What does this code mean???

[Mars]

I am reading the code of a program from the examples, but really don't
understand this one........

int i,j;

............... ............... ............... ........

while (scanf("%d",&i) ,j)
{
............... ............... ..........
}

So I think the question should become,
how does the compiler evaluate the expression, (int i,int j)??
(am I right?? 'cause scanf function returns an integer as I know)

[Ulrich Eckhardt]

Re: What does this code mean???

Mars wrote:[color=blue]
> I am reading the code of a program from the examples, but really don't
> understand this one........
>
> int i,j;
>
> ............... ............... ............... .......
>
> while (scanf("%d",&i) ,j)
> {
> ............... ............... .........
> }
>
> So I think the question should become,
> how does the compiler evaluate the expression, (int i,int j)??[/color]

That's a typical application of the comma-operator, look it up in your
books.
[color=blue]
> (am I right?? 'cause scanf function returns an integer as I know)[/color]

You really should not need to ask what scanf() returns and what that
returnvalue means, you should know where to find that info. You are right
with your assumption though.

Uli

[Jonathan Burd]

Re: What does this code mean???

Mars wrote:[color=blue]
> I am reading the code of a program from the examples, but really don't
> understand this one........
>
> int i,j;
>
> ............... ............... ............... .......
>
> while (scanf("%d",&i) ,j)
> {
> ............... ............... .........
> }[/color]

Break it down like this:

scanf("%d", &i), j

The value returned by scanf is discarded and the whole expression
evaluates to the value of j. A sequence point occurs
at the comma. (See Annex C of the C Standard).

The loop is reading in values and will continue executing
while j is not 0.

As an example, see what this does:

/* comma.c */
#include <stdio.h>

int
main (void)
{
int a, b = 6;
int ret;

ret = (a = 5, b);

printf ("a: %d; ret: %d\n", a, ret);

return 0;
}

Regards,
Jonathan.

--
"We must do something. This is something. Therefore, we must do this."
- Keith Thompson

[Nick Austin]

Re: What does this code mean???

On Sun, 30 Jan 2005 19:20:52 +0800, Mars <Mars@Mars> wrote:
[color=blue]
>I am reading the code of a program from the examples, but really don't
>understand this one........
>
>int i,j;
>
>.............. ............... ............... ........
>
>while (scanf("%d",&i) ,j)
>{
>.............. ............... ..........
>}[/color]

It's the comma operator. The call to scanf() occurs as part of
the loop but occurs before evaluating the end condition.

It's the same as either:

scanf("%d",&i);
while (j)
{
/*.............. ............... .......... */
scanf("%d",&i);
}

or:

while (1)
{
scanf("%d",&i);
if (!j)
break;
/*.............. ............... .......... */
}

Nick.

[Mars]

Re: What does this code mean???

icic, thx~


Jonathan Burd mentioned:[color=blue]
> Mars wrote:
>[color=green]
>>I am reading the code of a program from the examples, but really don't
>>understand this one........
>>
>>int i,j;
>>
>>............. ............... ............... .........
>>
>>while (scanf("%d",&i) ,j)
>>{
>>............. ............... ...........
>>}[/color]
>
>
> Break it down like this:
>
> scanf("%d", &i), j
>
> The value returned by scanf is discarded and the whole expression
> evaluates to the value of j. A sequence point occurs
> at the comma. (See Annex C of the C Standard).
>
> The loop is reading in values and will continue executing
> while j is not 0.
>
> As an example, see what this does:
>
> /* comma.c */
> #include <stdio.h>
>
> int
> main (void)
> {
> int a, b = 6;
> int ret;
>
> ret = (a = 5, b);
>
> printf ("a: %d; ret: %d\n", a, ret);
>
> return 0;
> }
>
> Regards,
> Jonathan.
>[/color]

[Mars]

Re: What does this code mean???

Thx~~


Nick Austin mentioned:[color=blue]
> On Sun, 30 Jan 2005 19:20:52 +0800, Mars <Mars@Mars> wrote:
>
>[color=green]
>>I am reading the code of a program from the examples, but really don't
>>understand this one........
>>
>>int i,j;
>>
>>............. ............... ............... .........
>>
>>while (scanf("%d",&i) ,j)
>>{
>>............. ............... ...........
>>}[/color]
>
>
> It's the comma operator. The call to scanf() occurs as part of
> the loop but occurs before evaluating the end condition.
>
> It's the same as either:
>
> scanf("%d",&i);
> while (j)
> {
> /*.............. ............... .......... */
> scanf("%d",&i);
> }
>
> or:
>
> while (1)
> {
> scanf("%d",&i);
> if (!j)
> break;
> /*.............. ............... .......... */
> }
>
> Nick.
>[/color]

[Jonathan Burd]

Re: What does this code mean???

Mars wrote:[color=blue]
> icic, thx~[/color]

<snip>

Please do not top-post in c.l.c.
Top-posting blurs the context and makes it
difficult to see what you are replying to.

Regards,
Jonathan.

--
"We must do something. This is something. Therefore, we must do this."
- Keith Thompson

[Richard Bos]

Re: What does this code mean???

Jonathan Burd <jonathan.burd@ REMOVEMEgmail.c om> wrote:
[color=blue]
> Mars wrote:[color=green]
> > while (scanf("%d",&i) ,j)[/color][/color]
[color=blue]
> Break it down like this:
>
> scanf("%d", &i), j
>
> The value returned by scanf is discarded and the whole expression
> evaluates to the value of j. A sequence point occurs
> at the comma. (See Annex C of the C Standard).
>
> The loop is reading in values and will continue executing
> while j is not 0.[/color]

True, but I must say that this is an inadvisable way of using the comma
operator. The scanf() call has nothing to do with whether or not the
loop continues at all. This use of the comma is mere obfuscation.

