operator

Question

The accepted answers to the questions here and here say that it's all about operator precedence and thus,

cout << i && j ;

is evaluated as

(cout << i) &&j ;

since the precedence of Bitwise-Operators is greater than that of Logical-Operators. (It's not the bitwise operator here, but it is the symbol itself which is setting the precedence.)

I wonder then why the following code doesn't output 1:

int x=2, y=1 ;
cout << x>y ;

since the precedence of Relational-Operators is greater than Bitwise-Operators; the last line should get treated as cout << (x>y) ;.

I got a warning that overloaded operator<< has higher precedence than a comparison operator.

What is the precedence of overloaded operator<< in comparison to all other existing operators?

Jason Liam · Answer

someone tell what is the precedence of overloaded operator << in comparison to all other existing operators?

The inserter << has higher precedence than the relational operator<, operator> etc. Refer to operator precedence.

This means that cout << x>y is grouped as(and not evaluated as):

(cout << x)>y;

Now, cout << x returns a reference to cout which is then used as the left hand operand of operator>. But since there is no overloaded operator> that takes cout as left hand operand and int as right hand operand you get the error saying exactly that:

 error: no match for ‘operator>’ (operand types are ‘std::basic_ostream’ and ‘int’)

Note

Additionally, in your question you've used the phrase "is evaluated as" which is incorrect. The correct phrase would be "is grouped as".

S.Ptr · Answer

According to the C++ reference (https://en.cppreference.com/w/cpp/language/operator_precedence), the precedence of the bitwise shift operator is greater than the precedence of relational operators. cout <<, despite being a I/O operation, is an overload of the bitwise shift operators.

operator<< with cout and precedence

2 Answers

Note