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Re: Operator precedence [message #185056 is a reply to message #185052] Tue, 25 February 2014 04:59 Go to previous messageGo to previous message
Jerry Stuckle is currently offline  Jerry Stuckle
Messages: 2598
Registered: September 2010
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Senior Member
On 2/24/2014 10:41 PM, Richard Damon wrote:
> 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.
>

No, it shows how it works with THIS compiler. Not even this system - a
different compiler may give different results. But then the same is
true for ANY compiler.

But then all I claimed was to try it and see how it works for THIS system.

For instance, what is the output of:

int main() {
int i = 1;
int j = i << 32;
printf("%d\n", j);
?>

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

IF you have the formal specification available, and are able to
understand it.

> 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.
>

That is true - and you will not find your example in the ANSI C spec -
which is why its output is not defined.

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

You tell me.

>
>
>>> 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.
>

I have - many times while working on updating C courses. So much, in
fact, that I completely wore out the ANSI spec I had (and which cost me
a bunch of $$$).

The spec specifies the order that operators are evaluated. It does NOT
specify the order in which operands are evaluated. That is why your
example has undefined results - the spec does not define what should happen.

>
>>
>>> 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.
>

PHP is defined as rigorously as other languages. Just in a more
reader-friendly way. (It took me a long time to figure out my way
through the ANSI C spec - and even longer for SQL).

But then since you seem to hate PHP so much, I suggest you just find
another language more to your liking and stop wasting people's time in a
PHP newsgroup. More trolls we don't need.

--
==================
Remove the "x" from my email address
Jerry Stuckle
jstucklex(at)attglobal(dot)net
==================

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