Richard

[Servé La]

Re: What does this code mean???


"Richard Bos" <rlb@hoekstra-uitgeverij.nl> schreef in bericht
news:41fdeebe.2 54654545@news.i ndividual.net.. .
[color=blue]
> True, but I must say that this is an inadvisable way of using the comma
> operator. The scanf() call has nothing to do with whether or not the
> loop continues at all. This use of the comma is mere obfuscation.[/color]

That's your opinion of course. I like this style because the return value of
scanf is often ignored anyway and you dont need to put in 2 calls to scanf
as in:

scanf();
while(j)
{
scanf();
}

[Lawrence Kirby]

Re: What does this code mean???

On Mon, 31 Jan 2005 14:49:30 +0100, Servé La wrote:
[color=blue]
>
> "Richard Bos" <rlb@hoekstra-uitgeverij.nl> schreef in bericht
> news:41fdeebe.2 54654545@news.i ndividual.net.. .
>[color=green]
>> True, but I must say that this is an inadvisable way of using the comma
>> operator. The scanf() call has nothing to do with whether or not the
>> loop continues at all. This use of the comma is mere obfuscation.[/color]
>
> That's your opinion of course. I like this style because the return value of
> scanf is often ignored anyway[/color]

Unfortunately true. Consider however there are very few situations where
ignoring the return value of scanf() isn't a bug.

Lawrence

[Eric Sosman]

Re: What does this code mean???



Servé La wrote:[color=blue]
> "Richard Bos" <rlb@hoekstra-uitgeverij.nl> schreef in bericht
> news:41fdeebe.2 54654545@news.i ndividual.net.. .
>
> [color=green]
>>True, but I must say that this is an inadvisable way of using the comma
>>operator. The scanf() call has nothing to do with whether or not the
>>loop continues at all. This use of the comma is mere obfuscation.[/color]
>
>
> That's your opinion of course. I like this style because the return value of
> scanf is often ignored anyway [...][/color]

What else do you ignore? NULL values from malloc() and
fopen()? Warning messages from compilers? STOP signs? Fire
alarms?

No wonder software sucks.

--
Eric.Sosman@sun .com

[CBFalconer]

Re: What does this code mean???

Lawrence Kirby wrote:[color=blue]
>[/color]
.... snip ...[color=blue]
>
> Unfortunately true. Consider however there are very few situations
> where ignoring the return value of scanf() isn't a bug.[/color]

I rarely use scanf so am not sure, but are there any situations in
which it isn't a bug?

--
"If you want to post a followup via groups.google.c om, don't use
the broken "Reply" link at the bottom of the article. Click on
"show options" at the top of the article, then click on the
"Reply" at the bottom of the article headers." - Keith Thompson

[parser]

Re: What does this code mean???

see the code below
x=(scanf("%d",& i),j,k,l,...... ,z); /*x will hav the value of z*/

u should that the comma separated expressions are evaluated from Left
to Right and thats why the whole expression i.e. x, woud have a value
of the rightmost expression i.e. z

now consider this example

printf("here is a value : %d", ( 1,2,3,4,5,6,7,8 ,9));

output:
here is a vlaue : 9
so now i think that the it is clear to u

thanx :-)

[Chris Croughton]

Re: What does this code mean???

On Tue, 01 Feb 2005 14:29:07 +0000, Lawrence Kirby
<lknews@netacti ve.co.uk> wrote:
[color=blue]
> On Mon, 31 Jan 2005 15:50:41 +0000, CBFalconer wrote:
>[color=green]
>> Lawrence Kirby wrote:[color=darkred]
>>>[/color]
>> ... snip ...[color=darkred]
>>>
>>> Unfortunately true. Consider however there are very few situations
>>> where ignoring the return value of scanf() isn't a bug.[/color]
>>
>> I rarely use scanf so am not sure, but are there any situations in
>> which it isn't a bug?[/color]
>
> Only situations where you didn't care if the read operation failed or
> encountered end-of-file, and the scanf() just readss characters without
> writing values. scanf(fp, " ") to skip white-space perhaps in some odd
> circumstances. Not very likely but it is difficult to say that there
> is NO situation where this isn't a bug. Well actually easy to say, maybe I
> should and just wait for others to come up with the counterexamples . :-)[/color]

I'm not aware of any for scanf, but sscanf can certainly be legitimately
used without checking the return value, because it leaves its arguments
alone if it has no data so they default to the latest value. Silly
example:

#include <stdio.h>
int main(void)
{
char buff[256];
int x = 0;
int y = 0;
while (fgets(buff, 256, stdin))
{
sscanf(buff, "%d%d", &x, &y);
printf("%d + %d = %d\n", x, y, x+y);
}
return 0;
}

I suppose one could use something similar with scanf() and check feof()
after each call.

Chris C

[Servé La]

Re: What does this code mean???


"Eric Sosman" <eric.sosman@su n.com> schreef in bericht
news:ctlio9$fmt $1@news1brm.Cen tral.Sun.COM...[color=blue][color=green]
>>True, but I must say that this is an inadvisable way of using the comma
>>operator. The scanf() call has nothing to do with whether or not the
>>loop continues at all. This use of the comma is mere obfuscation.[/color]
>
>
> That's your opinion of course. I like this style because the return value[/color]
of[color=blue]
> scanf is often ignored anyway [...][/color]

What else do you ignore? NULL values from malloc() and
fopen()? Warning messages from compilers? STOP signs? Fire
alarms?


As as advised in this group, scanf is best to be avoided except in quick
test programs. I haven't seen a lot of error checking in that kind of
programs.