variable allocated from stack/bss ??

[onkar]

Given the following code & variable i .

int main(int argc,char **argv){
int i;
printf("%d\n",i );
return 0;
}

here i is allocated from bss or stack ?

I think it is stack,because it prints garbage value; If it were
allocated from bss the default value would be 0
please tell me if I am right or wrong ??

[MQ]

Re: variable allocated from stack/bss ??


onkar wrote:

Given the following code & variable i .
>
int main(int argc,char **argv){
int i;
printf("%d\n",i );
return 0;
}
>
here i is allocated from bss or stack ?
>
I think it is stack,because it prints garbage value; If it were
allocated from bss the default value would be 0
please tell me if I am right or wrong ??

Local variables are stored on the stack. BSS is used for uninitialized
global or static variables.

[Richard Heathfield]

Re: variable allocated from stack/bss ??

MQ said:

<snip>

Local variables are stored on the stack.

C&V please.

BSS is used for uninitialized
global or static variables.

C&V please.

--
Richard Heathfield
"Usenet is a strange place" - dmr 29/7/1999

http://www.cpax.org.uk

email: rjh at the above domain, - www.

[Richard Heathfield]

Re: variable allocated from stack/bss ??

onkar said:

Given the following code & variable i .
>
int main(int argc,char **argv){
int i;
printf("%d\n",i );
return 0;
}
>
here i is allocated from bss or stack ?

The C Standard doesn't require your implementation to have a bss or a stack.
It does require, however, that you provide a proper prototype for printf,
and it also requires that you don't evaluate indeterminately-valued
objects.

--
Richard Heathfield
"Usenet is a strange place" - dmr 29/7/1999

http://www.cpax.org.uk

email: rjh at the above domain, - www.

[santosh]

Re: variable allocated from stack/bss ??

onkar wrote:

Given the following code & variable i .
>
int main(int argc,char **argv){
int i;
printf("%d\n",i );
return 0;
}
>
here i is allocated from bss or stack ?
>
I think it is stack,because it prints garbage value; If it were
allocated from bss the default value would be 0
please tell me if I am right or wrong ??

The C standard doesn't specify the exact places where variables of
different types are stored. It just specifies their scope,
implementations are free to do whatever they want behind the scenes, as
long as all the necessary semantics are maintained.

Coming to your off-topic question, the short answer is yes, on most
implementations .

[Martin Ambuhl]

Re: variable allocated from stack/bss ??

onkar wrote:

Given the following code & variable i .
>
int main(int argc,char **argv){
int i;
printf("%d\n",i );
return 0;
}
>
here i is allocated from bss or stack ?
>
I think it is stack,because it prints garbage value; If it were
allocated from bss the default value would be 0
please tell me if I am right or wrong ??

Neither 'bss' nor 'stack' have any meaning in C. All we can say is that
'i' is an uninitialized automatic variable, and so has an indeterminate
initial value. If your implementation has some segment of memory
initialized to zero -- and that is by no means a given -- then it may,
or may not, use parts of that for some automatic variables. The
mechanism is completely up to the implementation.

[Martin Ambuhl]

Re: variable allocated from stack/bss ??

MQ wrote:

onkar wrote:

>Given the following code & variable i .
>>
>int main(int argc,char **argv){
> int i;
> printf("%d\n",i );
> return 0;
>}
>>
>here i is allocated from bss or stack ?
>>
>I think it is stack,because it prints garbage value; If it were
>allocated from bss the default value would be 0
>please tell me if I am right or wrong ??

>
Local variables are stored on the stack. BSS is used for uninitialized
global or static variables.

There is no such requirement for an implementation of the C programming
language. There is, in fact, no requirement for things called 'stack'
or 'BSS' to even exist. Please don't give such foolish and incorrect
'answers'. They only show your limited experience and lack of
qualification to give meaningful answers.

[jacob navia]

Re: variable allocated from stack/bss ??

Martin Ambuhl wrote:

onkar wrote:
>

>Given the following code & variable i .
>>
>int main(int argc,char **argv){
> int i;
> printf("%d\n",i );
> return 0;
>}
>>
>here i is allocated from bss or stack ?
>>
>I think it is stack,because it prints garbage value; If it were
>allocated from bss the default value would be 0
>please tell me if I am right or wrong ??

>
>
Neither 'bss' nor 'stack' have any meaning in C. All we can say is that
'i' is an uninitialized automatic variable, and so has an indeterminate
initial value. If your implementation has some segment of memory
initialized to zero -- and that is by no means a given -- then it may,
or may not, use parts of that for some automatic variables. The
mechanism is completely up to the implementation.
>

Translation:

In some weird implementation that could exist somewhere there is no
stack and no bss. Since I want to be pedantic I will generalize then
to say that "The C language dfoesn't require a stack" even if I know
that 99% of all the implementations in work stations today use
exactly those.

[jacob navia]

Re: variable allocated from stack/bss ??

onkar wrote:

Given the following code & variable i .
>
int main(int argc,char **argv){
int i;
printf("%d\n",i );
return 0;
}
>
here i is allocated from bss or stack ?
>
I think it is stack,because it prints garbage value; If it were
allocated from bss the default value would be 0
please tell me if I am right or wrong ??
>

Yes, you are right

[Richard Heathfield]

Re: variable allocated from stack/bss ??

jacob navia said:

onkar wrote:

>Given the following code & variable i .
>>
>int main(int argc,char **argv){
> int i;
> printf("%d\n",i );
> return 0;
>}
>>
>here i is allocated from bss or stack ?
>>
>I think it is stack,because it prints garbage value; If it were
>allocated from bss the default value would be 0
>please tell me if I am right or wrong ??
>>

>
Yes, you are right

Chapter and verse, please.

