演習 2-7

無職の間にK&Rを再読。演習問題の解答をさらす。解く順番は適当。

演習 2-7
x のビット位置 p から n ビットを反転（1 を 0 、 0 を 1 にする）し、他のビットはそのままにした x を返す関数 invert(x, p, n) を書け。

Exercise 2-7
Write a function invert(x, p, n) that returns x with the n bits that begin at position p inverted (i.e., 1 changed into 0 and vice versa), leaving the others unchanged.

演習2-6とほとんど同じ。

#include <stdio.h>

/* Answer */
unsigned invert(unsigned x, int p, int n)
{
  n = (n > p + 1) ? p + 1 : n;
  unsigned mask = ~(~0 << n) << (p - n + 1);
  return x ^ mask;
}



void showbits(long num, int size);
void _showbits(long num, int remain);


int main(void)
{
  unsigned short a = 0x5678;
  unsigned short b;

  showbits(a, sizeof(a));

  b = invert(a, 11, 12);

  showbits(b, sizeof(b));
  
  return 0;
}



void showbits(long num, int size)
{
  _showbits(num, size * 2 - 1);
  putchar('\n');
}


void _showbits(long num, int remain)
{
  static const char bitpattern[16][5] = {
    "0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111",
    "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111",
  };
 
  if (remain > 0) {
    _showbits(num >> 4, remain - 1);
    putchar(',');
  }
  
  printf("%s", bitpattern[ num & 0x0F ]);
}