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author | zlg <zlg@zlg.space> | 2013-05-05 00:54:37 -0500 |
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committer | zlg <zlg@zlg.space> | 2013-05-05 00:54:37 -0500 |
commit | e6819292b0d0a45a5d3f74af9ef320e544f42c82 (patch) | |
tree | c980752b577ee065516738a74313060f1a99c786 | |
parent | Add exercise descriptions and answers for ch1 (diff) | |
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Solve Exercise 4-1: strrindex()
Diffstat (limited to '')
-rw-r--r-- | ch4/4-01_strrindex.c | 87 |
1 files changed, 87 insertions, 0 deletions
diff --git a/ch4/4-01_strrindex.c b/ch4/4-01_strrindex.c new file mode 100644 index 0000000..b6400c4 --- /dev/null +++ b/ch4/4-01_strrindex.c @@ -0,0 +1,87 @@ +#include <stdio.h> +#include <string.h> + +/* The C Programming Language: 2nd Edition + * + * Exercise 4-1: Write the function strrindex(s, t), which returns the position + * of the /rightmost/ occurrence of 't' in 's', or -1 if there is none. + * + * Answer: The directions are unclear. Is it asking for the rightmost index of + * the first result from the beginning of the string, or the first index of the + * first occurrence from the end of the string? I will assume the leftmost + * index of the first occurrence from the end of the string, since it seems + * more difficult and relevant. + * + * This exercise was made simpler by remembering the work we did in Chapter 3. + * Specifically, itoa2. We made a reverse(s) function which reverses strings. + * If we reverse our two strings, we can pass them to strindex(), do a little + * math, and get the rightmost equivalent, all without copying and modifying + * another function! This strengthens the concept of writing functions that + * don't need to be fully understood in order to use or expand on their + * functionality. + * + * In this instance, I reversed strindex's behavior without needing to know + * anything beyond the concept of what strings are. + * + * (For the record I understand how strindex works :P) + */ + +#define MAXLINE 1000 + +void reverse(char s[]) { + int c, i, j; + + for (i = 0, j = strlen(s)-1; i < j; i++, j--) { + c = s[i]; + s[i] = s[j]; + s[j] = c; + } +} + +int strindex(char s[], char t[]) { + int i, j, k; + + for (i = 0; s[i] != '\0'; i++) { + for (j = i, k = 0; t[k] != '\0' && s[j] == t[k]; j++, k++) { + } + if (k > 0 && t[k] == '\0') { + return i; + } + } + return -1; +} + +int strrindex(char s[], char t[]) { + /* Instead of starting at the end and going back, let's be super lazy + * and just reverse it, call our buddy strindex, and reverse the + * strings again. + */ + reverse(s); + reverse(t); + /* We add strlen(t) to point to the first index of t when it's + * reversed again */ + int o = strindex(s, t) + strlen(t); + reverse(s); + reverse(t); + + // Be sure to return the correct offset with some math :3 + return (strlen(s) - o); +} + +int main() { + char haystack[MAXLINE] = "charred and burned, ouch! My stomach churned... I hope I'll be okay!"; + char needle[MAXLINE] = "ned"; + int pos = strrindex(haystack, needle); + int i = 0; + + /* This is mostly just to provide a handy visual aid */ + printf("The phrase %s's rightmost occurrence is %d:\n", needle, pos); + printf("%s\n", haystack); + for (i = 0; i < pos; i++) { + putchar(' '); + } + for (i = 0; i < strlen(needle); i++) { + putchar('^'); + } + printf("\n"); +} |