gcc与字符串类型不一致

Question

我有以下测试程序:

#include <string>
#include <iostream>

int main()
{
    std::string s;
    std::string a = "sd";

    std::cout << a==s ? "y" : "n";

    return 0;
}

尝试编译它g++ test.cpp会产生以下神秘错误:

error: no match for 'operator==' (operand types are 'std::basic_ostream<char>' and 'std::string {aka std::basic_string<char>}')
  std::cout << a==s ? "y" : "n";
                ^

它似乎s被正确编译为类型std::string,而a被编译为std::basic_ostream<char>!？救命!!

Answer 1

((std::cout << a) == s) ? "y" : "n"; 由于运算符优先级,编译器解析了您的语句:您需要括号.

std::cout << (a==s ? "y" : "n");

Answer 2

编译器的错误消息在这里非常有用.它表示运营商的LHS属于类型,std::basic_ostream<char>而运营商的RHS属于类型std::string {aka std::basic_string<char>}.

即线

std::cout << a==s ? "y" : "n";

被解释为

(std::cout << a) == s ? "y" : "n";

要将编译器更改为为==运算符选择正确的对象,您必须使用parantheses来覆盖该行为.

std::cout << (a==s ? "y" : "n");