使用strsep在C中解析字符串（替代方法）

Question

I want to parse a string, and I use strsep function:

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

int main()
{

  char str[] = "Marco:Q:2F7PKC";
  char *token1, *token2, *token3;
  char *r = malloc(30);

  strcpy(r, str);

  token1 = strsep(&r, ":");
  token2 = strsep(&r, ":");
  token3 = strsep(&r, ":");

  printf("tok1 = %s\n", token1);
  printf("tok2 = %s\n", token2);
  printf("tok3 = %s\n", token3);

  free(r);

  return 0;

}

The function do its job well, but If I launch valgrind , the allocated string char * r does not freed correctly ( definitely lost: 30 bytes in 1 blocks ).

I'd like to know why and if there are alternative way to do the same thing, maybe without call strsep .

I call valgrind with valgrind --tool=memcheck --leak-check=full --show-reachable=yes ./a.out

Answer 1

strsep overwrites the target of its first (pointer-to-pointer) argument, so you lose the pointer to the malloc 'd buffer's base. In fact, if you were do put a printf("%p\\n", r); just before the free , you'd find out that you're freeing a null pointer, which has no effect.

The easy solution is to introduce an additional variable to keep that pointer around and free it when you're done. Idiomatic usage would be

char *r = strdup("Marco:Q:3F7PKC");
// check for errors

char *tok = r, *end = r;
while (tok != NULL) {
    strsep(&end, ":");
    puts(tok);
    tok = end;
}

free(r);

Answer 2

I would like to a bit simplify a good reply from Fred Foo:

char *end, *r, *tok;

r = end = strdup("Marco:Q:3F7PKC");
assert(end != NULL);

while ((tok = strsep(&end, ":")) != NULL) {
    printf("%s\n", tok);
}

free(r);

It gives the same result. But it is worth to say that strsep(3) stores next value after delimiter into end variable and returns current value (into tok variable).

Answer 3

The strsep function updates its first argument (so it points right after the token it found). You need to store the value returned by malloc in a separate variable and free this variable.