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Dan's Tests | Question from 'Conditional Operator'

 
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I have come across following question from Dan's operator section tests:
________________________________________________________________________

What is the result of attempting to compile and run the above program?
a. Prints: false,false,false
b. Prints: false,false,true
c. Prints: false,true,false
d. Prints: false,true,true
e. Prints: true,false,false
f. Prints: true,false,true
g. Prints: true,true,false
h. Prints: true,true,true
i. Runtime error
j. Compiler error
k. None of the above
__________________________________________________________________________
Now, based on the operator precedence and operator evaluation rules, '&&' has got a higher precedence than '||'. So, according to me, these should be the order of steps executed at runtime:
  • b is assigned a true value
  • Since b is true, the short-circuit &&operator has to evaluate second operand, i.e. c
  • c is assigned a true value
  • The result of && operand is true
  • Now, the || operator is evaluated. a is assigned a true value.
  • Since a is already true, it returns the value of expression as true
  • Now, x is assigned a true value.


  • Based on this logic, the answer to this question should be 'h' while the correct answer is 'e'. (I have checked it with compiler too). Now, it looks like the operators are being evaluated from left to right but that should only occur when two operators are of equal precedence.
    Why is it behaving this way? If its always supposed to behave this way, what does the higher precedence of && to || imply? When does it usually come into action? Please support with an example.
    Thanks,
    Nidhi.
     
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    The evaluation order is like on a mathematical equation where
    x = a + b * c
    can be converted to
    x = a + (b * c)
    without actually altering the order of evaluation.
    The same is true here:
    boolean x = (a = true) || (b = true) && (c = true)
    equals
    boolean x = (a = true) || ((b = true) && (c = true))
     
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    I agree. In this question, JVM decides not to evaluate the logical AND part of expression because logical OR is happy with the left part already. You can even insert division by zero to the right part, it doesn't throw anything as long as the left part is true.
    [ September 24, 2003: Message edited by: Vad Fogel ]
     
    Nidhi Bangur
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    Thanks both of you! Its clear now. I did a little bit of experimenting to make the flow clear in my mind. For anybody who is interested, I am pasting the code and results below:

    If you run this code, the results are as follows:

    i is4
    i is6
    i is2
    true,true,false,16

    Now, this makes it very clear. Once you resolve the precedence by putting brackets as suggested by Mika, the expression is evaluated from left-to-right. That's why, "i is4" is printed before "i is6" and "i is2".
    [ September 24, 2003: Message edited by: Nidhi Bangur ]
     
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