C 编程在随机字符串中查找所有单词

Question

Say I have a very random string such as:

"%^&%thank*(^  ^&* you&*^^guys"

What is the most efficient way to find all the words in the string? Without checking the string character by character?

Here I wrote how I would have done this due to request

int length(char *c) {
    int n = 0;

    while(*(c+n)){
        n++;
    }
    return n;
}

int main(int argc, char *argv[]) { 
    int n;
    int m=0;
    int count=1;

    if(argv[1]==NULL) {
        printf("%s","error" );
    }

    while(argv[count]!=NULL){
        n=length(argv[count]);
        while(m!=n){
            if('a'<argv[count][m]<'z'){
                //do stuff
            }
        }
        count++;
    }

    return 0;
}

Answer 1

You can just use strtok(3) to parse the string at multiple delimiters. In terms of making this work for random strings, you might need to have a collection of all the possible delimiters that could occur. Here is a very basic example of using strtok() :

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int main(void) {
    char str[] = "%^&%thank*(^  ^&* you&*^^guys";
    const char *delim = "%^*(& ";

    char *word = strtok(str, delim);
    while (word != NULL) {
        printf("%s\n", word);
        word = strtok(NULL, delim);
    }

    return 0;
}

UPDATE:

Here is a more useful method, which collects the delimiters from str :

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <ctype.h>

#define MAXCHAR 256

int main(void) {
    char str[] = "%^&%thank*(^  ^&* you&*^^guys";
    int count[MAXCHAR] = {0};
    char *word;
    unsigned char curr;
    size_t charcount = 0, numbytes = strlen(str);
    char delim[numbytes+1];

    for (size_t i = 0; str[i] != '\0'; i++) {
        curr = str[i];
        if (!isalpha(str[i]) && count[curr] == 0) {
            delim[charcount++] = str[i];
            count[curr] = 1;
        }
    }
    delim[charcount] = '\0';

    word = strtok(str, delim);
    while (word != NULL) {
        printf("%s\n", word);
        word = strtok(NULL, delim);
    }

    return 0;
}

This solution uses a hashing O(n) approach for only adding unique delimeters. This is a possible solution, but the approach of going through character by character is more efficient. This is because all you need is a temporary buffer to store the current word being processed, and once a non alpha character is seen, terminate the buffer and start again.