有人可以解释这个 C++ 逗号运算符短路示例吗？

Question

Can someone explain this C++ comma operator short-circuiting example?

bIsTRUE     = true, false, true;
bIsFALSE    = (true, false), true;
bIsAlsoTRUE = ((true, false), true);

Why does the second version short-circuit and return false (at least in MSVC++) and the other two versions do not but return true ?

Answer 1

The comma operator has lower precedence than assignment, so these are parsed as

(bIsTRUE     = true), false, true;     
(bIsFALSE    = (true, false)), true;   
(bIsAlsoTRUE = ((true, false), true)); 

The comma operator does not short-circuit. It evaluates its left operand, ignores the result, then evaluates its right operand.

bIsTRUE is true because the right operand of the assignment is true .

bIsFALSE is false because (true, false) evaluates true , ignores the result, then evaluates and yields false .

bIsAlsoTRUE is true because ((true, false), true) evaluates (true, false) , ignores the result, then evaluates and yields true .