Segmentation fault in c (C90), Whats the problem? - c

Question

Heres my main.c:
int main() {
char *x = "add r3,r5";
char *t;
char **end;
t = getFirstTok(x,end);
printf("%s",t);
}
And the function getFirstTok:
/* getFirstTok function returns a pointer to the start of the first token. */
/* Also makes *endOfTok (if it's not NULL) to point at the last char after the token. */
char *getFirstTok(char *str, char **endOfTok)
{
char *tokStart = str;
char *tokEnd = NULL;
/* Trim the start */
trimLeftStr(&tokStart);
/* Find the end of the first word */
tokEnd = tokStart;
while (*tokEnd != '\0' && !isspace(*tokEnd))
{
tokEnd++;
}
/* Add \0 at the end if needed */
if (*tokEnd != '\0')
{
*tokEnd = '\0';
tokEnd++;
}
/* Make *endOfTok (if it's not NULL) to point at the last char after the token */
if (endOfTok)
{
*endOfTok = tokEnd;
}
return tokStart;
}
Why do i get segmentation fault running this main program?
I'm programming a two pass aseembler and i need a function that get parse a string by a delimiter, In this case a white space. Is it better to use strtok instead for this purpose?
I need a command pasrer - So that it will extract "add", an operand parser (By , delimiter), To extract "r3" and "r5". I wanted to check if this getFirstTok function is good for this purpose but when i try to run it i get a segmentation fault:
Process finished with exit code 139 (interrupted by signal 11: SIGSEGV)
Thank you.

Answer 1

As pointed out in the comments, string literals are read-only, as they are baked into the compiled program. If you don't want to go with the suggested solution of making your "source program" a stack-allocated array of characters (char x[] = "add r3,r5"), you can use a function like strdup(3) to make a readable/writable copy like so:
#include <string.h>
[...]
char *rw_code = strdup(x);
t = getFirstTok(rw_code, end);
printf("%s", t);
free(rw_code); /* NOTE: invalidates _all_ references pointing at it! */
[...]
And as a little aside, I always make string literals constant const char *lit = "...", as the compiler will usually warn me if I attempt to write to them later on.