bit shifting in C

Question

int x = 2;

x = rotateInt('L', x, 1); // should return 4

x = rotateInt('R', x, 3); // should return 64

Here is the code, can someone check it and let me know what the error is?

Compilation is successful, but it says Segmentation Fault when I execute it.

int rotateInt(char direction, unsigned int x, int y)
{
  int i;

  for(i = 0; i < y; i++)
  {  

    if(direction == 'R')
    {
       if((x & 1) == 1)
       {
         x = x >> 1;
         x = (x ^ 128);     
       }
       else    
         x = x >> 1;
     }
     else if(direction == 'L')
     {
       if((x & 128) == 1)  
       {
         x = x << 1;
         x = (x ^ 1);     
       }  
       else
       x = x << 1;
     }
   }
   return x;   
 }

If you're done with your related question from 10 min ago stackoverflow.com/questions/3928659/… you might want to accept an answer before moving on. — Dusty
– Dusty, Commented Oct 14, 2010 at 0:20

user124493 · Accepted Answer · 2010-10-14 00:21:13Z

9

Start honing your debugging skills now. If you are going to be any form of an engineer, you'll need to write programs of some variety, and will thus be debugging all of your life.

A simple way to start debugging is to put print statements in to see how far your code makes it before it dies. I recommend you start by isolating the error.

answered Oct 14, 2010 at 0:21

user124493

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5 Comments

Tim Over a year ago

it prints nothing but segmentation fault, that is it.

Josh Over a year ago

That mens it's crashing before your first print statement. Move the print statement earlier and retry. Repeat until you know which line is causing the segfault

user124493 Over a year ago

@Josh that's exactly what I was going to say, thanks. Don't forget to put a \n in your printf statement to flush the stream.

bstpierre Over a year ago

For bonus points, learn how to use a debugger. You'll usually find the problem much faster than with printfs.

user124493 Over a year ago

@bstpierre true, but I figured that if we hadn't discovered printf-debugging yet that we weren't quite ready for the concept of a debugger :)

Andrew Cooper · Accepted Answer · 2010-10-14 00:24:34Z

2

Not sure about the seg fault, but I think

if((x & 128) == 1)

should be

if((x & 128) == 128)

or just

if(x & 128)

answered Oct 14, 2010 at 0:24

Andrew Cooper

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Comments

Antoine Trouve · Accepted Answer · 2010-10-14 00:29:27Z

1

I tried on my computer (MacBookPro / Core2Duo) and it worked. By the way, what's your target architecture ? Some (many) processors perform rotation instead of shifts when you use the C operators ">>" and "<<".

answered Oct 14, 2010 at 0:29

Antoine Trouve

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7 Comments

Antoine Trouve Over a year ago

OK, so I asmume it is an Intel Machine, which does not rotate (at least it is not the case of ny Intel Core2Duo). However it masks the argument with 0xff in the case of integer rotation (i.e. "0xff00 >> 16 = 0" but "0xff00 >> 40 = 0xff").

RBerteig Over a year ago

A compiler that did an actual bit rotation rather than a pure shift for the << and >> operators would be non-compliant. I don't see any wiggle-room in the definitions of the operators. They are shifts and not rotates, and have been since the dawn of the language. See C99, section 6.5.7 where it says "The result of E1 << E2 is E1 left-shifted E2 bit positions; vacated bits are filled with zeros." It goes on to describe the boundaries of the defined behavior in gory detail. Right shifts are described similarly, but with slightly different boundaries.

Antoine Trouve Over a year ago

It is not a matter of compiler, but processor. For instance PowerPC's shift instruction does rotation.

RBerteig Over a year ago

@Antoine, shifting by more bits than the number of bits in your integral type is explicitly Undefined Behavior. Anything can happen. The compiler is not obligated to be "reasonable" or to generate code that does something "reasonable".

RBerteig Over a year ago

@Antoine, rotation is simply not allowed as the implementation of the << operator. The language of the standard is extremely clear. It goes on to say that E1<<E2 is equivalent to multiplication by a power of 2, as long as the values are in bounds. It is also clear that shifting by a zero or negative amount is UB, as is shifting by more than the bits in the integral type.

|

No one in particular · Accepted Answer · 2010-10-14 00:25:48Z

0

When you use ^ don't you mean the or operator | ?

answered Oct 14, 2010 at 0:25

No one in particular

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