--
Richard Heathfield
"Usenet is a strange place" - dmr 29/7/1999

http://www.cpax.org.uk

email: rjh at the above domain, - www.

[jacob navia]

Re: variable allocated from stack/bss ??

Richard Heathfield wrote:

jacob navia said:
>
>

>>onkar wrote:
>>

>>>Given the following code & variable i .
>>>
>>>int main(int argc,char **argv){
>> int i;
>> printf("%d\n",i );
>> return 0;
>>>}
>>>
>>>here i is allocated from bss or stack ?
>>>
>>>I think it is stack,because it prints garbage value; If it were
>>>allocated from bss the default value would be 0
>>>please tell me if I am right or wrong ??
>>>

>>
>>Yes, you are right

>
>
Chapter and verse, please.
>

Since bss variables are initialized to zero (the standard does not
name them "bss" of course but it requires that uninitialized
variables be initilaized to zero), that variable CAN'T be an
uninitialized variable in that sense. That mens is an automatic
variable allocated in temporary storage. This storage is normally
the stack in most implementations

[Richard Heathfield]

Re: variable allocated from stack/bss ??

jacob navia said:

Richard Heathfield wrote:

>jacob navia said:
>>
>>

>>>onkar wrote:
>>>
>>>>Given the following code & variable i .
>>>>
>>>>int main(int argc,char **argv){
>>> int i;
>>> printf("%d\n",i );
>>> return 0;
>>>>}
>>>>
>>>>here i is allocated from bss or stack ?
>>>>
>>>>I think it is stack,because it prints garbage value; If it were
>>>>allocated from bss the default value would be 0
>>>>please tell me if I am right or wrong ??
>>>>
>>>
>>>Yes, you are right

>>
>>
>Chapter and verse, please.
>>

>
Since bss variables

Where does the Standard define these?

are initialized to zero (the standard does not
name them "bss" of course but it requires that uninitialized
variables be initilaized to zero),

It says no such thing.


#include <stdio.h>

int main(void)
{
int bss;

You seem to be claiming that this "bss variable" must be initialised to 0.
If so, you're wrong.

--
Richard Heathfield
"Usenet is a strange place" - dmr 29/7/1999

http://www.cpax.org.uk

email: rjh at the above domain, - www.

[jacob navia]

Re: variable allocated from stack/bss ??

Richard Heathfield wrote:

jacob navia said:
>
>

>>Richard Heathfield wrote:
>>

>>>jacob navia said:
>>>
>>>
>>>
>>>>onkar wrote:
>>>>
>>>>
>>>>>Given the following code & variable i .
>>>>>
>>>>>int main(int argc,char **argv){
>>>> int i;
>>>> printf("%d\n",i );
>>>> return 0;
>>>>>}
>>>>>
>>>>>here i is allocated from bss or stack ?
>>>>>
>>>>>I think it is stack,because it prints garbage value; If it were
>>>>>allocate d from bss the default value would be 0
>>>>>please tell me if I am right or wrong ??
>>>>>
>>>>
>>>>Yes, you are right
>>>
>>>
>>>Chapter and verse, please.
>>>

>>
>>Since bss variables

>
>
Where does the Standard define these?
>
>

>>are initialized to zero (the standard does not
>>name them "bss" of course but it requires that uninitialized
>>variables be initilaized to zero),

>
>
It says no such thing.
>
>
#include <stdio.h>
>
int main(void)
{
int bss;
>
You seem to be claiming that this "bss variable" must be initialised to 0.
If so, you're wrong.
>

What?
Can't you read?
I said that that wasn't a bss variable because (precisely) is not
cleared to zero.

[Richard Heathfield]

Re: variable allocated from stack/bss ??

jacob navia said:

Richard Heathfield wrote:

<snip>

>int main(void)
>{
> int bss;
>>
>You seem to be claiming that this "bss variable" must be initialised to
>0. If so, you're wrong.
>>

>
What?
Can't you read?

Sure. Can you?

I said that that wasn't a bss variable because (precisely) is not
cleared to zero.

No, the object's failure to get a default static initialiser value does not,
as you claim, imply that it's a "bss variable" (for which you have yet to
provide a reference to the Standard's definition); rather, it implies that
the object doesn't have static storage duration and is not a sub-member of
a partly-initialised aggregate object.

--
Richard Heathfield
"Usenet is a strange place" - dmr 29/7/1999

http://www.cpax.org.uk

email: rjh at the above domain, - www.

[Al Balmer]

Re: variable allocated from stack/bss ??

On Wed, 29 Nov 2006 13:54:23 +0000, Richard Heathfield
<rjh@see.sig.in validwrote:

>jacob navia said:
>

>Richard Heathfield wrote:

>
><snip>
>

>>int main(void)
>>{
>> int bss;
>>>
>>You seem to be claiming that this "bss variable" must be initialised to
>>0. If so, you're wrong.
>>>

>>
>What?
>Can't you read?

>
>Sure. Can you?
>

>I said that that wasn't a bss variable because (precisely) is not
>cleared to zero.

>
>No, the object's failure to get a default static initialiser value does not,
>as you claim, imply that it's a "bss variable" (for which you have yet to
>provide a reference to the Standard's definition); rather, it implies that
>the object doesn't have static storage duration and is not a sub-member of
>a partly-initialised aggregate object.

I'm only seeing one side of this discussion, but now I'm curious. Has
Jacob invented a new type of variable called "bss"? In my assembler
coding days, "bss" meant "block storage symbol" and was implemented by
simply incrementing the program counter by the requested amount, which
of course had no effect on memory thus skipped.

--
Al Balmer
Sun City, AZ