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Problem with Bit Shifting to isolate bits

Posted on 2004-10-04
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Last Modified: 2010-05-18
I'm using Visual Studio.NET specifically the C++ compiler, and I'm writing a mini-compiler of my own. I read in 16 bit numbers, and I need to break them down (isolate certain groups of bits).

I wrote a Get_Bits function (just now) that I will paste below. My problem is that when I do a left shift, instead of deleting the bits shifted out of the 16 bit scope, the integer simply gets extended. I need those bits to be deleted so I can extract certain sections, such as bit 6-8 or something like that.

int _tmain(int argc, _TCHAR* argv[])
{
      unsigned int temp;
      temp = 0x1FF3;
      cout<<hex<<temp<<endl;
      temp=Get_Bits(temp,5,5);
      cout<<temp<<endl;
      return 0;
}
unsigned int Get_Bits(unsigned int instruction, int lower, int upper)
{
      int left, right;
      for (left=0;left<(15-upper);left++)
      {
            instruction=instruction<<1;
            cout<<"<<"<<instruction<<endl;
      }
      for (right=0;right<(lower+left);right++)
      {
            instruction=instruction>>1;
            cout<<">>"<<instruction<<endl;
      }
      return instruction;
}

With the value I have assigned to temp in my main, and the parameters entered into the call to Get_Bits, I would like to clear everything but the fifth bit from the right (start counting with zero). Instead this is my resulting output.

1ff3
<<3fe6
<<7fcc
<<ff98
<<1ff30
<<3fe60
<<7fcc0
<<ff980
<<1ff300
<<3fe600
<<7fcc00
>>3fe600
>>1ff300
>>ff980
>>7fcc0
>>3fe60
>>1ff30
>>ff98
>>7fcc
>>3fe6
>>1ff3
>>ff9
>>7fc
>>3fe
>>1ff
>>ff
ff

I need some way to cap the unsigned int at 16 bits, so that the shifting will indeed just clear bits. Or if you have suggestions as to how I can do this more easily, I would appreciate those as well.
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Question by:drumrboy44
1 Comment
 
LVL 55

Accepted Solution

by:
Jaime Olivares earned 500 total points
Comment Utility
You don't need to shift to 'clear' some undesired bits, just use and operator.
I will describe you pseudo code:

1) Construct a mask. By example if you want bits 4 to 9, your mask will be 000000000000001111110000
2) Apply the mask to the original number by example to:  00001111000011110000111100001111
    Result will be (I hope): 00000000000000000000001100000000
3) Now shift to the right using the lower value (4 in this example)
    Result will be (I hope again): 00000000000000000000000000110000
4) Your desired bits are now in first 6 bits.

To build your mask (1):
int mask = (1<<(high-low+1))-1)<<low;

To apply the mask and shift (2) and (3):
int desiredBits = (myNumber & mask)>>low;

So, if you join all:
int desiredBits = (myNumber & ((1<<(high-low+1))-1)<<low))>>low;

I guess it can be reduce to:
int desiredBits = (myNumber>>low) & (1<<(high-low+1))-1);
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