undo bitshift

This is a discussion on undo bitshift within the C Programming forums, part of the General Programming Boards category; Is it possible to find possible parameters for a bitshift operation? that is, for Code: int literal = 0xf0; int ...

  1. #1
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    Question undo bitshift

    Is it possible to find possible parameters for a bitshift operation?

    that is, for
    Code:
    int literal = 0xf0;
    int result = literal << variable;
    can I apply some bitwise operation to find a value for "variable" given a result and literal.

    It's easy enough to do in my head (just count the least significant digits until I get a 1), but what's the best way to do this with code?

  2. #2
    DESTINY BEN10's Avatar
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    i dont know how u can do it by using bitwise opeartor but i just want to say that left shift is the same as multiplying "literal" by 2 raised to the power "variable". So u can divide "result" by "literal" and take its square root to get the value of "variable".
    HOPE YOU UNDERSTAND.......

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  3. #3
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    sounds right but when I try it, it doesn't work:

    Code:
    #include <math.h>
    #include <stdio.h>
    
    int main(int argc, char **argv) {
    
    int variable = 3;
    int literal = 0xf0;
    int result = literal << variable;
    int rl = result/literal;
    printf("literal: %x\n",literal);
    printf("result: %x\n",result);
    printf("result/literal %x \n",rl);
    printf("sqrt(result/literal): %x\n",sqrt(rl));
    
    }
    gives me

    Code:
    $./test
    literal: f0
    result: 780
    result/literal 8 
    sqrt(result/literal): 667f3bcd
    have I missed something?

  4. #4
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    sqrt is not the right thing to use here, I'm pretty sure. Also, it returns a floating point value, so using %x is definitely wrong (unless you actually want to see what the floating point value looks like as a integer - but that's unlikely).

    I guess you could do:
    Code:
    x = (int)(log(result / literal) / log(2));
    But it's probably quicker to do:
    Code:
    int c = 0;
    int x = result / literal;
    while (x) {  c++; x >>= 1; }
    c is the number of bits literal was shifted.

    --
    Mats
    Compilers can produce warnings - make the compiler programmers happy: Use them!
    Please don't PM me for help - and no, I don't do help over instant messengers.

  5. #5
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    ok, so i guess
    Code:
     variable = c-1;
    since we don't want count the last bit from >>= operation, which would be the first 1.

    thanks everyone!

  6. #6
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    Ok, yes, or use while(x > 1).

    --
    Mats
    Compilers can produce warnings - make the compiler programmers happy: Use them!
    Please don't PM me for help - and no, I don't do help over instant messengers.

  7. #7
    DESTINY BEN10's Avatar
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    Quote Originally Posted by young turk View Post
    sounds right but when I try it, it doesn't work:

    Code:
    #include <math.h>
    #include <stdio.h>
    
    int main(int argc, char **argv) {
    
    int variable = 3;
    int literal = 0xf0;
    int result = literal << variable;
    int rl = result/literal;
    printf("literal: %x\n",literal);
    printf("result: %x\n",result);
    printf("result/literal %x \n",rl);
    printf("sqrt(result/literal): %x\n",sqrt(rl));
    
    }
    gives me

    Code:
    $./test
    literal: f0
    result: 780
    result/literal 8 
    sqrt(result/literal): 667f3bcd
    have I missed something?
    sorry i was wrong with the square root part. But going in the similar manner and taking the log(with base 2) of 'rl' will give the value of 'variable'.
    I have modified your code by taking the above mentioned into consideration and changed various values of 'variable' and the result was satisfactory.
    Code:
    #include <math.h>
    #include <stdio.h>
    
    int main(void) {
    
    int variable = 5;//changed it to 2,3,4 etc.
    int literal = 4;
    int result = literal << variable;
    float rl = (float)result/literal;
    printf("literal: %d\n",literal);
    printf("result: %d\n",result);
    printf("result/literal %f \n",rl);
    printf("variable=sqrt(result/literal): %f\n",log(rl)/log(2));//log(with base 2) of rl. In C log with a base of 10 is given so i changed it accordingly
    getch();
    return 0;
    
    }
    HOPE YOU UNDERSTAND.......

    By associating with wise people you will become wise yourself
    It's fine to celebrate success but it is more important to heed the lessons of failure
    We've got to put a lot of money into changing behavior


    PC specifications- 512MB RAM, Windows XP sp3, 2.79 GHz pentium D.
    IDE- Microsoft Visual Studio 2008 Express Edition

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