how to put 8 "int" => 10100010 into one character of type "char"

[Anna]

I try to put 8 int bit for example 10100010 into one character of type
char(1 octet) with no hope . Could anyone propose a simple way to do
it? Thank you very much.

[Ben Bacarisse]

Re: how to put 8 "int&qu ot; => 10100010 into one character of type "char&quot ;

Anna <petitmouton@gm ail.comwrites:

I try to put 8 int bit for example 10100010 into one character of type
char(1 octet) with no hope . Could anyone propose a simple way to do
it? Thank you very much.

It would help is we saw what you did. I would write:

unsigned char c = 0xA2; /* Hex A2 is 1010 0010 */

Not that unsigned char is almost always safer for this sort of thing.

--
Ben.

[Anna]

Re: how to put 8 &quot;int&qu ot; =&gt; 10100010 into one character of type &quot;char&quot ;

On 15 juin, 11:52, Ben Bacarisse <ben.use...@bsb .me.ukwrote:

Anna <petitmou...@gm ail.comwrites:

I try to put 8 int bit for example 10100010 into one character of type
char(1 octet) with no hope . Could anyone propose a simple way to do
it? Thank you very much.

>
It would help is we saw what you did. I would write:
>
unsigned char c = 0xA2; /* Hex A2 is 1010 0010 */
>
Not that unsigned char is almost always safer for this sort of thing.
>
--
Ben.

thank you Ben. I have bit sequence 01010101.... that I want to
implement into a simulator, but in the simulator, they use a char as a
type of data since each int comprise between 2-4 octet and one char is
only 1 octet (=8 bits 0 or 1). I guess it will make the simulator run
faster, I'm not sure.

[Ben Bacarisse]

Re: how to put 8 &quot;int&qu ot; =&gt; 10100010 into one character of type &quot;char&quot ;

Anna <petitmouton@gm ail.comwrites:

On 15 juin, 11:52, Ben Bacarisse <ben.use...@bsb .me.ukwrote:

>Anna <petitmou...@gm ail.comwrites:

I try to put 8 int bit for example 10100010 into one character of type
char(1 octet) with no hope . Could anyone propose a simple way to do
it? Thank you very much.

>>
>It would help is we saw what you did. I would write:
>>
> unsigned char c = 0xA2; /* Hex A2 is 1010 0010 */
>>
>Not that unsigned char is almost always safer for this sort of thing.
>>
>--
>Ben.

Best not quote sig blocks.

thank you Ben. I have bit sequence 01010101.... that I want to
implement into a simulator, but in the simulator, they use a char as a
type of data since each int comprise between 2-4 octet and one char is
only 1 octet (=8 bits 0 or 1). I guess it will make the simulator run
faster, I'm not sure.

For your information, in C a char is not always an octet and there are
systems where the sizes of the various types can be quite surprising.
I don't think that affects your question, though.

I can't tell if I answered your question. If I did not, you need to
give much more information. Some actual code is often the best way
for people here to see what the problem is.

--
Ben.

[Anna]

Re: how to put 8 &quot;int&qu ot; =&gt; 10100010 into one character of type &quot;char&quot ;

On 15 juin, 12:20, Ben Bacarisse <ben.use...@bsb .me.ukwrote:

Anna <petitmou...@gm ail.comwrites:

On 15 juin, 11:52, Ben Bacarisse <ben.use...@bsb .me.ukwrote:

Anna <petitmou...@gm ail.comwrites:
I try to put 8 int bit for example 10100010 into one character of type
char(1 octet) with no hope . Could anyone propose a simple way to do
it? Thank you very much.

>

It would help is we saw what you did. I would write:

>

unsigned char c = 0xA2; /* Hex A2 is 1010 0010 */

>

Not that unsigned char is almost always safer for this sort of thing.

>

--
Ben.

