C++ Operator precedence and the return statement

Question

If I do something like return a ? b : c; return a ? b : c; or return a && a2 && a3;

Could it ever be evaluated as just return a and then the function just returns immediately before evaluating the rest?

Answer 1

return is a statement , not an expression . So it can never be misinterpreted the way you think.

The statement is always of the form return [some expression]; (and the expression is optional). The expression, if present, is evaluated first, and its value is bound to the return value of the function.

Answer 2

To make this clearer I'm going to restate the question a little:

return a ? b() : c();

return a && a2() && a3();

In the first case, one of either b or c will be called but not the other.

In the second case, if a is false then neither a2 nor a3 will be called. If a2 returns false, a3 won't be called.

Answer 3

In return a && a2 && a3; , if a is false, there's no need to evaluate the rest of the expression. The result will always be false. So a2 and a3 will not be evaluated. This is called "short circuiting".