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C question

Started by daniel_i_l Apr 1, 2009 at 6:13 PM 5 replies 700+ views

Original Post

Apr 1, 2009 at 6:13 PM

Let's say I have a pointer i_p and I want to check if it's true, and if it is, evaluate a = f(i_p) and then check that a > and a < 5. Can I do:


int a;
if(i_p && a=f(i_p) && a>0 && a<5){
...
}

I know that I can accomplish this with a nested if but I'm already somewhat nested so I'd rather do this in one line. Is this more efficient that calling f twice? Does this code have well defined behaviour or am I exploiting an undefined feature? Thanks. [Edited by - daniel_i_l on April 1, 2009 6:34:50 PM]

SiCrane

Apr 1, 2009 at 6:16 PM

&& introduces a sequence point, so it's defined behavior. I'd personally still put it on multiple lines.

LessBread

Apr 1, 2009 at 6:18 PM

There's no telling if a=f(i_p) will be executed first.

Palidine

Apr 1, 2009 at 6:21 PM

Quote:
Original post by LessBread
There's no telling if a=f(i_p) will be executed first.

What? That can't be true. Order of operations in a boolean comparison sequence is defined left to right ordering. Otherwise early out code would never work:

This, for example, is guaranteed to never have a NULL pointer exception:

if ( pointer && pointer->SomeBooleanFunction() ){    //do some stuff}

-me

SiCrane

Apr 1, 2009 at 6:22 PM

Quote:
Original post by LessBread
There's no telling if a=f(i_p) will be executed first.

No, && is a sequence point. Unless && is overloaded, and you can't overload && on primitives like int and bool, then you're guaranteed that all side effects of evaluations will be completed on the left hand side of an && before evaluation begins on the right hand side.

daniel_i_l

Apr 1, 2009 at 6:23 PM

Quote:
Original post by LessBread
There's no telling if a=f(i_p) will be executed first.

Why is it possible that a < 0 will be evaluated before a=f(i_p) but not before i_p (the first expression)?
[Edit: I see that this has already been talked about above - thanks everyone]

DevFred

Apr 1, 2009 at 7:09 PM

If you make a unsigned, you can eliminate one check. Although that's probably implementation defined or something :)

unsigned a;if(i_p && a=f(i_p) && a<5){...}

Oh, and if you don't need a inside the loop, you can simplify further:

if(i_p && f(i_p) < 5U){...}

Oh wait, that doesn't check for zero.

BTW, is the variable name "i_p" an april's fool joke?

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