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I was playing the Javascript game with somebody and we were having fun making ridiculous and absurd expressions to make our inputs get a particular output.
This little charming one
!a!=!!b^!!-!a||!+!a|!c always seemed to return 1. I tried to reason it out, but I gave up after losing track of all the !s.
Are there any values for a, b, and c which do not return 1? If not, why does it always return 1?
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No, you can not return multiple values like this in C. A function can have at most one single return value.
No, you can't return multiple types.
The report_ratio function has a void return type, so it doesn't need to explicitly return a value. Execution of report_ratio "falls off the bottom" and returns no value to the caller.
return 1 in the main function means that the program does not execute successfully and there is some error. return 0 means that the user-defined function is returning false. return 1 means that the user-defined function is returning true.
Short answer, yes. a = false, b = false, c = true is a counter-example because your equation is identical to (!!a || !!b || !c).
Long answer:
!a!=!!b^!!-!a||!+!a|!c is
(((!a) != (!!b)) ^ (!!(-!a))) || ((!+!a)|!c) which reduces to
((Boolean(a) == Boolean(b)) ^ (!a)) || (Boolean(a) | !c) so all of a, b and c are only dealt with as truthy/falsey values and the result must be a 1 or 0 since | and ^ both coerce booleans to numbers.
So obviously (from inspection of the right of the ||) if either a is truthy or c is falsey, you get 1.
If a is falsey and c is truthy, you have two possibilities,
b is truthy in which case the ^ clause is 1 so the right of the || is never reached.b is falsey, in which case the ^ clause is 0 so the right of the || dominates to produce 0.
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How about this:
var a = undefined, b=undefined, c=!a alert(!a!=!!b^!!-!a||!+!a|!c) // Output: 0 Live demo.
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