How can I set a integer array all bits to zero?

[I wish]

#include <string.h>

int a[ 100 ];

memset( a, 0, sizeof(a) );

Does that guarantee all bits zero?

--
|
___
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<| |>---------------------------------- ShepJeng.twbbs. org -------------
/ plum.cs.nccu.ed u.tw

[Jirka Klaue]

I wish wrote:

#include <string.h>

int a[ 100 ];

memset( a, 0, sizeof(a) );

Does that guarantee all bits zero?

Depends, C90 no, C99 yes.

Both: int a[100] = {0};

Jirka

[Bertrand Mollinier Toublet]

Jirka Klaue wrote:

I wish wrote:

#include <string.h>

int a[ 100 ];

memset( a, 0, sizeof(a) );

Does that guarantee all bits zero?

Depends, C90 no, C99 yes.

You piqued my curiosity. What's up with C99 that the memset guarantees
all bits zero ?
Both: int a[100] = {0};

--
Bertrand Mollinier Toublet
"Lead developer. Iron developer. Copper developer." -- Sean Egan

[Eric Sosman]

Jirka Klaue wrote:

I wish wrote:

#include <string.h>

int a[ 100 ];

memset( a, 0, sizeof(a) );

Does that guarantee all bits zero?

Depends, C90 no, C99 yes.

There's no "depends" about it: under all versions of
the Standard, `a' is set to all bits zero by this code.

What *does* depend on the Standard version is whether
all bits zero is the same as value zero. In C89/90, an
`int' with all bits zero has the value zero. In C99,
where an `int' is permitted to contain bits that are not
part of its value ("padding bits"), it is possible that a
zero-valued `int' might contain non-zero bits, and it
is also possible that an `int' with all bits zero might
be a "trap representation. "

--
Er*********@sun .com

[Jirka Klaue]

Eric Sosman wrote:

Jirka Klaue wrote: ....

int a[ 100 ];
memset( a, 0, sizeof(a) );

Does that guarantee all bits zero?

Depends, C90 no, C99 yes.

There's no "depends" about it: under all versions of
the Standard, `a' is set to all bits zero by this code.

What *does* depend on the Standard version is whether
all bits zero is the same as value zero.

That's what I meant. Sorry.
In C89/90, an
`int' with all bits zero has the value zero. In C99,
where an `int' is permitted to contain bits that are not
part of its value ("padding bits"), it is possible that a
zero-valued `int' might contain non-zero bits, and it
is also possible that an `int' with all bits zero might
be a "trap representation. "

I thought, it is the other way round. Hmm...

Jirka

[cody]

> What *does* depend on the Standard version is whether

all bits zero is the same as value zero. In C89/90, an
`int' with all bits zero has the value zero. In C99,
where an `int' is permitted to contain bits that are not
part of its value ("padding bits"), it is possible that a
zero-valued `int' might contain non-zero bits, and it
is also possible that an `int' with all bits zero might
be a "trap representation. "

Padding bits in int values? What reason could made an implementation do
this?

--
cody

[Freeware, Games and Humor]
www.deutronium.de.vu || www.deutronium.tk

[Christian Bau]

In article <bm************ @ID-176797.news.uni-berlin.de>,
"cody" <do************ *********@gmx.d e> wrote:

cody

[Freeware, Games and Humor]

But definitely not funny.

[Irrwahn Grausewitz]

"cody" <do************ *********@gmx.d e> wrote:

Padding bits in int values? What reason could made an implementation do
this?

Parity bits, for example.
--
Irrwahn
(ir*******@free net.de)

[Jun Woong]

"Eric Sosman" <Er*********@su n.com> wrote in message news:3F******** *******@sun.com ...

Jirka Klaue wrote:

I wish wrote:

#include <string.h>

int a[ 100 ];

memset( a, 0, sizeof(a) );

Does that guarantee all bits zero?
Depends, C90 no, C99 yes.

There's no "depends" about it: under all versions of
the Standard, `a' is set to all bits zero by this code.

The representations of integer types in C90 are vague because C90
doesn't say much. But we can know that unsigned char in C90 can have
padding bits from a relevant DR, even if it's not the committee's real
intent. Of course, with lack of enough description I think it's
useless to discuss the integer representations in C90.

What *does* depend on the Standard version is whether
all bits zero is the same as value zero. In C89/90, an
`int' with all bits zero has the value zero.

If the reason for this is because C90 doesn't mention padding bits for
integer types, the story changed after the DR that I mentioned above
and on which C99's model was based was published.
--
Jun, Woong (mycoboco at hanmail.net)

[Jack Klein]

On Wed, 15 Oct 2003 11:49:53 -0400, Eric Sosman <Er*********@su n.com>
wrote in comp.lang.c:

Jirka Klaue wrote:

I wish wrote:

#include <string.h>

int a[ 100 ];

memset( a, 0, sizeof(a) );

Does that guarantee all bits zero?

Depends, C90 no, C99 yes.

There's no "depends" about it: under all versions of
the Standard, `a' is set to all bits zero by this code.

What *does* depend on the Standard version is whether
all bits zero is the same as value zero. In C89/90, an
`int' with all bits zero has the value zero. In C99,
where an `int' is permitted to contain bits that are not
part of its value ("padding bits"), it is possible that a
zero-valued `int' might contain non-zero bits, and it
is also possible that an `int' with all bits zero might
be a "trap representation. "

Chapter and verse, please, that C90 guaranteed that all bits 0 is a
valid representation of the value 0 for _ANY_ type.

--
Jack Klein
Home: http://JK-Technology.Com
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