How can i tokenize a sentence starting with spaces in C

Question

I am trying to tokenize a sentence that starts with spaces but I get segmentation fault. I am trying to skip the spaces and store the words. Is there an alternative built-in function to achieve this?

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

[...]

    char *buff = "  push me", *token;
    
    token = strtok(buff, " \n");
    while (token) {
        printf("%s", token);
        token = strtok(NULL, " \n");
    }

[...]

Answer 1

change to this

char buff[100], *token;

strcpy(buff,"  push me");

Before you were pointing to a string constant with the buff variable. The compiler will put this in a segment that is read only. If you try to write to that segment you get a segmentation fault. So in the code above we allocate space in read/write memory to store the string and then copy it in.

Answer 2

buff points to a string literal that usually cannot be modified.
strspn and strcspn can be used to parse the sub-strings.

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

int main ( void) {
    char *buff = "  push me";
    char *token = buff;
    size_t span = 0;

    if ( ! buff) {
        return 1;
    }

    while ( *token) {
        token += strspn ( token, " \r\n\t\f\v"); // count past whitespace
        span = strcspn ( token, " \r\n\t\f\v"); // count non-whitespace
        if ( span) {
            printf ( "%.*s\n", (int)span, token); // precision field to print token
        }
        token += span; // advance token pointer
    }
}

Answer 3

strtok() , like others, are "library functions", not "built-in".

Very intelligent people wrote these functions expecting they would be used in clean, elegant and terse ways. I believe this is what you want:

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

int main() {
    char buf[] = "  push me"; // don't point at a "string literal'

    for( char *tok = buf; (tok = strtok( tok, " \n") ) != NULL; tok = NULL )
        printf( "%s\n", tok ); // added '\n' for clarity

    return 0;
}

Output

push
me