Thread: Short circuit evaluation

  1. #1
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    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:
    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);
    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?

  2. #2
    and the hat of int overfl Salem's Avatar
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    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.
    If you dance barefoot on the broken glass of undefined behaviour, you've got to expect the occasional cut.
    If at first you don't succeed, try writing your phone number on the exam paper.

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    Quote Originally Posted by Salem View Post
    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.
    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.

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    It doesn't. Assignment will overwrite the previous value of the left hand side.
    It is too clear and so it is hard to see.
    A dunce once searched for fire with a lighted lantern.
    Had he known what fire was,
    He could have cooked his rice much sooner.

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    Now it makes perfect sense. Thank you!

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