!( ( 1 || 0 ) && 0 )
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!( ( 1 || 0 ) && 0 )
Treat 1 as true and 0 as false, then evaluate the boolean expression.
yeh thats what i need help with lol. im doing the tutorial on c++ i got the other two down but this last one is tricking me.
Which part is tricking you?
gg
Do what with it?
!( ( that part what do i do with the !
The ! symbol means logical not.
yes i know that much but im stuck figuring it out...can you just show me how you would go about doing it?
Well, how would you evaluate (1 || 0)?
as 1
So far so good. So how would you evaluate (1 && 0)?
as 0.
Excellent. Just evaluate !0 and your problem is solved.
EDIT:
Strictly speaking, the result of (1 || 0) is true, not 1, and the result of (true && 0) is false, not 0.
See i just didnt know the order of it all. Thanks alot!
No problem. Actually, with all those parentheses to group subexpressions the order of evaluation is no different from the arithmetic you used since elementary/primary school.
More tricky would be:
... which evaluates to false instead.Code:!( 1 || 0 && 0 )
that one was actually easier for me lol...i look at things wierd i guess.