static allocation question...

[FMorales]

-- I didn't notice anything about this specifically
-- in the c.l.c. FAQ, nor could i get a strait answere
-- from my copy of the C standard. Hopeing someone
-- here could shed some light on it :)...

6.2.1
"If the declarator or type specifier that declares
the identifier appears inside a block or within the
list of parameter declarations in a function definition,
the identifier has block scope, which terminates at the
end of the associated block."

I believe this talks about what we learned early on :

int main(void)
{
{
int a;
}
printf("%d\n", a); /* ERROR - Invalid */
return 0;
}

Once outside of the closing } a is no longer accessible.
It's out of our current "scope". Which brings me to my
question. Since that data is static, is it automatically
allocated at program start up?

int main(void)
{
int i;
return 0;
}

As "i" would be in this example ? What if there's a conditional
statement there ? As in :

int main(void)
{
int i = 0;
if(!i) {
int a;
}
return 0;
}

If "a" is /NOT/ automatically allocated with "i", what is
the compiler doing to make sure it gets allocated only when,
and if it's ever needed (like if "i" changes before the
"if", or if it waits for input from a user which we'll assume
is unknown untill runtime will "a" still get allocated ?)

Any help is greatly appreciated, just something i started
wondering about :) FMorales...

[Pushkar Pradhan]

Did you try this?
In C you have to declare all variables first and then start your code.
So you would have to do
int a;
int i;

What you did is allowed in c++. I mean this code:

int main(void)
{
int i = 0;
if(!i) {
int a;
}
return 0;
}
Won't compile.

FMorales wrote: -- I didn't notice anything about this specifically
-- in the c.l.c. FAQ, nor could i get a strait answere
-- from my copy of the C standard. Hopeing someone
-- here could shed some light on it :)...

6.2.1
"If the declarator or type specifier that declares
the identifier appears inside a block or within the
list of parameter declarations in a function definition,
the identifier has block scope, which terminates at the
end of the associated block."

I believe this talks about what we learned early on :

int main(void)
{
{
int a;
}
printf("%d\n", a); /* ERROR - Invalid */
return 0;
}

Once outside of the closing } a is no longer accessible.
It's out of our current "scope". Which brings me to my
question. Since that data is static, is it automatically
allocated at program start up?

int main(void)
{
int i;
return 0;
}

As "i" would be in this example ? What if there's a conditional
statement there ? As in :

int main(void)
{
int i = 0;
if(!i) {
int a;
}
return 0;
}

If "a" is /NOT/ automatically allocated with "i", what is
the compiler doing to make sure it gets allocated only when,
and if it's ever needed (like if "i" changes before the
"if", or if it waits for input from a user which we'll assume
is unknown untill runtime will "a" still get allocated ?)

Any help is greatly appreciated, just something i started
wondering about :) FMorales...

[Paulo Pereira]

On Sat, 04 Oct 2003 18:09:53 -0500, Pushkar Pradhan wrote:

Did you try this?
In C you have to declare all variables first and then start your code.
only in c89
in c99, variables can be declared anywhere.
So you would have to do
int a;
int i;

What you did is allowed in c++. I mean this code:

> int main(void)
> {
> int i = 0;
> if(!i) {
> int a;
> }
> return 0;
> }

Won't compile.

FMorales wrote:

-- I didn't notice anything about this specifically
-- in the c.l.c. FAQ, nor could i get a strait answere
-- from my copy of the C standard. Hopeing someone
-- here could shed some light on it :)...

6.2.1
"If the declarator or type specifier that declares
the identifier appears inside a block or within the
list of parameter declarations in a function definition,
the identifier has block scope, which terminates at the
end of the associated block."

I believe this talks about what we learned early on :

int main(void)
{
{
int a;
}
printf("%d\n", a); /* ERROR - Invalid */
return 0;
}

Once outside of the closing } a is no longer accessible.
It's out of our current "scope". Which brings me to my
question. Since that data is static, is it automatically
allocated at program start up?

int main(void)
{
int i;
return 0;
}

As "i" would be in this example ? What if there's a conditional
statement there ? As in :

int main(void)
{
int i = 0;
if(!i) {
int a;
}
return 0;
}
any good compiler won't generate code for your conditional statement
unless you set i to be "volatile".

