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? — user1233508
– user1233508, Commented May 26, 2012 at 20:29

Richard J. Ross III · Accepted Answer · 2012-05-26 20:30:53Z

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, 2012 at 20:30

Richard J. Ross III

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1 Comment

Daniel Fischer Over a year ago

Technically, a string literal isn't const (6.4.5(6)), it's just undefined behaviour if one tries to modify it.

fhtuft · Accepted Answer · 2012-05-26 21:06:14Z

0

#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';
}

answered May 26, 2012 at 21:06

fhtuft

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