Is short-circuiting logical operators mandated? And evaluation order?

Question

Does the ANSI standard mandate the logical operators to be short-circuited, in either C or C++?

I\'m confused for I recall the K&R book saying your code

Answer 1

Your question comes down to C++ operator precedence and associativity. Basically, in expressions with multiple operators and no parentheses, the compiler constructs the expression tree by following these rules.

For precedence, when you have something like A op1 B op2 C, you could group things as either (A op1 B) op2 C or A op1 (B op2 C). If op1 has higher precedence than op2, you'll get the first expression. Otherwise, you'll get the second one.

For associativity, when you have something like A op B op C, you could again group thins as (A op B) op C or A op (B op C). If op has left associativity, we end up with the first expression. If it has right associativity, we end up with the second one. This also works for operators at the same precedence level.

In this particular case, && has higher precedence than ||, so the expression will be evaluated as (a != "" && it == seqMap.end()) || isEven.

The order itself is "left-to-right" on the expression-tree form. So we'll first evaluate a != "" && it == seqMap.end(). If it's true the whole expression is true, otherwise we go to isEven. The procedure repeats itself recursively inside the left-subexpression of course.

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Interesting tidbits, but the concept of precedence has its roots in mathematic notation. The same thing happens in a*b + c, where * has higher precedence than +.

Even more interesting/obscure, for a unparenthasiszed expression A1 op1 A2 op2 ... opn-1 An, where all operators have the same precedence, the number of binary expression trees we could form is given by the so called Catalan numbers. For large n, these grow extremely fast. d