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So I am trying to solve this home assignment and I have been stuck with this one particular problem for a couple of hours and can't figure it out. I feel like I am so close! But then i change something in the code and something else isn't right..

/*
 * logicalShift - shift x to the right by n, using a logical shift
 *   Can assume that 0 <= n <= 31
 *   Examples: logicalShift(0x87654321,4) = 0x08765432
 *   Legal ops: ! ~ & ^ | + << >>
 *   Max ops: 20
 *   Rating: 3
 */
int logicalShift(int x, int n) {

    int move;
    int y;
    y = x >> n;
    y = ~y << 1;
    move = (y & (x >> n));

    return move;
}

What is missing here? I get 0x80000000 >> 31 as 0 but should be 1 - But other than that I don't know..

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2 Answers 2

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0

0x80000000 >> 31 = 1 if it's a logical shift.

0x80000000 >> 31 = -1 if it's an arithmetic shift.

In C++, if the value being shifted is unsigned, it is logical shift.

In Java, >> is arithmetic shift. >>> is logical shift.

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3 Comments

0x80000000 should be 1 if shifted with the logical shift 31.
Yeah. sorry.. miscalculate.
right shift in C it's implementation defined, so you can't rely on unsigned types being logical shift on all platforms
0

C's >> operator already performs a logical right shift on an unsigned integer. Does this do what you want?

#include <stdio.h>

unsigned long int logicalShift(unsigned long int x, unsigned int n) {
  return x >> n;
}

int main() {
  unsigned long int value = 0x80000000UL;
  unsigned int shift_amt = 31;
  unsigned long int result = logicalShift(value, shift_amt);
  printf("0x%lx >> %d = 0x%lx\n", value, shift_amt, result);
  return 0;
}

Result:

0x80000000 >> 31 = 0x1

If you are not allowed to cast to an unsigned data type, then the result of right shifting a signed value in C is implementation defined, according to this answer by Ronnie which cites K&R Second Edition. Even if this possible homework assignment were amended to allow the division operator, the alternate solution involving an elaboration on `x / (1 << n)' is also problematic because the rounding for division involving a negative number is also implementation-defined prior to C99. So unless you can tell us which C implementation your instructor is using and which ABI it implements, this question would appear to have no answer that is both easy and portable.

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6 Comments

I am not allowed to change the data type :/
@drleifz: Well that's inconvenient. What else are you not allowed to do? Can you use a cast?
I am only allowed to use the bit-wise operators (! ~ & ^ | + << >>) . Not allowed to use casting, if-statements, functions or anything like that :)
You've already used assignment (=) which is not in your list of permitted operators. Might want to review those requirements and make sure you're on the right track.
= is not considered a bit-wise operator according to the teacher. But I am allowed to use those bit-wise operators above including (=) but excluding (-, &&, ||, if-statements, functions)
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