c++ - 为什么我收到 'operator^' 不匹配的错误

Question

我遇到了一个错误

10:13: error: no match for 'operator^' (operand types are 'std::basic_ostream<char>' and 'int')
10:13: note: candidates are:
In file included from /usr/include/c++/4.9/ios:42:0,
             from /usr/include/c++/4.9/ostream:38,
             from /usr/include/c++/4.9/iostream:39,
             from 2:
/usr/include/c++/4.9/bits/ios_base.h:161:3: note: std::_Ios_Iostate std::operator^(std::_Ios_Iostate, std::_Ios_Iostate)
operator^(_Ios_Iostate __a, _Ios_Iostate __b)
^

代码是

// Example program
#include <iostream>
#include <string>

int main()
{
int a=1;
int b=2;

std::cout<<a^b;
}

operator ^ 可以使用哪些操作数？

Answer 1

根据Operator Precedence , operator<<优先级高于 operator^ .所以std::cout<<a^b;等同于 (std::cout<<a)^b; ; (std::cout<<a)将返回 std::cout通过引用，这是一个 std::basic_ostream<char> ;正如错误信息所说，你不能调用operator^与 std::cout ( std::basic_ostream<char> ) 和 int .

您可以使用括号来指定操作数应如何绑定(bind)到运算符的优先级。

std::cout << (a^b);
//           ~   ~