>
Best not quote sig blocks.
>

thank you Ben. I have bit sequence 01010101.... that I want to
implement into a simulator, but in the simulator, they use a char as a
type of data since each int comprise between 2-4 octet and one char is
only 1 octet (=8 bits 0 or 1). I guess it will make the simulator run
faster, I'm not sure.

>
For your information, in C a char is not always an octet and there are
systems where the sizes of the various types can be quite surprising.
I don't think that affects your question, though.
>
I can't tell if I answered your question. If I did not, you need to
give much more information. Some actual code is often the best way
for people here to see what the problem is.
>
--
Ben.

I think I'll have to try to understand the simulator first. it's a
very complicate network simulator with lots of code :-(. I will try to
implement what you suggested and will post the code if I have further
information. thank you very much.

[viza]

Re: how to put 8 &quot;int&qu ot; =&gt; 10100010 into one character of type&quot;char& quot;

Hi

On Sun, 15 Jun 2008 03:00:50 -0700, Anna wrote:

On 15 juin, 11:52, Ben Bacarisse <ben.use...@bsb .me.ukwrote:

>Anna <petitmou...@gm ail.comwrites:

I try to put 8 int bit for example 10100010 into one character of
type char(1 octet) with no hope . Could anyone propose a simple way
to do it? Thank you very much.

>>
>It would help is we saw what you did. I would write:
>>
> unsigned char c = 0xA2; /* Hex A2 is 1010 0010 */
>>
>Not that unsigned char is almost always safer for this sort of thing.
>>

thank you Ben. I have bit sequence 01010101.... that I want to implement
into a simulator, but in the simulator, they use a char as a type of
data since each int comprise between 2-4 octet and one char is only 1
octet (=8 bits 0 or 1). I guess it will make the simulator run faster,
I'm not sure.

Do you mean that you have eight integers but that each one has either the
value 1 or 0? In that case, do:

Est-ce que vous-voulez dire que vous avez huite integers et que chaqun
d'entre eux a le valeur 1 ou 0? En ce cas, faitez:

int integers[8];
unsigned char character;

character= ( integers[0] ? (1 << 0) : 0 )
| ( integers[1] ? (1 << 1) : 0 )
| ( integers[2] ? (1 << 2) : 0 )
| ( integers[3] ? (1 << 3) : 0 )
| ( integers[4] ? (1 << 4) : 0 )
| ( integers[5] ? (1 << 5) : 0 )
| ( integers[6] ? (1 << 6) : 0 )
| ( integers[7] ? (1 << 7) : 0 );

Actually this works if the integers have the value 0 or non-0. If you
are certain that they can only be 1 or 0, the following may be faster:

Ceci marche si les integers ont les valeurs 0 ou non-0. Si vous etez
certain que ils n'ont que 1 ou 0, le ci-dessus peut-etre plus vite:

character= (integers[0] << 0)
| ( integers[1] << 1)
| ( integers[2] << 2)
| ( integers[3] << 3)
| ( integers[4] << 4)
| ( integers[5] << 5)
| ( integers[6] << 6)
| ( integers[7] << 7);

HTH
viza

[CBFalconer]

Re: how to put 8 &quot;int&qu ot; =&gt; 10100010 into one character of type&quot;char& quot;

viza wrote:

Anna wrote:

>Ben Bacarisse <ben.use...@bsb .me.ukwrote:

>>Anna <petitmou...@gm ail.comwrites:
>>>
>>>I try to put 8 int bit for example 10100010 into one character
>>>of type char(1 octet) with no hope . Could anyone propose a
>>>simple way to do it? Thank you very much.
>>>
>>It would help is we saw what you did. I would write:
>>>
>> unsigned char c = 0xA2; /* Hex A2 is 1010 0010 */
>>>
>>Not that unsigned char is almost always safer for this sort of
>>thing.

>>
>thank you Ben. I have bit sequence 01010101.... that I want to
>implement into a simulator, but in the simulator, they use a
>char as a type of data since each int comprise between 2-4 octet
>and one char is only 1 octet (=8 bits 0 or 1). I guess it will
>make the simulator run faster, I'm not sure.

