Thread: Evaluationof expressions

  1. 10-28-2007 #1
    plan7
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    Evaluationof expressions

    Hi all

    I'm just re-reading K&R and there's something I don't understand, Appendix A7. Expressions:
    "The precedence and associativity of operators is fully specified, but the order of evaluation of expressions is, with certain exceptions, undefined, even if the subexpressions involve side effects. That is, unless the definition of an operator guarantees that its operands are evaluated in a particular order, the implementation is free to evaluate operands in any order, or even interleave their evaluation."
    Looking at this:
    Code:
    int a = 5;
int b = ++a;
    How can I be sure that 'b' takes the value of 6, if the evaluation of the expression '++a' could be interleaved?...I think I misunderstand something here, but I don't see what. I also searched on Google and on the board, read a lot, but it is still not really clear :/
    Hopefully somebody can explain me
  2. 10-28-2007 #2
    Salem
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    Statements are not interleaved - http://c-faq.com/expr/seqpoints.html

    What K&R are referring to would be something like this
    Code:
    a = f() + g() * h() - i();
    If you consider only precedence, you might conclude that g() and h() would be called first. This is simply not the case. Nor could you assume that g() and h() would be called consecutively.

    This is really important since it makes all such multiple side effect expressions such as this as being undefined.
    b = a++ + ++a;
    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.
  3. 10-28-2007 #3
    plan7
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    Hm...ok, that makes sense now. Many thanks!
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