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]

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]

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]

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]

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]

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]

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]

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]

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]

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.