>
Do you mean that you have eight integers but that each one has
either the value 1 or 0? In that case, do:
>
Est-ce que vous-voulez dire que vous avez huite integers et que
chaqun d'entre eux a le valeur 1 ou 0? En ce cas, faitez:
>
int integers[8];
unsigned char character;
>
character= ( integers[0] ? (1 << 0) : 0 )
| ( integers[1] ? (1 << 1) : 0 )
| ( integers[2] ? (1 << 2) : 0 )
| ( integers[3] ? (1 << 3) : 0 )
| ( integers[4] ? (1 << 4) : 0 )
| ( integers[5] ? (1 << 5) : 0 )
| ( integers[6] ? (1 << 6) : 0 )
| ( integers[7] ? (1 << 7) : 0 );
>
Actually this works if the integers have the value 0 or non-0.
If you are certain that they can only be 1 or 0, the following
may be faster:
>
Ceci marche si les integers ont les valeurs 0 ou non-0. Si vous
etez certain que ils n'ont que 1 ou 0, le ci-dessus peut-etre
plus vite:
>
character= (integers[0] << 0)
| ( integers[1] << 1)
| ( integers[2] << 2)
| ( integers[3] << 3)
| ( integers[4] << 4)
| ( integers[5] << 5)
| ( integers[6] << 6)
| ( integers[7] << 7);

What gives you the idea that Anna speaks French?

--
[mail]: Chuck F (cbfalconer at maineline dot net)
[page]: <http://cbfalconer.home .att.net>
Try the download section.


** Posted from http://www.teranews.com **

[Joachim Schmitz]

Re: how to put 8 &quot;int&qu ot; =&gt; 10100010 into one character of type&quot;char& quot;

CBFalconer wrote:

What gives you the idea that Anna speaks French?

Her email address petitmuoton?

[Joe Wright]

Re: how to put 8 &quot;int&qu ot; =&gt; 10100010 into one character of type&quot;char& quot;

CBFalconer wrote:

viza wrote:

>Anna wrote:

>>Ben Bacarisse <ben.use...@bsb .me.ukwrote:
>>>Anna <petitmou...@gm ail.comwrites:
>>>>
>>>>I try to put 8 int bit for example 10100010 into one character
>>>>of type char(1 octet) with no hope . Could anyone propose a
>>>>simple way to do it? Thank you very much.
>>>It would help is we saw what you did. I would write:
>>>>
>>> unsigned char c = 0xA2; /* Hex A2 is 1010 0010 */
>>>>
>>>Not that unsigned char is almost always safer for this sort of
>>>thing.
>>thank you Ben. I have bit sequence 01010101.... that I want to
>>implement into a simulator, but in the simulator, they use a
>>char as a type of data since each int comprise between 2-4 octet
>>and one char is only 1 octet (=8 bits 0 or 1). I guess it will
>>make the simulator run faster, I'm not sure.

>Do you mean that you have eight integers but that each one has
>either the value 1 or 0? In that case, do:
>>
>Est-ce que vous-voulez dire que vous avez huite integers et que
>chaqun d'entre eux a le valeur 1 ou 0? En ce cas, faitez:
>>
>int integers[8];
>unsigned char character;
>>
>character= ( integers[0] ? (1 << 0) : 0 )
> | ( integers[1] ? (1 << 1) : 0 )
> | ( integers[2] ? (1 << 2) : 0 )
> | ( integers[3] ? (1 << 3) : 0 )
> | ( integers[4] ? (1 << 4) : 0 )
> | ( integers[5] ? (1 << 5) : 0 )
> | ( integers[6] ? (1 << 6) : 0 )
> | ( integers[7] ? (1 << 7) : 0 );
>>
>Actually this works if the integers have the value 0 or non-0.
>If you are certain that they can only be 1 or 0, the following
>may be faster:
>>
>Ceci marche si les integers ont les valeurs 0 ou non-0. Si vous
>etez certain que ils n'ont que 1 ou 0, le ci-dessus peut-etre
>plus vite:
>>
>character= (integers[0] << 0)
> | ( integers[1] << 1)
> | ( integers[2] << 2)
> | ( integers[3] << 3)
> | ( integers[4] << 4)
> | ( integers[5] << 5)
> | ( integers[6] << 6)
> | ( integers[7] << 7);

