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&& operator commutative

Started by Arcamot Oct 16, 2005 at 5:31 AM 4 replies 900+ views

Original Post

Oct 16, 2005 at 5:31 AM

Hello have a look at this code: char data[256]; if (!f(data) && (strcmp(data, "") != 0)) { ... } f is a function which modifies the string in some way. We want first the function f to be evaluated and then the chcek whether the string is empty or not. My question is: Can I assume that the evaluation of this is always left to right? Or do I have to write: if (!f(data)) if (strcmp(data, "") != 0) { ... } the first code works, however I dont think that it is 100% right. thanks Sebastian

ToohrVyk

Oct 16, 2005 at 5:36 AM

&& is not commutative in C++ as it evaluates lazily. This means that, for the code a && b, b will be evaluated if and only if a is ~~false~~ true(and so, a will have to be evaluated first), so you are indeed right in your assumption. The same is also true for ||.

IndyJones

Oct 16, 2005 at 5:41 AM

Quote:
Original post by ToohrVyk
b will be evaluated if and only if a is false ...

Isn't b only evaluated if a is true (if a is false the && expression will return false - no matter what b may be).

Peregrin

Oct 16, 2005 at 5:44 AM

See this thread.

ToohrVyk

Oct 16, 2005 at 5:50 AM

Quote:
Original post by BRooksWackerman
Quote:
Original post by ToohrVyk
b will be evaluated if and only if a is false ...

Isn't b only evaluated if a is true (if a is false the && expression will return false - no matter what b may be).

Oops, you are perfectly right. Silly me.

haegarr

Oct 16, 2005 at 5:55 AM

Yes (at least in C++). In (a && b) the b is evaluated only if a is true, since otherwise the entire expression is already determined: (false && any) is always false. Similarly, in (a || b) b is evaluated only if a is false.

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