What will happen when the size of a local variable length array turns out to be 0 (zero)?

[Xiangliang Meng]

Hi.

void setValue(int n)
{
int size = getValueLength( );
int buffer_p[size];

if (buffer_p)
{
....
}
}
When the local variable size is 0, will buffer_p be a null pointer? I doubt
it.

What will happen when the size of a local variable length array turns out to
be 0 (zero)? Is it an undefined behavior? Or an implementation-defined one?

Thank you.

Best Regards,

Xiangliang Meng

[Xiangliang Meng]

Hi, all.

It's all my fault putting this both in comp.lang.c and comp.lang.c++. Very
sorry for that.

This fragment code is extracted out from a class member function. In
addition, those codes are compiled by gcc. I'm not familar with C++
standard. As far as I know, the variable length array is introduced in C99,
but I do NOT know whether C++ has this. So I cross-posted it.

It should be
void Componet::setVa lue(int n)
{
int size = getValueLength( );
int buffer_p[size];

if (buffer_p)
{
...
}
}

ATTENTION:
=============== =============== =============== =============== =============== =
=============== ====
Any one who answers my question, please post ONLY in comp.lang.c from now
on. Thank all of you.
=============== =============== =============== =============== =============== =
=============== ====

Best Regards,

Xiangliang Meng

"Xiangliang Meng" <xi************ *@hotmail.com> wrote in message
news:ca******** **@zcars0v6.ca. nortel.com...

Hi.

void setValue(int n)
{
int size = getValueLength( );
int buffer_p[size];

if (buffer_p)
{
....
}
}
When the local variable size is 0, will buffer_p be a null pointer? I doubt it.

What will happen when the size of a local variable length array turns out to be 0 (zero)? Is it an undefined behavior? Or an implementation-defined one?
Thank you.

Best Regards,

Xiangliang Meng

[Richard Bos]

"Xiangliang Meng" <xi************ *@hotmail.com> wrote:

It's all my fault putting this both in comp.lang.c and comp.lang.c++. Very
sorry for that.

This fragment code is extracted out from a class member function.
That should have told you; C does not have, never has had, and by the
grace of the Committee never will have classes.
It should be
void Componet::setVa lue(int n)
This is _not_ C. Whether it is valid C++ I do not know, but it is not C
- neither C89 nor C99 nor even Ganuck.
{
int size = getValueLength( );
int buffer_p[size];
Therefore, talking about this code as if it is guaranteed to have C99
semantics is misleading.
Any one who answers my question, please post ONLY in comp.lang.c from now
on. Thank all of you.

No, do _not_ answer this in comp.lang.c. Whatever it is, it is some kind
of (possibly Ganuck++-specific) C++. It is certainly not C, therefore it
does not belong in comp.lang.c. Follow-ups set the other way.

Richard

[ Learn to snip, btw. ]

[Alf P. Steinbach]

* Xiangliang Meng:

It's all my fault putting this both in comp.lang.c and comp.lang.c++. Very
sorry for that.

This fragment code is extracted out from a class member function. In
addition, those codes are compiled by gcc. I'm not familar with C++
standard. As far as I know, the variable length array is introduced in C99,
but I do NOT know whether C++ has this. So I cross-posted it.

It should be
void Componet::setVa lue(int n)
{
int size = getValueLength( );
int buffer_p[size];

if (buffer_p)
{
...
}
}

Above you're using a C99 feature in a C++ program.

C99 does not have classes, and C++ does not have variable length arrays.

So the mix of the two features is something not allowed in either
language.

However, in C++ you can use the standard library's vector class to
achieve the same effect, as follows:

void Componet::setVa lue( int n )
{
int size = getValueLength( );
std::vector<int > buffer( size );
...
}

and in C99 you can rewrite the thing as e.g.

void Componet_setval ue( Component* self, int n )
{
int size = self->getValueLength ();
if( size == 0 ) { return; }
int buffer[size];
...
}

But first of all you need to decide which language you're using.

--
A: Because it messes up the order in which people normally read text.
Q: Why is it such a bad thing?
A: Top-posting.
Q: What is the most annoying thing on usenet and in e-mail?

[Jorge Rivera]

Jowtte wrote:

Consider this code:

int testAlloc()
{
int size = 0;
int* buffer = new int [ size ];
int* buffer2 = new int [ 1 ];
...
}

It does allocate memory for buffer, yet I don't know how large it is.

Using my compiler, before "new" operating, buffer = buffer2 = 0xcccccccc
and after "new", buffer = 0x0x00430060, buffer2 = 0x00430030

I don't know why.

