[r.stiltskin]

You can't modify a string literal. In fact the correct syntax (in C++) for the first 2 lines you posted would be

        const char *s1 = "Brockport";
        const char *t1 = "SUNY";

In C, it's just char *s1 == ... as you wrote it, but they are still treated as constants.

If you want to be able to modify them, you have to either define them as arrays:

        char s1[] = "Brockport";
        char t1[] = "SUNY";

and then you can still manipulate them using pointer notation as you are doing in your strcpy function.

Alternatively, you can declare the two pointers and use malloc to allocate memory to them, but then you have to copy the strings char-by-char into that memory -- you can't just initialize the dynamic memory with "=".

This post has been edited by r.stiltskin: 22 April 2010 - 08:56 AM

[r.stiltskin]

Exactly what does your assignment say regarding the strcpy function?

[eZACKe]

1. int my_strcpy(char s[], char t[]) – this function overwrites string s by a copy of characters in string t. The copy should succeed only if length of s is at least as great as t and the function should return a 1. If s is shorter than t then there should be no change to s and the function should return a -1

That's all it says. We then have to write in in Pointer notation.

This post has been edited by eZACKe: 22 April 2010 - 09:08 AM

[r.stiltskin]

s2 (the "source" string) can be a literal, but s1 (the "destination" string) MUST be a char array. There's no getting around that, so if you think that you are required to use those two string literals you should contact your professor.

If you have to convince him/her (or yourself) about this, here is a 3-line program that will prove it:

int main() {
  char *s1 = "hello";
  *s1 = 'w';
}

Try running that -- you can't, it is guaranteed to crash.

[eZACKe]

Thank you very much.

[r.stiltskin]

What you can do is use a char array as a temporary container for the destination string, and after the strcpy function is finished, you can assign the s1 pointer to point to the temporary array, like this:

#include <stdio.h>
#include "your_string_header.h"

int main() {
  char *s1 = "Brockport";
  char *t1 = "SUNY";

  char s3[strlen(s1) + 1]; // do you know why I made this "+ 1"?
  strcpy( s3, t1 );
  
  s1 = s3;

  printf("%s\n", s1);
}

or if your compiler won't allow that (because of the way I defined the array) since you know the length of s1 you can do this:

#include <stdio.h>
#include "your_string_header.h"

int main() {
  char *s1 = "Brockport";
  char *t1 = "SUNY";

  char s3[10]; // do you know why I made this array length 10?
  strcpy( s3, t1 );
  
  s1 = s3;

  printf("%s\n", s1);
}

This post has been edited by r.stiltskin: 22 April 2010 - 10:33 AM

[NickDMax]

Topic split. The original conversation would continue here