I know () has higher precedence than <<, and << has higher precedence than ==, but I want to know why I can't write cout<<a==b; yet can write cout<<(a==b); in C++.
How the compiler translates cout<<a==b; and then shows error?
Why can't I write cout<<a==b; but can write cout<<(a==b);
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3 Answers
<< has higher precedence than ==
as you can see here.
The statement
cout<<a==b
is equivalent to
(cout<<a)==b
The expression
cout<<a
returns a stream. This stream is compared to b. If there is no left shift operator for a stream and a or no comparing operator for a stream and b this causes a compiler error
cout<<a==b is similar to (cout<<a) == b as << has higher precedence over ==. Now cout<<a will be syntactically incorrect if the type of a is not supported for <<. Next, if a has an overload for the << operator, it will again be syntactically wrong as the == operator can't operate with std::stream and type of b unless b overloads this compare operator.
But in case of cout<<(a==b), a==b will result in a boolean value. As the << operator support boolean value it is a valid operation.
<< priority is higher than == so it's interpreted as (cout<<a)==b
but = has lower so you can do :
bool t = a == b
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