#include <string>

void hello(std::string s) {
    std::cout << "STRING FUNCTION" << std::endl;
}

void hello(char* c) {
    std::cout << "CHAR FUNCTION" << std::endl;
}

int main(int argc, char* argv[]) {
    hello("ambiguous");
    hello((std::string)"string");
    hello((char*)"charp");

    return 0;
}

When I compile this program, I get the warning:

test.cpp:14: warning: deprecated conversion from string constant to ‘char*’

regarding the first call to hello . Running the program gives:

CHAR FUNCTION
STRING FUNCTION
CHAR FUNCTION

showing that the first call to hello is matching the signature hello(char* c) .

My question is, if, as a c++ program, a string literal, ( "ambiguous" ) is a std::string, why would it be cast to a char* and then match the function hello(char* c) rather than staying as a std::string and matching hello(std::string s) ?

I know I can pragma or -Wsomething out the warning (and that I can cast char* to string without concern), but I want to know why the compiler would even bother doing this cast and if there is a way to tell it not to. I am compiling with g++ 4.4.3.

Thank you.

Answer 1

String literals like "ambiguous" are not of type std::string . std::string is a library only type, with no language magic whatsoever. The type of a string literal is actually const char[N] , where N is the length of the literal.

For historical reasons (backward compatibility) string literals will implicitely convert to char* (violating const-correctness). This builtin conversion is prefered to the "user defined" conversion to std::string , which is why it calls the char* function and gives you the warning.

If you change the signature of hello to hello(const char* c) it will probably not give you a warning anymore (but still won't call the std::string version, to do that you need a manual cast).

Answer 2

您错了：“作为c ++程序，字符串文字（“模糊”）是std :: string”

Answer 3

("ambiguous") is not a std::string, it is an array of characters.

That's why its calling void hello(char* c).