how to best achieve string to number mapping in a c program - c

Question

I have a definite set of strings and its corresponding numbers:
kill -> 1
live -> 2
half_kill -> 3
dont_live -> 4
List is of 30 such strings and their number mapping.
If user enters "kill", I need to return 1 and if he enters "dont_live" I need to return 4.
How should I achieve this in c program? I am looking for an efficient solution because this operation needs to be done 100s of times.
should I put them in #define in my .h file?
Thanks in advance.

Answer 1

Sort your table, and use the standard library function bsearch to perform a binary search.
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
struct entry {
char *str;
int n;
};
/* sorted according to str */
struct entry dict[] = {
"dont_live", 4,
"half_kill", 3,
"kill", 1,
"live", 2,
};
int compare(const void *s1, const void *s2)
{
const struct entry *e1 = s1;
const struct entry *e2 = s2;
return strcmp(e1->str, e2->str);
}
int
main (int argc, char *argv[])
{
struct entry *result, key = {argv[1]};
result = bsearch(&key, dict, sizeof(dict)/sizeof(dict[0]),
sizeof dict[0], compare);
if (result)
printf("%d\n", result->n);
return 0;
}
Here's what you get when you run the program.
$ ./a.out kill
1
$ ./a.out half_kill
3
$ ./a.out foo
<no output>
PS: I reused portions of sidyll's program. My answer should now be CC BY-SA compliant :p

Answer 2

A possible solution:
#include <stdio.h>
#include <string.h>
struct entry {
char *str;
int n;
};
struct entry dict[] = {
"kill", 1,
"live", 2,
"half_kill", 3,
"dont_live", 4,
0,0
};
int
number_for_key(char *key)
{
int i = 0;
char *name = dict[i].str;
while (name) {
if (strcmp(name, key) == 0)
return dict[i].n;
name = dict[++i].str;
}
return 0;
}
int
main (int argc, char *argv[])
{
printf("enter your keyword: ");
char s[100]; scanf("%s", s);
printf("the number is: %d\n", number_for_key(s));
return 0;
}

Answer 3

Here's one approach:
int get_index(char *s)
{
static const char mapping[] = "\1.kill\2.live\3.half_kill\4.dont_live";
char buf[sizeof mapping];
const char *p;
snprintf(buf, sizeof buf, ".%s", s);
p = strstr(mapping, buf);
return p ? p[-1] : 0;
}
The . mess is to work around kill being a substring of half_kill. Without that issue you could simply search for the string directly.

Answer 4

If it is a very short list of strings then a simple block of ifs will be more than sufficient
if (0 == strcmp(value, "kill")) {
return 1;
}
if (0 == strcmp(value, "live")) {
return 2;
}
...
If the number approach 10 I would begin to profile my application though and consider a map style structure.

Answer 5

if you have a fixed set of strimgs, you have two options: generate a perfect hashing function (check gperf or cmph) or create a trie so that you never have to check charcters more than once.
Compilers usually use perfect hashes to recognize a language keyword, in your case I would probably go with the trie, it should be the fastest way (but nothing beats direct measurement!)

Answer 6

Is it really a bottleneck? You should worry about efficiency only if the simple solution proves to be too slow.
Having said that, possible speed improvements are checking the lengths first:
If it's 4 characters then it could be "kill" or "live"
If it's 9 characters then it could be "half_kill" or "dont_live"
or checking the first character in a switch statement:
switch (string[0]) {
case 'k':
if (strcmp(string, "kill") == 0)
return 1;
return 0;
case 'l':
...
default:
return 0;
}

Answer 7

Use hashmap/ hashtable i think this would be the best solution.

Answer 8

Can you use an Enumunerator?
int main(void) {
enum outcome { kill=1, live, half_kill, dont_live };
printf("%i\n", kill); //1
printf("%i\n", dont_live); //4
printf("%i\n", half_kill); //3
printf("%i\n", live); //2
return 0;
}

Answer 9

Create a list of const values:
const int kill = 1;
const int live = 2;
const int half_kill = 3;
etc