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0 pointsbnegreve7y ago0 comments

> However, the compiler ('s optimization step) is not magic and produces suboptimal code sometimes.

I agree that compilers are not always perfect, but in this particular case the two expressions are trivially equivalent from the associativity of the multiplication so the distinction had to be intentional.

But as gnuvince pointed out, the two expressions are not equivalent when you consider integer overflow.

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isbvhodnvemrwvn7y ago

Multiplication of floating-point numbers is not associative. Take for instance:

<smallest possible positive number> * 0.5 * 2

When evaluated like that:

(<SPPN> * 0.5) * 2

the result is 0 * 2 = 0

And when evaluated the other way around:

<SPPN> * (0.5 * 2)

the result is <SPPN> * 1 = <SPPN>

    double sppn = Double.MIN_VALUE;

    double first = sppn * (0.5 * 2);
    double second = (sppn * 0.5) * 2;

    System.out.println(sppn);
    System.out.println(first);
    System.out.println(second);

BeeOnRope7y ago

The two expressions are equivalent, even under overflow. With a few exceptions (eg division, right shift) fixed size integer math follows the same associative rules as "standard math".

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