Why do i get an error no match for 'operator^'

Question

I am getting an error

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) ^

the code is

// Example program #include <iostream> #include <string> int main() { int a=1; int b=2; std::cout<<a^b; }

What are the operands that can be used with operator ^ ?

^ ha a low operator precedence: Use parentheses std::cout << (a^b); — user2249683
– user2249683, Commented Oct 8, 2016 at 10:01
The actual question "What are the operands that can be used with operator ^ ?" is to broad. — user2249683
– user2249683, Commented Oct 8, 2016 at 10:18

songyuanyao · Accepted Answer · 2016-10-08 10:13:31Z

According to the Operator Precedence, operator<< has higher precedence than operator^. So std::cout<<a^b; is equivalent with (std::cout<<a)^b;; (std::cout<<a) will return std::cout by reference, which is a std::basic_ostream<char>; Just as the error message said, you can't call operator^ with std::cout(std::basic_ostream<char>) and int.

You could use parentheses to specify the precedence how the operands should be bound to operators.

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

Collectives™ on Stack Overflow

Why do i get an error no match for 'operator^'

1 Answer 1

Comments

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

Comments

Related