I don't get how the following expression gets simplified

I was told the following

Code:

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if ( (a && b) || (!a && !b) )

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was equivalent to

Code:

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if(!!a == !!b)

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Can someone explain to me how they arrive at this answer? I really don't see the steps involved.

Quote:

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Originally Posted by Overworked_PhD

I was told the following

Code:

---

if ( (a && b) || (!a && !b) )

---

was equivalent to

Code:

---

if(!!a == !!b)

---

Can someone explain to me how they arrive at this answer? I really don't see the steps involved.

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The first expression evaluates to true if both a and b are true, or if neither a nor b are true. Since any non-zero value is true, then you cannot just condense this to "a == b". But if you double-negate, then a true value negates to zero, then negates to one, and then the comparison will work.

If only we had a logical XOR (as opposed to bitwise), we could say:

Code:

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if( ! ( a ^^ b ) )

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But alas we don't.

Okay. I get it.

It could be further simplified to either:

!a == !b

- or -

!( !a ^ !b )

Quote:

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Originally Posted by Sebastiani

It could be further simplified to either:

!a == !b

- or -

!( !a ^ !b )

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The first one is going to be the faster of those two, I'm pretty sure.

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Mats