If "a" is /NOT/ automatically allocated with "i", what is
the compiler doing to make sure it gets allocated only when,
and if it's ever needed (like if "i" changes before the
the compiler can't assume that unless you provide the keyword static
"if", or if it waits for input from a user which we'll assume
is unknown untill runtime will "a" still get allocated ?)

Any help is greatly appreciated, just something i started
wondering about :) FMorales...

[Bertrand Mollinier Toublet]

Pushkar Pradhan wrote:

Did you try this?
In C you have to declare all variables first and then start your code.
So you would have to do
int a;
int i;

What you did is allowed in c++. I mean this code:

> int main(void)
> {
> int i = 0;
> if(!i) {
> int a;
> }
> return 0;
> }

Pushkar, you must have suffered a temporary access of madness. There is
nothing wrong with declaring variables at the beginning of a block. The
code above is perfectly valid C89 (as well as C99, of course).

--
Bertrand Mollinier Toublet
closity = 1.0 / farthitude
-- Arthur J. O'Dwyer

[Artie Gold]

FMorales wrote:

-- I didn't notice anything about this specifically
-- in the c.l.c. FAQ, nor could i get a strait answere
-- from my copy of the C standard. Hopeing someone
-- here could shed some light on it :)...

6.2.1
"If the declarator or type specifier that declares
the identifier appears inside a block or within the
list of parameter declarations in a function definition,
the identifier has block scope, which terminates at the
end of the associated block."

I believe this talks about what we learned early on :

int main(void)
{
{
int a;
}
printf("%d\n", a); /* ERROR - Invalid */
return 0;
}

Once outside of the closing } a is no longer accessible.
It's out of our current "scope". Which brings me to my
Correct.
question. Since that data is static, is it automatically
allocated at program start up?
It's *not* static. It is an *automatic variable*.

int main(void)
{
int i;
return 0;
}

As "i" would be in this example ? What if there's a conditional
statement there ? As in :

int main(void)
{
int i = 0;
if(!i) {
int a;
}
`a' is now out of scope.
return 0;
}

If "a" is /NOT/ automatically allocated with "i", what is
the compiler doing to make sure it gets allocated only when,
and if it's ever needed (like if "i" changes before the
"if", or if it waits for input from a user which we'll assume
is unknown untill runtime will "a" still get allocated ?)
What a particular compiler does is irrelevant; it is free to do
whatever it wants as long as the program semantics are correct.

Any help is greatly appreciated, just something i started
wondering about :) FMorales...

Fair enough.

HTH,
--ag


--
Artie Gold -- Austin, Texas
Oh, for the good old days of regular old SPAM.

[Pushkar Pradhan]

There is no need to get nasty, I tried it on my system and then only
sent my post. Somebody already pointed out I was wrong.
No need for you to show your frustration here.
Bertrand Mollinier Toublet wrote:

Pushkar Pradhan wrote:

Did you try this?
In C you have to declare all variables first and then start your code.
So you would have to do
int a;
int i;

What you did is allowed in c++. I mean this code:

> int main(void)
> {
> int i = 0;
> if(!i) {
> int a;
> }
> return 0;
> }

Pushkar, you must have suffered a temporary access of madness. There is
nothing wrong with declaring variables at the beginning of a block. The
code above is perfectly valid C89 (as well as C99, of course).

[Barry Schwarz]

On Sat, 04 Oct 2003 22:45:29 GMT, "FMorales" <al****@comcast .net>
wrote:

-- I didn't notice anything about this specifically
-- in the c.l.c. FAQ, nor could i get a strait answere
-- from my copy of the C standard. Hopeing someone
-- here could shed some light on it :)...

6.2.1
"If the declarator or type specifier that declares
the identifier appears inside a block or within the
list of parameter declarations in a function definition,
the identifier has block scope, which terminates at the
end of the associated block."

