int a=printf("%d\n" ,a);

[sugaray]

does the expression int a=printf("%d\n" ,a); implementation dependent ?
I supposed it would produced the result 2, but i got 4206596 (result
may vary on your machine), i wonder if the value produced was a memory
address and anything special with this expression ?

[Dan Pop]

In <sl************ *******@ekoi.cl .cam.ac.uk> Jeremy Yallop <je****@jdyallo p.freeserve.co. uk> writes:

Minti wrote:

Jeremy Yallop <je****@jdyallo p.freeserve.co. uk> wrote in message news:<bn******* *****@ID-114079.news.uni-berlin.de>...

Minti wrote:
> Jeremy Yallop <je****@jdyallo p.freeserve.co. uk> wrote in message news:<bn******* *****@ID-114079.news.uni-berlin.de>...
>> sugaray wrote:
>> > does the expression int a=printf("%d\n" ,a); implementation dependent ?
>>
>> No, it's undefined, since `a' hasn't been initialized at the time you
>> pass its value to printf(). Don't do that.
>
> By undefined do you mean undefined "value" or undefined "behaviour" ?

In this case there is no difference. The value of `a' is
indeterminate, so the behaviour when it is used is undefined.

But given the 'declaration'

int a = printf("%d", a);

Does the standard gaurantee that "a" would have been 'defined' before
its value is used by the printf.

Yes. The scope of `a' begins at the end of its declarator.

So, the answer is actually "no": if `a' was already defined prior to our
definition, the `a' in the printf call refers to the "new" `a', currently
uninitialised and not to the "old" `a'.

Dan
--
Dan Pop
DESY Zeuthen, RZ group
Email: Da*****@ifh.de

[Jeremy Yallop]

Dan Pop wrote:

In <sl************ *******@ekoi.cl .cam.ac.uk> Jeremy Yallop <je****@jdyallo p.freeserve.co. uk> writes:

Minti wrote:

Jeremy Yallop <je****@jdyallo p.freeserve.co. uk> wrote in message news:<bn******* *****@ID-114079.news.uni-berlin.de>...
Minti wrote:
> Jeremy Yallop <je****@jdyallo p.freeserve.co. uk> wrote in message news:<bn******* *****@ID-114079.news.uni-berlin.de>...
>> sugaray wrote:
>> > does the expression int a=printf("%d\n" ,a); implementation dependent ?
>>
>> No, it's undefined, since `a' hasn't been initialized at the time you
>> pass its value to printf(). Don't do that.
>
> By undefined do you mean undefined "value" or undefined "behaviour" ?

In this case there is no difference. The value of `a' is
indeterminate, so the behaviour when it is used is undefined.

But given the 'declaration'

int a = printf("%d", a);

Does the standard gaurantee that "a" would have been 'defined' before
its value is used by the printf.
Yes. The scope of `a' begins at the end of its declarator.

So, the answer is actually "no":

Well, that depends on what the question is :-). I understood it as
"Does the standard guarantee that "a" is in scope at the time its
value is used by printf?". I think you read it as "Does the standard
guarantee that if "a" is already defined its value is used by printf?"
(where "its" refers to "the already-defined a's"). We don't disagree
on the behaviour, anyway:
if `a' was already defined prior to our definition, the `a' in the
printf call refers to the "new" `a', currently uninitialised and not
to the "old" `a'.

Right.

Jeremy.

[Micah Cowan]

Jeremy Yallop <je****@jdyallo p.freeserve.co. uk> writes:

Micah Cowan wrote:

Jeremy Yallop <je****@jdyallo p.freeserve.co. uk> writes:

sugaray wrote:
> does the expression int a=printf("%d\n" ,a); implementation dependent ?

No, it's undefined, since `a' hasn't been initialized at the time you
pass its value to printf(). Don't do that.

How do you know that? He gave us an isolated expression: there is
no way to know whether or not it was initialized.

Read it again. It's a declaration, not an expression (despite the
words "the expression").

Doh! :-)

--
Micah J. Cowan
mi***@cowan.nam e