Thread: Quick question regardin bit shifting

So what value does 'squashed' have in it first? At any rate:

01100101 >> 1 = 00110010
00110010 >> 1 = 00011001
00011001 >> 1 = 00001100
00001100 >> 1 = 00000110
00000110 >> 1 = 00000011
00000011 >> 1 = 00000001

Seven shifts.

10000111 << 1 = 00001110

One shift.

00001110 | 00000001 = 00001111


See?


Quzah.

Originally Posted by quzah

Seven shifts.

Looks like 6.

Hm. Could be. I was thinking when I was working it out that their end result was wrong. Looks like I was right after all.

So yeah, anyway, your:
a) Input is wrong, or...
b) Your output is wrong, as to what you're telling is you're getting.

Because if you did shift that seven places, and not 6, your result would be zero for the first portion.


Quzah.

well.

im inputting a string of hexcharacters "1234567890" into a function which "squashes" them so that

Code:

squashdum[0] = "00010010" = "12"
squashdum[1] = "11000010" = "34"

and pairs of numbers/hex thereafter.

the MSB in squashdum[0] must be Or'd with squashdum[1]. at the moment, im doing only ONE shift.

im starting to think that its maybe to do with the numbering of the indexes of the array and the arrangement of the binary stack within it.

i.e. im thinking that the string in contains the values as...

Code:

squashdum[2] squashdum[1] squashdum[0]

by my logic would equate to a "binary" of within the array

Code:

01100101 ||| 01000011 ||| 00100001

the aim is to insert a zero or maybe 5 zero's at the start of the string "squashdum[0]" and watch the bits move along. if they fall off the edge then they just move to the next byte.

is this correct or have i confused myself?

I'm having trouble with:

Code:

squashdum[1] = "11000010" = "34"

Wouldn't that binary number represent the two 4-bit integers 12 (0xC) and 2?

1 and 2 in squashdum[0] looks right, but that one looks weird. 3 and 4 should be 00110100.

You can express binary numbers either way, but you have to express them consistently or you're going to get confused. squashdum[0] shows MSB -> LSB, but squashdum[1] shows LSB -> MSB. Typically I've seen MSB -> LSB so you have:

1 - 0001
2 - 0010
3 - 0011
etc.

I get the right answer. I'm not sure what you're doing wrong...

Code:

itsme@itsme:~/C$ ./shift
10000111 - 87
00001110 - 0E
itsme@itsme:~/C$ cat shift.c
#include <stdio.h>

void show_it(int num)
{
  int i;

  for(i = 7;i >= 0;--i)
    printf("%c", (num >> i) & 1 ? '1' : '0');
  printf(" - %02X\n", num);
}

int main(void)
{
  unsigned char num = 0x87;

  show_it(num);
  num <<= 1;
  show_it(num);

  return 0;
}

Code:

itsme@itsme:~/C$ ./shift
10000111 - 87
00001110 - 0E
itsme@itsme:~/C$

how do i set the LSB to zero? was thinking of an AND mask with 0xFE but would that work?

Yes, but you shouldn't need it. I don't see how that 1 is slipping in.