Re: Possible type with bitwise operator (&)

3 Aug 2020

      Em seg., 3 de ago. de 2020 às 12:18, Panagiotis Koutsourakis <
panagiotis.koutsourakis@monetdbsolutions.com> escreveu:
...
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.
int main ()
{
    bool r = true;
    bool key = false;
    const char * ptr = "not null";
    void * pvoid = NULL;
r = (ptr != pvoid);
    if (r)
       printf("true\n");
    else
       printf("false\n");

    r = ((ptr != pvoid) & key);
    if (r)
       printf("true");
    else
       printf("false");
    return 0;
}

output:
true
true

output:
true
false

Always output the key value, be true or false:
try:
key = true or key = false

Again (&) bitwise is not shortcut and only with C++ that bool is guaranted
whatever 0 or 1.

regards,
Ranier Vilela