>
What gives you the idea that Anna speaks French?
>

Calling herself 'petitmouton' is a clue.

--
Joe Wright
"Everything should be made as simple as possible, but not simpler."
--- Albert Einstein ---

[Malcolm McLean]

Re: how to put 8 &quot;int&qu ot; =&gt; 10100010 into one character of type &quot;char&quot ;


"viza" <tom.viza@gmil. comwrote in message

On Sun, 15 Jun 2008 03:00:50 -0700, Anna wrote:
>
Est-ce que vous-voulez dire que vous avez huite integers et que chaqun
d'entre eux a le valeur 1 ou 0? En ce cas, faitez:
>

Crosspost French to comp.lang.c.fr

--
Free games and programming goodies.

http://www.personal.leeds.ac.uk/~bgy1mm

[Malcolm McLean]

Re: how to put 8 &quot;int&qu ot; =&gt; 10100010 into one character of type &quot;char&quot ;


"Anna" <petitmouton@gm ail.comwrote in message

thank you Ben. I have bit sequence 01010101.... that I want to
implement into a simulator, but in the simulator, they use a char as a
type of data since each int comprise between 2-4 octet and one char is
only 1 octet (=8 bits 0 or 1). I guess it will make the simulator run
faster, I'm not sure.

>>

char inputchar = 0x55; will set the variable inputchar to 01010101

--
Free games and programming goodies.

http://www.personal.leeds.ac.uk/~bgy1mm

[Antoninus Twink]

Re: how to put 8 &quot;int&qu ot; =&gt; 10100010 into one character of type &quot;char&quot ;

On 15 Jun 2008 at 15:27, Malcolm McLean wrote:

"viza" <tom.viza@gmil. comwrote in message

>Est-ce que vous-voulez dire que vous avez huite integers et que chaqun
>d'entre eux a le valeur 1 ou 0? En ce cas, faitez:
>>

Crosspost French to comp.lang.c.fr

Describing that as French is more than a little generous...

[Anna]

Re: how to put 8 &quot;int&qu ot; =&gt; 10100010 into one character of type &quot;char&quot ;

On 15 juin, 13:33, viza <tom.v...@gmil. comwrote:

Hi
>
>
>
On Sun, 15 Jun 2008 03:00:50 -0700, Anna wrote:

On 15 juin, 11:52, Ben Bacarisse <ben.use...@bsb .me.ukwrote:

Anna <petitmou...@gm ail.comwrites:
I try to put 8 int bit for example 10100010 into one character of
type char(1 octet) with no hope . Could anyone propose a simple way
to do it? Thank you very much.

>

It would help is we saw what you did. I would write:

>

unsigned char c = 0xA2; /* Hex A2 is 1010 0010 */

>

Not that unsigned char is almost always safer for this sort of thing.

>

thank you Ben. I have bit sequence 01010101.... that I want to implement
into a simulator, but in the simulator, they use a char as a type of
data since each int comprise between 2-4 octet and one char is only 1
octet (=8 bits 0 or 1). I guess it will make the simulator run faster,
I'm not sure.

