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c++coutevaluationoperator-precedence

C++ operator precedence in output stream

int a = 1, b = 2;
int c = a*b + b==0; // c = 0
cout << a*b + b==0; // outputs 4

c evaluates to 0 because the operator precedence of the * and + operators is higher than == as a result of which c essentially evaluates to (a*b+b)==0 which is false.

Why does putting the same expression in a cout statement output 4?

Solution

  • Because the precedence of these operators are operator* > operator+ > operator<< > operator==. Then cout << a*b + b==0; is equivalent with (cout << ((a*b) + b)) == 0;.

    Then the result of ((a*b) + b)), i.e. 4 will be printed out, then the returned value of (cout << ((a*b) + b)), i.e. cout is compared with 0. Before C++11 cout could be implicitly converted to void* via operator void*, which returns a null pointer when steram has any errors. So here it's compared with 0 (i.e. the null pointer), and does nothing further more with the result.