Short circuit evaluation

I'm currently studying C and I'm having a hard time wrapping my head around how exactly short circuit evaluation works.
Two examples provided to me were:

Code:

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int a = 1, b = 1, c = -1;
c = --a && b--;
printf("%d %d %d", a, b, c);

a = 0, b = 0, c = -1;
c = a++ || ++b;
printf("%d %d %d", a, b, c);

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I know from my notes that the first evaluation returns 0 1 0
and the second evaluation returns 1 1 1. What I can't quite seem to get is why?

How are you on figuring out say
--a ; c = a && b;

The only tricky part is working out whether the RHS gets evaluated at all. If it doesn't, then the side effect doesn't happen either.

Quote:

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Originally Posted by Salem

How are you on figuring out say
--a ; c = a && b;

The only tricky part is working out whether the RHS gets evaluated at all. If it doesn't, then the side effect doesn't happen either.

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So I think in that case 'a' would = 0, therefore c = 0 because the statement is false? I'm mostly confused by c = -1. So what I've been thinking is that -1 = 0 && 1
or
-1 = 0 || 1

I'm starting to think that the -1 has nothing to do with the evaluation.

It doesn't. Assignment will overwrite the previous value of the left hand side.

Now it makes perfect sense. Thank you!