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.

[Joachim Schmitz]

CBFalconer wrote:

What gives you the idea that Anna speaks French?

Her email address petitmuoton?

[Joe Wright]

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 ---

[Anna]

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]

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