!(1 && !(0 || 1)) Help

If this is true, !(1 && !(0 || 1)) why exactly?
Can anyone explain how this is to be read?

0 || 1
!(0 || 1)
1 && !(0 || 1)
!(1 && !(0 || 1))

actually, go from left to right. parentheses do not change the direction AND operates in, but parentheses can still effect evaluation of either side:

Code:

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int * p = 0;
if ( !( p && !( *p || 0 ) ) )
  printf ( "citizen is right: p == %p\n", ( void * )p );

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Quote:

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Originally Posted by scope

If this is true, !(1 && !(0 || 1)) why exactly?
Can anyone explain how this is to be read?

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!(1 && !(0 || 1))
= !(1 && !(1))
= !(1 && 0)
= !(0)
= 1

But note that, the individual terms are evaluated from left to right.