kdeng00
85 lines
1.6 KiB
C
85 lines
1.6 KiB
C
/*
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*
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* Exercise 2-9. In a two's complement number system, x &= (x-1) deletes the
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* rightmost 1-bit in x. Explain why. Use this observation to write a faster
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* version of bitcount.
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* rotated to the right by n positions.
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*
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* Observation:
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* In a two's complement number system, x &= (x-1) deletes the rightmost
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* 1-bit in x. The reason for that is simple once things are broken down.
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* Using an example, let's say x is 19. 19 in binary is 10011.
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* (x-1) is 18 or 10010 in binary. Using the bitwise AND operator the result
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* is:
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* 10011
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* & 10010
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* -------
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* 10010
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*
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* 10010 is 18 and 10011 with the rightmost bit removed.
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*
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* Author: Kun Deng
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*/
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#include <stdio.h>
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#include <stdlib.h>
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// Counts the number of 1-bits in a number
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int bitcount(unsigned int x);
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// Counts the number of 1-bits in a number but faster than bitcount()
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int bitcount_improved(unsigned int x);
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int main(int argc, char **argv)
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{
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unsigned int x = 511;
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if (argc == 2)
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{
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x = atoi(argv[1]);
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}
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int bit_amount = bitcount(x);
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printf("%d has %d bits\n", x, bit_amount);
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int bit_amount_improved = bitcount_improved(x);
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printf("%d has %d bits - Improved\n", x, bit_amount_improved);
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return 0;
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}
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int bitcount(unsigned int x)
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{
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int b;
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for (b = 0; x != 0; x>>= 1)
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{
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if (x & 01)
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{
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++b;
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}
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}
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return b;
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}
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int bitcount_improved(unsigned int x)
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{
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int b;
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// x &= (x-1) is always removing the rightmost bit. With this it's easier
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// to count the amount of 1-bits simply by incrementing b.
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for (b = 0; x != 0; x &= (x-1))
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{
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++b;
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}
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return b;
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}
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