I believe this talks about what we learned early on :

int main(void)
{
{
int a;
}
printf("%d\n", a); /* ERROR - Invalid */
return 0;
}

Once outside of the closing } a is no longer accessible.
It's out of our current "scope". Which brings me to my
question. Since that data is static, is it automatically
allocated at program start up?

int main(void)
{
int i;
return 0;
}

As "i" would be in this example ? What if there's a conditional
statement there ? As in :

int main(void)
{
int i = 0;
if(!i) {
int a;
}
return 0;
}

If "a" is /NOT/ automatically allocated with "i", what is
the compiler doing to make sure it gets allocated only when,
and if it's ever needed (like if "i" changes before the
"if", or if it waits for input from a user which we'll assume
is unknown untill runtime will "a" still get allocated ?)

Any help is greatly appreciated, just something i started
wondering about :) FMorales...

You don't have any static objects in your sample code. While you
rethink the problem, consider that scope and duration (or life span)
have different meanings.
<<Remove the del for email>>

[Sheldon Simms]

On Sat, 04 Oct 2003 21:23:08 -0500, Pushkar Pradhan wrote:

Bertrand Mollinier Toublet wrote:

Pushkar Pradhan wrote:

Did you try this?
In C you have to declare all variables first and then start your code.
So you would have to do
int a;
int i;

What you did is allowed in c++. I mean this code:
> int main(void)
> {
> int i = 0;
> if(!i) {
> int a;
> }
> return 0;
> }

Pushkar, you must have suffered a temporary access of madness. There is
nothing wrong with declaring variables at the beginning of a block. The
code above is perfectly valid C89 (as well as C99, of course).

There is no need to get nasty, I tried it on my system and then only
sent my post. Somebody already pointed out I was wrong.
No need for you to show your frustration here.

Are you claiming that the compiler you use on your system failed
to compile the code?

[Pushkar Pradhan]

Actually I did it without the block {}, that's why it failed. I mean
if(a)
int i;

Sheldon Simms wrote:

On Sat, 04 Oct 2003 21:23:08 -0500, Pushkar Pradhan wrote:

Bertrand Mollinier Toublet wrote:

Pushkar Pradhan wrote:
Did you try this?
In C you have to declare all variables first and then start your code.
So you would have to do
int a;
int i;

What you did is allowed in c++. I mean this code:
> int main(void)
> {
> int i = 0;
> if(!i) {
> int a;
> }
> return 0;
> }
Pushkar, you must have suffered a temporary access of madness. There is
nothing wrong with declaring variables at the beginning of a block. The
code above is perfectly valid C89 (as well as C99, of course).

There is no need to get nasty, I tried it on my system and then only
sent my post. Somebody already pointed out I was wrong.
No need for you to show your frustration here.

Are you claiming that the compiler you use on your system failed
to compile the code?

[Pushkar Pradhan]

I think it is allocated at runtime, I used gdb on:

int i = 0;
if(!i) {
int a;
}
return 0;
} When I tried to print a outside the loop gdb said:
No symbol "a" in current context
So it must have been created when the block is reached.

FMorales wrote: -- I didn't notice anything about this specifically
-- in the c.l.c. FAQ, nor could i get a strait answere
-- from my copy of the C standard. Hopeing someone
-- here could shed some light on it :)...

6.2.1
"If the declarator or type specifier that declares
the identifier appears inside a block or within the
list of parameter declarations in a function definition,
the identifier has block scope, which terminates at the
end of the associated block."

I believe this talks about what we learned early on :

int main(void)
{
{
int a;
}
printf("%d\n", a); /* ERROR - Invalid */
return 0;
}

Once outside of the closing } a is no longer accessible.
It's out of our current "scope". Which brings me to my
question. Since that data is static, is it automatically
allocated at program start up?

int main(void)
{
int i;
return 0;
}

As "i" would be in this example ? What if there's a conditional
statement there ? As in :

int main(void)
{
int i = 0;
if(!i) {
int a;
}
return 0;
}

If "a" is /NOT/ automatically allocated with "i", what is
the compiler doing to make sure it gets allocated only when,
and if it's ever needed (like if "i" changes before the
"if", or if it waits for input from a user which we'll assume
is unknown untill runtime will "a" still get allocated ?)

Any help is greatly appreciated, just something i started
wondering about :) FMorales...