segmentation fault with copying string in c

Question

The following code shows a segmentation error. How to solve the problem? What is the problem with the code?

#include <stdio.h>

void stcp (char *, char *);

int
main ()
{
  char *s = "This is first string";
  char *t = "string to be copied";
  stcp (s, t);
  printf ("%s", s);
  getch ();
}

void
stcp (char *s, char *t)
{
  while ((*s++ = *t++) != '\0');

}

possible duplicate of [Why do I get a segmentation fault when writing to a string?](http://stackoverflow.com/questions/164194/why-do-i-get-a-segmentation-fault-when-writing-to-a-string) — DCoder, May 26 '12 at 20:29

score 1 · Accepted Answer · answered May 26 '12 at 20:30

1

A string literal is constby default. To make it un-const, you must make it an array:

char s[] = "this is my string";
char t[] = "another string";

answered May 26 '12 at 20:30

Richard J. Ross III

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Technically, a string literal isn't `const` (6.4.5(6)), it's just undefined behaviour if one tries to modify it. – Daniel Fischer May 26 '12 at 20:42

score 0 · Answer 2 · answered May 26 '12 at 21:06

#include <stdio.h>

void stcp (char *s, char *t);

int main (void)
{
  int i;
  char s[] = "This is first string";
  char t[] = "string to be copied        ";
  stcp (s, t);
  printf ("%s\n", s);
  printf ("%s\n", t);
  //getch ();
  return 0;
}

void stcp (char *s, char *t)
{
  int i;
  for (i=0;  (s[i]  != '\0')  &&  (t[i] != '\0') ;i++) {
    printf("%c  %c\n",s[i],t[i]);
    s[i] = t[i];
  }
  s[i] ='\0';
}

segmentation fault with copying string in c

2 Answers2