I think the standard dictates that calls to new with size 0 will return
a valid pointer, therefore this is well-defined behavior (although what
really happens is likely to be an allocation of 1, or sizeof(T);

[Andrew]

On Thu, 10 Jun 2004 10:08:51 GMT, rl*@hoekstra-uitgeverij.nl (Richard
Bos) wrote:

This is _not_ C. Whether it is valid C++ I do not know, but it is not C
- neither C89 nor C99 nor even Ganuck.

<OT>
Ganuck? What is this?
</OT>

--
Andrew

[Branimir Maksimovic]

"Xiangliang Meng" <xi************ *@hotmail.com> wrote in message
news:ca******** **@zcars0v6.ca. nortel.com...
............... ...........

What will happen when the size of a local variable length array turns

out to

be 0 (zero)? Is it an undefined behavior? Or an implementation-defined one?

This is g++ specific extension in this case.
If size is 0 then nothing will happen, though you can't access
array elements 'cause there are none.
In order to avoid this situation use this:
void Componet::setVa lue(int n)
{
int size = getValueLength( );
int buffer_p[size];
//> if (buffer_p)
// buffer_p would be always non null pointer
if(sizeof buffer_p) // sizeof would be evaluated in run time {
...
}
}

Finally, I have question for C newsgroup.
Given,

void func(size_t size)
{
assert(size != 0);
int buf[size];
size_t tmp= sizeof (buf);
}

what would be result of sizeof? Some compile time constant
or run time evaluated value?

Greetings, Bane.

[CBFalconer]

Jorge Rivera wrote:

Jowtte wrote:

int testAlloc()
{
int size = 0;
int* buffer = new int [ size ];
int* buffer2 = new int [ 1 ];
...
}

It does allocate memory for buffer, yet I don't know how large it is.

Using my compiler, before "new" operating, buffer = buffer2 = 0xcccccccc
and after "new", buffer = 0x0x00430060, buffer2 = 0x00430030

I don't know why.

I think the standard dictates that calls to new with size 0 will return
a valid pointer, therefore this is well-defined behavior (although what
really happens is likely to be an allocation of 1, or sizeof(T);

This is not valid C code, and is OT in c.l.c. F'ups set.

--
Chuck F (cb********@yah oo.com) (cb********@wor ldnet.att.net)
Available for consulting/temporary embedded and systems.
<http://cbfalconer.home .att.net> USE worldnet address!

[CBFalconer]

Xiangliang Meng wrote:

It's all my fault putting this both in comp.lang.c and
comp.lang.c++. Very sorry for that.
.... snip ...
ATTENTION:
=============== ====
Any one who answers my question, please post ONLY in comp.lang.c
from now on. Thank all of you.

You have a simple mechanism available for that, which should have
been applied to the original post. Set followups.

--
Chuck F (cb********@yah oo.com) (cb********@wor ldnet.att.net)
Available for consulting/temporary embedded and systems.
<http://cbfalconer.home .att.net> USE worldnet address!

[Jack Klein]

On Thu, 10 Jun 2004 15:41:00 +0200, "Branimir Maksimovic"
<bm***@eunet.yu > wrote in comp.lang.c:

"Xiangliang Meng" <xi************ *@hotmail.com> wrote in message
news:ca******** **@zcars0v6.ca. nortel.com...
............... ..........

What will happen when the size of a local variable length array turns

out

to

be 0 (zero)? Is it an undefined behavior? Or an implementation-defined

one?

This is g++ specific extension in this case.
If size is 0 then nothing will happen, though you can't access
array elements 'cause there are none.
In order to avoid this situation use this:

void Componet::setVa lue(int n)
{
int size = getValueLength( );
int buffer_p[size];

//> if (buffer_p)
// buffer_p would be always non null pointer
if(sizeof buffer_p) // sizeof would be evaluated in run time

{
...
}
}

Finally, I have question for C newsgroup.
Given,

void func(size_t size)
{
assert(size != 0);
int buf[size];
size_t tmp= sizeof (buf);
}

what would be result of sizeof? Some compile time constant
or run time evaluated value?

Greetings, Bane.

The sizeof operator in C99 is evaluated at run time when it is applied
to a VLA. In all other cases it is the same as before, a compile time
operation that yields a compile time constant.

--
Jack Klein
Home: http://JK-Technology.Com
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http://www.contrib.andrew.cmu.edu/~a...FAQ-acllc.html