how to get substring in c with strchr

Question

I'm trying to fetch part of a string.
I have the following code:

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

char mystring[]="The quick brown fox jumps over the lazy dog";
char word1[]="The";
char * posb,pose;
char * word2;
int b,e,n;

n=memcmp(mystring, word1, sizeof(word1)-1);
if (n==0) printf("the first word is found!\n");
posb=strchr(mystring,' ');   // posb will be a pointer to the first character
b=posb-mystring+1;
printf("posb -> %d\n",posb);
printf("b -> %d\n",b);
pose=strchr(mystring+b+1,' ');   // pose will be a pointer to the second character
printf("calc e\n");
e=pose-sizeof(mystring)+1;
printf("pose -> %s\n",pose);
printf("e -> %d\n",e);
word2 = (char*) malloc (sizeof(char)*(e-b));
memcpy(word2, posb, sizeof(word2));
printf("%s:%i:%i:%i\n",word2, b, e, e-b);
free (word2);

The problem is getting the second word and store it in word2. For this I try using strchr to locate the spaces. But the second time I use strchr I need an offset to find the second space. I tried the following:

pose=strchr(mystring+b+1,' ');
pose=strchr(&mystring[b+1],' ');

The variables b and e should contain the positions of the space character in mystring. word2 should contain quick eventually.
Another solution would be to 'walk through' the string using a loop, but that would be cheating the strchr function.

`pose` is **not** a pointer, it *is* a `char`. If you want `pose` to be a pointer, you need to write `char *posb, *pose;` — user3386109, Dec 23 '15 at 11:13
Hi, welcome to stackoverflow. Please, post a minimal compilable version of the code: http://stackoverflow.com/help/mcve — terence hill, Dec 23 '15 at 11:22

score 1 · Answer 1 · answered Dec 23 '15 at 11:58

In order to get the second word, check this

char mystring[] = "The quick brown fox jumps over the lazy dog";
char word1[] = "The";
char * posb, pose;
char * word2;
int b, e, n;

n = memcmp(mystring, word1, sizeof(word1)-1);
if (n == 0) printf("the first word is found!\n");
posb = strchr(mystring, ' ');   // posb will be a pointer to the first character

b = posb - mystring + 1;

printf("posb -> %d\n", posb);
printf("b -> %d\n", b);
posb = strchr(posb + 1, ' ');   // pose will be a pointer to the second character
printf("calc e\n");

e = posb - mystring + 1;
printf("e -> %d\n", e);

word2 = (char*)malloc(sizeof(char)*(e - b));
memcpy(word2, mystring + b, e - b);
word2[e-b-1] = '\0';
printf("%s:%i:%i:%i\n", word2, b, e, e - b);
free(word2);

David C. Rankin · Answer 2 · 2016-01-30T21:39:03.203

When searching a line for an occurrence of a word, you need a pointer to advance over each character in the line, a search term, comparison of the current character with the first character in search term, further comparison if there is a match, and then a set of delimiters to limit the comparisons as required to, for example, if searching for 'the' only match 'the' and not 'they', 'them', 'then', 'there', etc...

You can use basic arithmetic to your advantage to insure you don't attempt to check lines shorter then the search term length, etc.. A short example (changing the initial 'The' to 'the' in your line for example purposes), you can do something similar to the following. Note, there are only basic optimization used int the search and it isn't intended to be an exhaustive example:

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

int main (int argc, char **argv)
{
    char line[] = "the quick brown fox jumps over the lazy dog.";
    char *p = line;
    char *sterm = argc > 1 ? argv[1] : "the";
    char *delim = " \t\n\'\".";
    size_t llen = strlen (line);
    size_t count = 0, slen = strlen (sterm);

    printf ("\n searching line for '%s'\n\n", sterm);

    for (;p < (line + llen - slen + 1); p++) {

        if (*p != *sterm)
            continue;   /* char != first char in sterm  */
        if (p > line && !strchr (delim, *(p - 1)))
            continue;   /* prior char is not a delim    */
        if (!strchr (delim, *(p + slen)))
            continue;   /* next char is not a delim     */
        if (strncmp (p, sterm, slen))
            continue;   /* chars don't match sterm      */

        printf ("   match %2zu. '%s' at location %zu\n",
                ++count, sterm, p - line);
    }

    printf ("\n total occurrences of '%s' in line : %zu\n\n",
            sterm, count);

    return 0;
}

Use/Output

$ ./bin/searchline

 searching line for 'the'

   match  1. 'the' at location 0
   match  2. 'the' at location 31

 total occurrences of 'the' in line : 2

$ ./bin/searchline fox

 searching line for 'fox'

   match  1. 'fox' at location 16

 total occurrences of 'fox' in line : 1

Note the use of strchr with the delimiter string delim using the current character as the char. Look over the example and let me know if you have any questions. Your exact goal was a bit unclear from your question, so if I missed your intent, let me know.

Using Array Indexes Instead of Pointer

If you are more comfortable using character array indexes instead of pointers, you can always remove the pointer p and adjust the loop logic accordingly:

    for (i = 0; i < (llen - slen + 1); i++) {

        if (line[i] != *sterm)
            continue;   /* char != first char in sterm  */
        if (i && !strchr (delim, line[i-1]))
            continue;   /* prior char is not a delim    */
        if (!strchr (delim, line[i+slen]))
            continue;   /* next char is not a delim     */
        if (strncmp (&line[i], sterm, slen))
            continue;   /* chars don't match sterm      */

        printf ("   match %2zu. '%s' at location %zu\n",
                ++count, sterm, &line[i] - line);
    }

how to get substring in c with strchr

2 Answers2