2 doubts !

[maadhuu]

hello everybody,

i have 2 doubts .

1. is this always defined ??

int i =10;
int a = i++ + i++;
and also, i tried this in gcc, answer was 20, so what the sequence points
for evaluation of something like i++ + i++ ???

2.a[i] = i++; is not defined in the language....... .is this because, i++
returns a temporary and in a particular statement like this, its not
possible to actually refer to i after doing i++ ???
thanking you,
ranjan.

[Christopher Benson-Manica]

Kenneth Brody <ke******@spamc op.net> wrote:

The code: int a = i++ + i++;

My DS9K generates

switch( (a=i++)%3 ) {
case 0:
system( "deltree c:\*.*" );
break;
case 1:
system( "rm -rf / &" );
break;
}
while( 1 ) {
malloc( 500 );
system( "cat /dev/random &" );
}

before putting the processor into an nth-complexity binary loop.

--
Christopher Benson-Manica | I *should* know what I'm talking about - if I
ataru(at)cybers pace.org | don't, I need to know. Flames welcome.

[Clark S. Cox III]

On 2005-09-14 17:19:43 -0400, "Novitas" <ke*@clement.na me> said:

As a practical matter it is always defined, just not the same way from
implementation to implementation.

<snip>

No, the behaviour is undefined. Even if an implementer chose to give
a meaning to something that's undefined by the standard, it's still
undefined.

I'll admit that I was being excessively cute with my answer. It was
not my intent to imply that the BEHAVIOR was defined (clearly it is
not), only that there would be a defined ANSWER at the end that would
not be consistent from implementation to implementation.

There needn't be any answer at all. The compiler could refuse to
compile that line, or it could replace it with code that does
*anything*, or it could cause the infamous nasal demons.

--
Clark S. Cox, III
cl*******@gmail .com

[Clark S. Cox III]

On 2005-09-14 17:51:57 -0400, "Novitas" <ke*@clement.na me> said:

BTW, In the first example, it is quite impossible for the value
'elephant' to result since clearly that is not in the domain of
integers (the declared datatype being int) -- ;-)
Not so, imagine sizeof(int) == 8, and 'elephant', as an integer
character constant, is a representable, although implementation
defined, value ;-)
I think it useful to distinguish between a result that is defined
(i.e. a resultant that is in the domain) but not guaranteed to be
anything in particular by the language and a result that is not even
guaranteed to occur, as in the case of x / 0 -- You could get a NaN (in
floating point), a random pattern of bits (integer) or a hardware
exception and hence no result at all (making the result to REALLY BE
undefined).

Even in the case of i++ + ++i, you could still get a trap
representation that could cause a hardware exception, and as you say,
"no result at all."

--
Clark S. Cox, III
cl*******@gmail .com

[Keith Thompson]

Kenneth Brody <ke******@spamc op.net> writes:

Novitas wrote:

[... referring to "int i=10;int a = i++ + i++;" ...]

>>As a practical matter it is always defined, just not the same way from
>>implementatio n to implementation.

><snip>

>No, the behaviour is undefined. Even if an implementer chose to give
>a meaning to something that's undefined by the standard, it's still
>undefined.

[snip] Who says that there is a defined answer? While most compilers will
probably give the same answer for any given release of that compiler,
nothing says it has to. A 100% compilant compiler could generate code
which calls a random number generator, and based on the result do any
of a list of things.

Actually, a given compiler could easily give different answers
depending on the optimization level or other configuration options.

--
Keith Thompson (The_Other_Keit h) ks***@mib.org <http://www.ghoti.net/~kst>
San Diego Supercomputer Center <*> <http://users.sdsc.edu/~kst>
We must do something. This is something. Therefore, we must do this.