>
Do you mean that you have eight integers but that each one has either the
value 1 or 0? In that case, do:
>
Est-ce que vous-voulez dire que vous avez huite integers et que chaqun
d'entre eux a le valeur 1 ou 0? En ce cas, faitez:
>
int integers[8];
unsigned char character;
>
character= ( integers[0] ? (1 << 0) : 0 )
| ( integers[1] ? (1 << 1) : 0 )
| ( integers[2] ? (1 << 2) : 0 )
| ( integers[3] ? (1 << 3) : 0 )
| ( integers[4] ? (1 << 4) : 0 )
| ( integers[5] ? (1 << 5) : 0 )
| ( integers[6] ? (1 << 6) : 0 )
| ( integers[7] ? (1 << 7) : 0 );
>
Actually this works if the integers have the value 0 or non-0. If you
are certain that they can only be 1 or 0, the following may be faster:
>
Ceci marche si les integers ont les valeurs 0 ou non-0. Si vous etez
certain que ils n'ont que 1 ou 0, le ci-dessus peut-etre plus vite:
>
character= (integers[0] << 0)
| ( integers[1] << 1)
| ( integers[2] << 2)
| ( integers[3] << 3)
| ( integers[4] << 4)
| ( integers[5] << 5)
| ( integers[6] << 6)
| ( integers[7] << 7);
>
HTH
viza

Bonjour Viza,
Bien joué, je parle français aussi :-) merce beaucoup pour ta réponse.
Je trouve très intéressant et je pense que je vais essayer
d'implementer ce ci sur mon programme. Merci, merci, merci
Anna

[Anna]

Re: how to put 8 &quot;int&qu ot; =&gt; 10100010 into one character of type &quot;char&quot ;

On 15 juin, 13:33, viza <tom.v...@gmil. comwrote:

Hi
>
>
>
On Sun, 15 Jun 2008 03:00:50 -0700, Anna wrote:

On 15 juin, 11:52, Ben Bacarisse <ben.use...@bsb .me.ukwrote:

Anna <petitmou...@gm ail.comwrites:
I try to put 8 int bit for example 10100010 into one character of
type char(1 octet) with no hope . Could anyone propose a simple way
to do it? Thank you very much.

>

It would help is we saw what you did. I would write:

>

unsigned char c = 0xA2; /* Hex A2 is 1010 0010 */

>

Not that unsigned char is almost always safer for this sort of thing.

>

thank you Ben. I have bit sequence 01010101.... that I want to implement
into a simulator, but in the simulator, they use a char as a type of
data since each int comprise between 2-4 octet and one char is only 1
octet (=8 bits 0 or 1). I guess it will make the simulator run faster,
I'm not sure.

>
Do you mean that you have eight integers but that each one has either the
value 1 or 0? In that case, do:
>
Est-ce que vous-voulez dire que vous avez huite integers et que chaqun
d'entre eux a le valeur 1 ou 0? En ce cas, faitez:
>
int integers[8];
unsigned char character;
>
character= ( integers[0] ? (1 << 0) : 0 )
| ( integers[1] ? (1 << 1) : 0 )
| ( integers[2] ? (1 << 2) : 0 )
| ( integers[3] ? (1 << 3) : 0 )
| ( integers[4] ? (1 << 4) : 0 )
| ( integers[5] ? (1 << 5) : 0 )
| ( integers[6] ? (1 << 6) : 0 )
| ( integers[7] ? (1 << 7) : 0 );
>
Actually this works if the integers have the value 0 or non-0. If you
are certain that they can only be 1 or 0, the following may be faster:
>
Ceci marche si les integers ont les valeurs 0 ou non-0. Si vous etez
certain que ils n'ont que 1 ou 0, le ci-dessus peut-etre plus vite:
>
character= (integers[0] << 0)
| ( integers[1] << 1)
| ( integers[2] << 2)
| ( integers[3] << 3)
| ( integers[4] << 4)
| ( integers[5] << 5)
| ( integers[6] << 6)
| ( integers[7] << 7);
>
HTH
viza

Bonjour Viza,
Oui, je parle français aussi :-) merci beaucoup pour ton aide. Ton
programme me semble très intéressant, je vais essayer d'implementer ce
ci sur le simulateur. Merci beaucoup.
Anna