string return

>minimalist< · 01-07-2006, 08:53 PM

forgive me for my ignorance, but what is wrong with the following code?

Code:

#include <stdio.h>

int main()
{
char *str;
str = getstring();
printf("%s\n", str);
return 0;
}

char* getstring()
{
char *string;
strcpy(string,"text");
return string;
}

paulsm4 · 01-07-2006, 09:00 PM

The problem is simply that you didn't allocate any space for "string" - you just declared a pointer. Here's one possible solution:

Code:

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

char * getstring (char *buffer, int buflen);

int
main()
{
  char *s = malloc (10);
  printf("%s\n",
    getstring(s, 10));
  free (s);
  return 0;
}

char*
getstring(char *buffer, int buflen)
{
  strncpy(buffer,"text",buflen);
  return buffer;
}

You certainly don't need the "malloc()" - you could just as easily have declared "char s[10]" and accomplished the same thing (and eliminated the call to "free()" at the same time).

Similarly, strncpy() copy is arguably "safer" than "strcpy()" in preventing buffer overruns - but unfortunately, it, too, has its own problems.

Bottom line: allocate string space, and your program is fine!

'Hope that helps ..

>minimalist< · 01-07-2006, 09:06 PM

Thank you, and yes, that helps quite a bit.

vetrimani · 01-08-2006, 04:49 AM

#include <stdio.h>

int main()
{
char *str;
str = getstring(4);
printf("%s\n", str);
return 0;
}

char* getstring(int len)
{
char *string = malloc(len+1); /*One extra for null char at end*/
strcpy(string,"text");
return string;
}