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# && and || -- precedence evaluation doubt

 P: n/a According to the precedence rules && has a higher precedence than || and both are 'left to right' associative. So in an expression like ++i || ++j && ++k where i, j, k are 0, -1, 0 (before the expression is evaluated), shouldn't the && part be evaluated before the || part? In which case will the evaluation be stopped short if the right side evaluates to 1 (true)? Or is the evaluation of this expression implementation dependent? -- Roshan Mathews Dr. MGR Deemed University Chennai Apr 18 '06 #1
8 Replies

 P: n/a Roshan Mathews wrote On 04/18/06 11:54,: According to the precedence rules && has a higher precedence than || and both are 'left to right' associative. So in an expression like ++i || ++j && ++k where i, j, k are 0, -1, 0 (before the expression is evaluated), shouldn't the && part be evaluated before the || part? In which case will the evaluation be stopped short if the right side evaluates to 1 (true)? Precedence is not evaluation order. (Don't feel bad; many others have confused the two.) Precedence says that the expression above means the same as ++i || (++j && ++k) /* RIGHT */ rather than (++i || ++j) && ++k /* RWONG */ Once the meaning is determined (by precedence or by explicit grouping or by a combination), the evaluation proceeds. For this expression, ++i is evaluated first. Since it turns out to be non-zero (for the given values), the rest of the expression is not evaluated at all, and the result of the expression is 1. Or is the evaluation of this expression implementation dependent? No. -- Er*********@sun.com Apr 18 '06 #2

 P: n/a "Roshan Mathews" wrote: According to the precedence rules && has a higher precedence than || and both are 'left to right' associative. So in an expression like ++i || ++j && ++k where i, j, k are 0, -1, 0 (before the expression is evaluated), shouldn't the && part be evaluated before the || part? No. This higher precendence means nothing more than that a || b && c is equal to a || (b && c), not to (a || b) && c. It says nothing about the order in which any of the operands are evaluated. This is just as true of * and +, or of [] and ==. [] having higher precedence than == means that x == y[z] does evaluate to x == (y[z]), not to the (nonsensical) (x == y)[z]. It does not mean that y and z are evaluated before x. There is a special case, however, for the && and || operators, which guarantees the opposite of your what you suppose. They are said to short-circuit; that is, they are required to be evaluated left-to-right, and when the value of the entire expression is known, that value is returned and no other operands are even looked at. This means that ++i || ++j && ++k is handled like this: Evaluate ++i. If this is true (non-zero), the whole expression is true (to be precise, it is 1); stop here. Ok, we now know that i used to be -1, and ++i was 0 (and since both || and && also introduce a sequence point, so is i, now; this is not important in this example, but can be in other circumstances; e.g., it means that --i && a[i] does not invoke undefined behaviour). Evaluate ++j. If it is zero, the sub-expression ++j && ++k is zero, and since ++i is also zero if we've got this far, so is the entire expression; if so, stop here. Now we know that ++i was 0 and ++j was non-zero. Evaluate ++k. If it's false, the expression is false (0); if it's true (non-zero), the expression is true (specifically 1). Richard Apr 18 '06 #3

 P: n/a Roshan Mathews wrote: According to the precedence rules && has a higher precedence than || and both are 'left to right' associative. So in an expression like ++i || ++j && ++k where i, j, k are 0, -1, 0 (before the expression is evaluated), shouldn't the && part be evaluated before the || part? In which case will the evaluation be stopped short if the right side evaluates to 1 (true)? Or is the evaluation of this expression implementation dependent? Just because && has higher precedence than "||" doesn't mean that it must be executed first, merely that the entire right-side must be evaluated before or-ing it in. The entire boolean sequence still needs to be evaluated left-to-right. Your statement is equivalent to: ++i || ( ++j && ++k ) The "++i" is evaluated first. Because "++i" is non-zero, the rest of the expression is not evaluated. -- +-------------------------+--------------------+-----------------------------+ | Kenneth J. Brody | www.hvcomputer.com | | | kenbrody/at\spamcop.net | www.fptech.com | #include | +-------------------------+--------------------+-----------------------------+ Don't e-mail me at: Apr 18 '06 #4

 P: n/a Richard Bos wrote: There is a special case, however, for the && and || operators, which guarantees the opposite of your what you suppose. They are said to short-circuit; that is, they are required to be evaluated left-to-right, Boring pedantic question: Are they required to really be evaluated left-to-right, or must the implementation merely behave "as-if" the arguments had been evaluated left-to-right? -- Christopher Benson-Manica | I *should* know what I'm talking about - if I ataru(at)cyberspace.org | don't, I need to know. Flames welcome. Apr 19 '06 #5

 P: n/a In article Christopher Benson-Manica writes: Richard Bos wrote: There is a special case, however, for the && and || operators, which guarantees the opposite of your what you suppose. They are said to short-circuit; that is, they are required to be evaluated left-to-right, Boring pedantic question: Are they required to really be evaluated left-to-right, or must the implementation merely behave "as-if" the arguments had been evaluated left-to-right? Always as-if. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ Apr 19 '06 #6

 P: n/a Christopher Benson-Manica wrote: Richard Bos wrote: There is a special case, however, for the && and || operators, which guarantees the opposite of your what you suppose. They are said to short-circuit; that is, they are required to be evaluated left-to-right, Boring pedantic question: Are they required to really be evaluated left-to-right, or must the implementation merely behave "as-if" the arguments had been evaluated left-to-right? There is no 'merely' about it for a conforming implementation. :-) An implementation can replace... i || j++ && 0; ....with... if (!i) j++; However... i || 0 && j++; ....must be a no-op (assuming i is not volatile.) the more classic case is something like... if (p && *p) Search for Chris Torek's posts on how an implementation can 'silently' trap by dereferencing a potential null pointer in parallel to the null test, but it must 'cope' with that trapping and behave as if it worked correctly. There are circumstances involving volatile objects where a practical conforming implementation must still be very careful about how it handles the evaluation order of certain expressions. -- Peter Apr 19 '06 #7

 P: n/a Christopher Benson-Manica wrote: Richard Bos wrote: There is a special case, however, for the && and || operators, which guarantees the opposite of your what you suppose. They are said to short-circuit; that is, they are required to be evaluated left-to-right, Boring pedantic question: Are they required to really be evaluated left-to-right, or must the implementation merely behave "as-if" the arguments had been evaluated left-to-right? Well, of course, everything is as-if. However, that as-if had also better involve not reading any volatiles, not calling rand(), not causing any FP exceptions, and so on, and so forth. Your program really must not be able to tell the difference in any reliable way. (Difference in execution time, for example, does not count as reliable.) Richard Apr 19 '06 #8

 P: n/a On Wed, 19 Apr 2006 00:08:08 +0000 (UTC), Christopher Benson-Manica wrote: Richard Bos wrote: There is a special case, however, for the && and || operators, which guarantees the opposite of your what you suppose. They are said to short-circuit; that is, they are required to be evaluated left-to-right,Boring pedantic question: Are they required to really be evaluatedleft-to-right, or must the implementation merely behave "as-if" thearguments had been evaluated left-to-right? It's required that you not know or care ;-) -- Al Balmer Sun City, AZ Apr 19 '06 #9

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