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Re: Operator precedence [message #185052 is a reply to message #185017] Tue, 25 February 2014 03:41 Go to previous messageGo to previous message
Richard Damon is currently offline  Richard Damon
Messages: 58
Registered: August 2011
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On 2/24/14, 7:00 AM, Jerry Stuckle wrote:

> What language gives every possible way for you to screw up your
> programming?
>
> If you want to find out how some screwed-up code works in ANY language,
> you try it and find out.
>

Herein lies the path to madness.

Take the following C code

#include <stdio.h>

int main() {
int i = 4;
i = ++i + i++;
printf("%i\n", i);
return 0;
}

Question, you want to find out what this is supposed to do. By your
statement, you run it on a system. The problem is that the answer you
get is only valid for THAT system, and possible for that EXACT piece of
code.

The REAL answer is you compare the code to the formal specification of
the language, and you get the answer.

Too many people use your answer, and then complain because they get a
different answer on another machine, or in different circumstances, and
then complain that the compiler is "broken", when the real answer is
that THEIR program was the one that was broken.

(For those not familiar with C, the answer in this case is that the
program has performed "undefined behavior" by modifying i twice without
a sequence point, and thus *ANY* resultant behavior is acceptable.

Here is a question, does PHP's documentation promise how the equivalent
PHP statement will execute?

....


>> The assignment operators in PHP have, in effect, different precedences
>> on the left and the right (a feature that some languages flaunt) but
>> that is not common and certainly not part of what I'd call basic
>> knowledge. The documentation gives one example of this peculiarity, but
>> that does not clear the matter up conclusively. For example, in some
>> languages permit assignment to the result of a conditional expression,
>> but I think it's hard to tell from the documentation what these two do:
>>
>> $bool ? $a : $b = 42;
>> ($bool ? $a : $b) = 42;
>>
>
> Show me what language provides you with examples of every possible
> combination of operators and expressions.

Read the C Grammer definition. It precisely describes how ANY expression
is to be parsed.

It does give the compiler freedom to choose the order to execute the
parse tree, so things without inherent order have no promise of order,
but this is made explicit.


>
>> The operator precedence table suggests that both mean the same thing,
>> but they don't. I don't think the meaning of either is easily
>> determined from the documentation.
>>
>> <snip>
>>
>
> So you try it out, as you do in any language. But just because you can
> come up with some weird combination of operators which isn't listed in
> the doc does not mean the language is not rigorously defined.
> with
> But it does mean you are trolling.
>

Rigorously defined would mean "precisely or exactly defined", which,
yes, does say that if you can come up with a "weird" combination, that
is not documented as to what is to happen, the definition is not
rigorous. A Formal definition, will be complete, as that is part of
being formal.
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