Set bits

[s]

If I have:

unsigned char value = 0xBD;
unsigned char 2bits = 0x02;

How do I set the two MSB in value to the two LSB in 2bits without
changing any other bits in value?

<mytry>
value &= 0x3F; //set those two bits to zero
value |= (2bits<<6);
</mytry>

Do I have to clear(or set) those two bits first?

[Frank Roland]

"s" <youshouldbe@ho me> schrieb im Newsbeitrag
news:<3F85BE28. 1020901@home>.. .

If I have: unsigned char value = 0xBD; unsigned char 2bits = 0x02;
Your identifiert may not start with a number. You may use two_bits instead.
How do I set the two MSB in value to the two LSB in 2bits without changing any other bits in value? <mytry> value &= 0x3F; //set those two bits to zero value |= (2bits<<6); </mytry>

Use the now identifier in your try and it should work.

Kind regards,

[s]

Thanks for the reply and sorry for the bad identifier!

Here's a better explanation of what I'm really trying to do:

unsigned char target = 0xBD;
unsigned char small_value = 2;
unsigned position = 0xC0;

I need to put the two bits in small_value into target at the place
indicated by the set bits in position.

thanks

Frank Roland wrote:

Your identifiert may not start with a number. You may use two_bits instead.

value |= (2bits<<6);

Use the now identifier in your try and it should work.

Kind regards,

[Tim Hagan]

s wrote:

If I have:

unsigned char value = 0xBD;
unsigned char 2bits = 0x02;
2bits is an invalid identifier. The first character must be a letter or
an underscore, followed by any sequence of digits and/or upper or lower
case letters.
How do I set the two MSB in value to the two LSB in 2bits without
changing any other bits in value?

<mytry>
value &= 0x3F; //set those two bits to zero
value |= (2bits<<6);
</mytry>
Looks good to me. However, it isn't very interesting, since the two
bits that you are trying to set in 'value' are already set you way
want them. Observe:

0xBD = 10111101
^^
0x02 = 00000010
Do I have to clear(or set) those two bits first?

You must clear them first, as you did in your example.

--
Tim Hagan

[Tim Hagan]

s wrote:

Here's a better explanation of what I'm really trying to do:

unsigned char target = 0xBD;
unsigned char small_value = 2;
unsigned position = 0xC0;

I need to put the two bits in small_value into target at the place
indicated by the set bits in position.

Here's one way to do it. It is not a very elegant solution, nor is the
output particularly interesting.

#include <stdio.h>

int main(void)
{
unsigned char target = 0xBD;
unsigned char small_value = 2;
unsigned char position = 0xC0; /* must be non-zero */
unsigned char temp = position;
int i = 0;

while (!(temp & 1))
{
temp >>= 1;
i++;
}
target &= ~position;
target |= (small_value << i);
printf("%#2X\n" , target);
return 0;
}

--
Tim Hagan

[Tim Hagan]

Tim Hagan wrote:

s wrote:

If I have:

unsigned char value = 0xBD;
unsigned char 2bits = 0x02;
2bits is an invalid identifier. The first character must be a letter or
an underscore, followed by any sequence of digits and/or upper or lower
case letters.

I tried to paraphrase K&R2 (a *big* mistake) and flubbed it. I should
have mentioned that underscores are considered to be letters and can
appear anywhere in identifier names. However, it is not recommended that
identifiers *begin* with an underscore since C library routines usually
use such names and conflicts may result. Also, some identifiers are
reserved as keywords (auto, break, case, char, ... ).

How do I set the two MSB in value to the two LSB in 2bits without
changing any other bits in value?

<mytry>
value &= 0x3F; //set those two bits to zero
value |= (2bits<<6);
</mytry>

Looks good to me. However, it isn't very interesting, since the two
bits that you are trying to set in 'value' are already set you way
want them.

Gak! I need a better proofreader.
Observe:

0xBD = 10111101
^^
0x02 = 00000010

Do I have to clear(or set) those two bits first?

You must clear them first, as you did in your example.

--
Tim Hagan