Evaluating Operators

This is a discussion on Evaluating Operators within the C++ Programming forums, part of the General Programming Boards category; Hi, I am new to the message board here, and brand new to C++. I'm trying to learn how to ...

  1. #1
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    Evaluating Operators

    Hi, I am new to the message board here, and brand new to C++. I'm trying to learn how to do this on my own, to better my life. Anyhow, here is my question:

    the statement !(1 && !(0 || 1)) is a true statement.

    Why?

    Could someone explain this statement to me and how its evaluated? Thanks in advance!

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    What do you know about these operators? How would you go about solving it?

    Do you know that in this case 1 is true and 0 is false?

    Do you know what order of operations means?

    Do you know what 1 || 0 evaluates to? How about 1 && 0? How about !1?

    Do you know how parentheses affect the order of evaluation?

  3. #3
    Registered User hk_mp5kpdw's Avatar
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    Start with what's in the innermost set of parenthesis (the OR):
    !(1 && !(0 || 1)) = !(1 && !(1))

    Next, evaluate the NOT(!):
    !(1 && !(1)) = !(1 && 0)

    Next evaluate the AND:
    !(1 && 0) = !(0)

    Last, evaluate the NOT(!):
    !(0) = 1
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    and the hat of wrongness Salem's Avatar
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    Remember that && and || use short-circuit evaluation. The RHS might not even be evaluated.

    If you had !(0 && !(0 || 1)) for example, the red part of the expression would not even be evaluated ( despite the extra () ). The LHS side of the && is false, therefore the whole thing is false regardless of what the RHS is.

    This is to allow you to test and dereference a pointer for example, with complete safety.
    char *p;
    if ( p && *p ) doStuff(*p);

    First checks that the pointer is not NULL, then checks that the char being pointed at isn't zero.

    Boolean expressions are read L->R, not starting with the most deeply nested ( ) sub-expression.
    If you dance barefoot on the broken glass of undefined behaviour, you've got to expect the occasional cut.
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    Quote Originally Posted by hk_mp5kpdw View Post
    Start with what's in the innermost set of parenthesis (the OR):
    !(1 && !(0 || 1)) = !(1 && !(1))

    Next, evaluate the NOT(!):
    !(1 && !(1)) = !(1 && 0)

    Next evaluate the AND:
    !(1 && 0) = !(0)

    Last, evaluate the NOT(!):
    !(0) = 1





    Thank you for your help hk. Very useful explanation, it helped A LOT.

  6. #6
    Kiss the monkey. CodeMonkey's Avatar
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    Well, Salem, that allows me to simplify a great deal of my code.
    "If you tell the truth, you don't have to remember anything"
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