Bit shifts
You are encouraged to solve this task according to the task description, using any language you may know.
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This is a basic data operation. It represents a fundamental action on a basic data type.
You may see other such operations in the Basic Data Operations category, or:
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Perform the following bit shifts, if supported in your language:
- Logical / arithmetic left shift - left shift where vacated bits are filled with 0; usually equivalent to multiplying integer by 2^N
- Arithmetic right shift - right shift where vacated bits are filled with the sign bit; usually equivalent to dividing signed integer by 2^N, rounding downwards towards negative infinity
- Logical right shift - right shift where vacated bits are filled with 0; usually equivalent to dividing unsigned integer by 2^N
If possible, perform the right shifts on a negative number, so one can distinguish between the arithmetic and logical shifts.
C
The C standard does not specifies what happens if you try to shift a negative signed integer. The behavior is therefore implementation-dependent. Typically, it performs an arithmetic shift in this case.
<c>#include <stdio.h> int main() {
int x = -3; int n = 1;
/* convert the signed integer into unsigned, so it will perform logical shift */ unsigned int y = x;
printf("x << n: %d\n", x << n); /* left shift */
printf("x >> n: %d\n", x >> n); /* arithmetic right shift */
printf("y >> n: %d\n", y >> n); /* logical right shift */
return 0;
}</c>
The output of this program on a typical platform is:
x << n: -6 x >> n: -2 y >> n: 2147483646
Haskell
The first operand can be Int, Integer, and any of the sized integer and word types. The second operand must be of type Int.
import Data.Bits x :: Integer x = -3 n :: Int n = 1 main = do print $ shiftL x n -- left shift print $ shiftR x n -- arithmetic right shift print $ shift x n -- You can also use the "unified" shift function; positive is for left shift, negative is for right shift print $ shift x (-n)
The output of this program on a typical platform is:
-6 -2 -6 -2
Java
<java>public static void main(String[] args) {
int x = -3; int n = 1;
System.out.println("x << n: " + (x << n)); // left shift
System.out.println("x >> n: " + (x >> n)); // arithmetic right shift
System.out.println("x >>> n: " + (x >>> n)); // logical right shift
}</java>
The output of this program is:
x << n: -6 x >> n: -2 x >>> n: 2147483646
JavaScript
<javascript>var x = -3; var n = 1; alert("x << n: " + (x << n)); // left shift alert("x >> n: " + (x >> n)); // arithmetic right shift alert("x >>> n: " + (x >>> n)); // logical right shift</javascript>
The output of this program is:
x << n: -6 x >> n: -2 x >>> n: 2147483646
OCaml
<ocaml>let x = -3;; let n = 1;;
Printf.printf "x lsl n: %d\n" (x lsl n);; (* left shift *) Printf.printf "x asr n: %d\n" (x asr n);; (* arithmetic right shift *) Printf.printf "x lsr n: %d\n" (x lsr n);; (* logical right shift *)</ocaml>
The output of this program is:
x lsl n: -6 x asr n: -2 x lsr n: 1073741822
Note that "int" in OCaml is 31 bits wide.
Perl
<perl>$x = -3; $n = 1; print '$x >> $n: ', $x >> $n, "\n"; # logical right shift
use integer; # "use integer" enables bitwise operations to return signed ints print "after use integer:\n"; print '$x << $n: ', $x << $n, "\n"; # left shift print '$x >> $n: ', $x >> $n, "\n"; # arithmetic right shift</perl>
Output:
$x >> $n: 2147483646 after use integer: $x << $n: -6 $x >> $n: -2
Python
<python>x = -3 n = 1 print 'x << n:', x << n # left shift print 'x >> n:', x >> n # arithmetic right shift</python>
The output of this program is:
x << n: -6 x >> n: -2