From 06c2a9d69b05cc717299b0325f5001ff0303eecf Mon Sep 17 00:00:00 2001 From: zlg Date: Sun, 31 Mar 2013 05:38:14 -0500 Subject: Solve Exercise 2-6: setbits() --- ch2/2-06_setbits.c | 44 ++++++++++++++++++++++++++++++++++++++++++++ 1 file changed, 44 insertions(+) create mode 100644 ch2/2-06_setbits.c diff --git a/ch2/2-06_setbits.c b/ch2/2-06_setbits.c new file mode 100644 index 0000000..67c3efc --- /dev/null +++ b/ch2/2-06_setbits.c @@ -0,0 +1,44 @@ +#include + +/* The C Programming Language: 2nd Edition + * + * Exercise 2-6: Write a function setbits(x,p,n,y) that returns x with the n + * bits that begin at position p set to the rightmost bits of y, leaving the + * other bits unchanged. + * + * Answer: The book hands you the function you need to get this done: getbits. + * To be honest I don't fully understand the function, and this portion of the + * chapter is very poorly explained. For instance, what is a lower order bit? + * What is a right-adjusted bit field? Further, why are octal numbers used for + * binary operations when each octal bit is only 3 binary bits long? It would + * make more sense to use hexadecimal. + * + * At any rate, getbits()'s p parameter is the position from the right side of + * a binary number. The far right bit is position 0 (zero), and n is the number + * of binary fields it should take, reading to the right. This behavior wasn't + * entirely obvious. + * + * Anyway, setbits() clears the right-most n bits of y and replaces them with + * the result of getbits(). It took a little clever thought, but it works. :) + */ + +unsigned getbits(unsigned x, int p, int n) { + return (x >> (p + 1 - n)) & ~(~0 << n); +} + +unsigned setbits(unsigned x, unsigned p, unsigned n, unsigned y) { + /* ~0 << n sets the rightmost n fields to zero. Using & on that clears them + * from the end of y. Then using | on the result of getbits() sets them to + * the end of y, where they belong. */ + return y & (~0 << n) | getbits(x, p, n); +} + +int main() { + printf("With x as 74, p as 5, n as 4, and y as 0, setbits returns %d\n", setbits(74, 5, 4, 0)); + printf("1[0010]00 to 000[0000], creating 0000010\n\n"); + printf("With x as 90, p as 3, n as 2, and y as 20, setbits returns %d\n", setbits(90, 3, 2, 20)); + printf("101[10]10 to 00101[00], creating 0010110\n\n"); + printf("With x as 256, p as 8, n as 1, and y as 14, setbits returns %d\n", setbits(256, 8, 1, 14)); + printf("[1]00000000 to 00000111[0], creating 000001111\n"); + return 0; +} -- cgit v1.2.3-54-g00ecf msg=1&follow=1'>Expand)AuthorFilesLines 2013-07-15Solve Exercise 4-12: Recursive itoa()zlg1-0/+72