C Boolean logic... not being very logical

[zanza]

can someone explain to me why these two statements are different?

while(!(choice=='u'||choice=='U'||choice=='d'||choice=='D'))
while(choice!='u'||choice!='U'||choice!='d'||choice!='D')

sorry to be asking such a moronic question, but i swear ive used similar statements to the second one before, but it doesnt work now.

žÅNžÅ

 

[Salem]

You also need to change your || into &&

[tt]choice!='u'||choice!='U'[/tt]
is equivalent to
[tt]!(choice=='u'[highlight]&&[/highlight]choice=='U')[/tt]

But such a statement is always true, so maybe that's the problem.

#include <stdio.h>

int main( void ) {
    int a, b;
    for ( a = 0 ; a <= 1 ; a++ ) {
        for ( b = 0 ; b <= 1 ; b++ ) {
            printf("%d %d %d %d\n", a, b, ( a!=0 || b!=0 ), !( a==0 && b==0 ) );
        }
    }
    return 0;
}

--

 

[JavaDude32]

The first is a set of ors and then the negation of that set.

In the first, if the character is equal to any of those chars then the result is false.

The second is a set of not equal ors.

This one is always true.

Hope that helps.
-JavaDude32

 

[victorv]

hi,

> C-Language speaking:

!( a==b || c==d ) is equivalent to ( a!=b && c!=d )


> Mathematical (Logic bracnh) speaking:

you can find this in the "De Morgan's Laws"

( Google it and look for some simple article)


> In everydaylife you can say:

A doctor suggests: you can eat

vegetables OR fruit

Another suggests you the opposit : you can NOT eat

vegetables AND fruit



bye

 

[ArkM]

No such animal as 'special' C logic; it's a common logic. Negation of Predicate1 or Predicate2 is not Predicate1 and not Predicate2.
Negation of A is equal to B is A is not equal to B...