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| -rw-r--r-- | ch2/2-07_invert.c | 38 |
1 files changed, 38 insertions, 0 deletions
diff --git a/ch2/2-07_invert.c b/ch2/2-07_invert.c new file mode 100644 index 0000000..244f854 --- /dev/null +++ b/ch2/2-07_invert.c @@ -0,0 +1,38 @@ +#include <stdio.h> + +/* The C Programming Language: 2nd Edition + * + * Exercise 2-07: 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. + * + * Answer: The trick is to start with ~0 to get all 1s, then left-shift n + * times. After that, flip again to get n number of 1s. Next, left shift the + * appropriate number of times needed to line up the mask of 1s to the proper + * location. Apply that mask to x with XOR and bam, job done! + * + * Here's a step by step illustration of invert(205, 5, 4), assuming 16-bit + * unsigned integers: + * + * 1111111111111111 ~0 + * 1111111111110000 << 4 + * 0000000000001111 ~ + * 0000000000011110 << (5 - 4 == 1) + * + * Mask created. Now let's take our original number... + * 0000000011001101 205 + * #### ^ ...and apply the mask! + * 0000000011010011 End Result (211) + */ + +unsigned invert(unsigned x, unsigned p, unsigned n) { + return x ^ ((~(~0 << n)) << (p - n)); +} + +int main() { + printf("invert(205, 5, 4) returns %d\n", invert(205, 5, 4)); + printf("11001101 flips 110[0110]1 to produce...\n11010011\n"); + printf("invert(1876, 7, 3) returns %d\n", invert(1876, 7, 3)); + printf("11101010100 flips 1110[101]0100 to produce...\n11100100100\n"); + return 0; +} |