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# !! operator

Posted on 2003-11-17
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edit - - i get it
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Question by:jini555
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LVL 19

Expert Comment

ID: 9766495
jini555:

> int i = !! (SHORT) j;

! = Boolean NOT
!! = Boolean NOT NOT

It's silly, but that's what it means.  You should avoid using it.  It is the same as saying this:
int i = (j != 0);

Hope That Helps,
Dex*
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LVL 45

Accepted Solution

Kdo earned 50 total points
ID: 9766534

Depending on the compiler, optimization, and the mood of the guy implementing this, the two "!" operators may cancel each other out.

But in real life, they don't.  The first "!" operator (actually, the righter-most of the two) implicitly converts (short)j to a boolean value so that the result will be 0 or 1.  The second "!" operator then complements the value so that the net effect is the same as:

int i = ((short)j) ? 1 : 0;

Kent
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LVL 45

Expert Comment

ID: 9766540

Isn't it amazing how many "other ways" there are to do things in C?

:)

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LVL 19

Expert Comment

ID: 9766594
Kent:

Yeah it is.  These two are equivilant:
int i = ((short)j) ? 1 : 0;

And:
int i = (j != 0);

Dex*
0

LVL 45

Expert Comment

ID: 9766631

Actually, they're not.

if *j* is a variable type that is longer than a short, then your test MAY fail depending on the value of *j*.  You need to recast it as a short (to truncate the upper bits) before testing its value.

But with that small change, we do have three ways to do the same thing that I'd do in two lines of assembler.  :)

Kent
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