Rotate 8 bits for 3 to right in p. language C

[8051help]

Hi.

I need to rotate for an example x=0XFF for 3 to right.

I cant do that only with >> because that is shifting.
When you want rotate you need to "memory" all bit not to lose such as in case for shifting (i think).
So that is not the same.

Can I use this (x & 0xFF)>>3 ; for rotate ??

 

[Wade_Hassler]

Your compiler might do what you want if the number is declared as signed
Or if it is unsigned....

if( x & 128)  x = ( x>>3 )| 128;
else          x = ( x>>3);

 

[Ian Rogers]

If you are rotating you need to know the condition of bit 0

shiftbits = 3;
while(shiftbits--)
   {
   if (rotatebyte && ox1)
      {
      rotatebyte>>= 1;
      rotatebyte+=128;
      }
   else
       rotatebyte>>=1;
   }

 

[8051help]

when i done like

if( x & 128){  x = ( x>>3 )| 128;
          display=x;   }
else  x = ( x>>3);
                 display=x; // display is memory mapped at adress 0x8000

the result is not good..


ANOTHER post with rotatebyte doesnt work

Thank you guys but doesnt work

i working in micro vision 3 atmel 8051

 

[8051help]

May I use this

(x>>3) | (x<<( (-3) & 0xFF ));

 

[8051help]

When I use (x>>3) | (x<<( (-3) & 0xFF )); and classic shifting I got the same result . Maybe because all diodes are 1111111???

 

[8051help]

I have result

x=0xFF;
(((x>>3)&0x1f)|(x<<5));

 

[Ian Rogers]

I have result

x=0xFF;
(((x>>3)&0x1f)|(x<<5));

That is one way, but its not portable..... If you want to rotate 4 times you'll need more code

I had a typo in my code sorry

shiftbits = 3;
while(shiftbits--)
   {
   if (rotatebyte & 0x1) // Should have been bit wise!!!
      {
      rotatebyte>>= 1;
      rotatebyte+=128;
      }
   else
       rotatebyte>>=1;
   }

 

[8051help]

when I write that code in interrupt rutine i have problem with execution...

if (++counter=20) {  //i have error c213 : left side of asn-op not an lvalue THAT ERROR SHOWS  THIS LINE OF CODE
shiftbits = 3;
while(shiftbits--)
  {
  if (x & 0x1) // Should have been bit wise!!!
  {
  x>>= 1;
  x+=128;
    display=x;
  }
  else
  x>>=1;
  }
   counter=0;        }
   
}

i have error c213 : left side of asn-op not an lvalue

 

[Ian Rogers]

Have you declared "shiftbits" and "x" ???

I write two functions... ROTR and ROTL both can take byte and word arguments!!

 

[8051help]

Have you declared "shiftbits" and "x" ???

I write two functions... ROTR and ROTL both can take byte and word arguments!!

Yes, I did.

shiftbits is byte and x is byte

 

[8051help]

I have one more question .

if (!((--counter )+=2)) { a=1}

in that case , what would be value of counter for a=1???

 

[Ian Rogers]

Yes, I did.

shiftbits is byte and x is byte

Ah!! Remember that 8051 has limited "bit access" memory!!

if NOT ( pre increment counter) +2 .... Will never be true unless it started out negative..

Assume counter = 1.....
pre increment counter (before evaluation ) counter = 2
counter += 2 counter now contains 4 ....
if NOT.... well!! the only time this will be true is if the count was -3 to start with

The only time a = 1 is when counter = -3...

 

[NorthGuy]

int8_t rotate_byte_right(int8_t x, int8_t pos) {
  return (x >> pos) | (x << (8 - pos));
}

int16_t rotate_word_right(int16_t x, int8_t pos) {
  return (x >> pos) | (x << (16 - pos));
}

any_int_type rotate_anything_right(any_int_type x, int8_t pos) {
  return (x >> pos) | (x << (sizeof(x)*8 - pos));
}

It'll be totally different if you do it in assembler.

 

[Mikebits]

If you are rotating you need to know the condition of bit 0

I feel silly for asking, but I don't understand what you guys mean by rotating a byte, could someone explain what you mean by rotate a byte?

Thanks

 

[Les Jones]

Hi Mike,
To complicate things further there is rotate and rotate through carry. This is my undersanding of "rotate". (Right) Bit 7 to bit 6 and so on. Bit 0 to bit 7. A rotate through carry is carry bit to bit 7, bit 7 to bit 6 and so on bit 0 to carry bit.

Les.

 

[misterT]

I feel silly for asking, but I don't understand what you guys mean by rotating a byte, could someone explain what you mean by rotate a byte?

Thanks

Rotation is same as shifting, except that the bits that drop out at one end appears at the other end.

int8_t rotate_byte_right(int8_t x, int8_t pos) {
  return (((uint8_t)x) >> pos) | (x << (8 - pos));
}

You have to be careful when shifting signed integers. Arithmetic shift to the right add 1's instead of zeros to the left. I edited the code to cast right shift to unsigned.

 

[NorthGuy]

You have to be careful when shifting signed integers. Arithmetic shift to the right add 1's instead of zeros to the left. I edited the code to cast right shift to unsigned.

That's a good point. Many compilers do this because the geniuses who wrote the standard didn't define the behaviour.

An easy work around is to re-define the arguments as unsigned. The "pos" argument should also be unsigned:

uint8_t rotate_byte_right(uint8_t x, uint8_t pos) {
  return (x >> pos) | (x << (8 - pos));
}

uint16_t rotate_word_right(uint16_t x, uint8_t pos) {
  return (x >> pos) | (x << (16 - pos));
}

any_unsigned_int_type rotate_anything_right(any_unsigned_int_type x, uint8_t pos) {
  return (x >> pos) | (x << (sizeof(x)*8 - pos));
}