Re: Possible type with bitwise operator (&)

Hi Ranier, On 8/3/20 4:59 PM, Ranier Vilela wrote:

Em seg., 3 de ago. de 2020 às 10:53, Gregory Burd escreveu:

...

Sjoerd,

Nice explanation, so if they are both correct and one avoids unnecessary work (even just a tiny bit) why not adjust the code to add your explanation as a comment and use the more efficient version?

I'm not sure that is correct. It is the equivalent of writing: if (b->key)

To see that the code is correct, you can write down the truth table of the *bitwise* AND operator: | b->ttype != TYPE_void | b->tkey || (b->ttype != TYPE_void) & b->tkey | |-----------------------+--------------++-----------------------------------| | false (== 0) | false (== 0) || 0 (== false) | | true (== 1) | false (== 0) || 0 (== false) | | false (== 0) | true (== 1) || 0 (== false) | | true (== 1) | true (== 1) || 1 (== true) | With a little thought you can see that this is the same as the *logical* AND operator. It works because both operands are of type bool, so their values are 0 for false and 1 for true in C99 if you cast them to integral types. (see for example the answers to this Stack Overflow question: https://stackoverflow.com/questions/4767923/c99-boolean-data-type) As Sjoerd wrote in his email the only difference might be a performance penalty in case the first operand is true, if the second operand is expensive to evaluate. This is not the case here because b->tkey is a bool value so its evaluation is cheap. In any case, if you find a particular set of inputs where this (or any other) part of the code produces wrong results we would be interested to see a bug report. Hope this helps, Panos.

Was that the intention of the original author of the code?

regards, Ranier Vilela

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