How to assign an array of characters to a string literal

Question

I have an array of char and I'm trying to have a string literal with the same chars in the array.

I tried strcpy, and try =, and I tried what I did in the following code. But it doesn't seem to work or I'm understanding something.

char s1[10]="Youssef";
char *s2
while(*s2!='\0')
    *s2++=*s1++;
printf("%s",s2);

Process doesn't return.

Answer 1

String literals are read only.

In any case, what you are trying to do seems you are confused.

A string literal: char *sl = "string literal";

An uninitialized char pointer: char *s2;

In order to do the copy you like, you first need to allocate memory for the string.

Moreover, you cannot do pointer arithmetics with an array. Arrays and pointers are not the same thing!

Furthermore, you should remember the origin of s2 pointer, since after incrementing it until the copy is complete, you would then need to reset the pointer.. Exercise: Think what would happen if you did the copy in a function (preferably named mystrcpy`)...

Complete example:

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

int main(void) {
  char s1[10]="Youssef";
  char *s2 = malloc(sizeof(char) * (strlen(s1) + 1)); // +1 for the NULL-terminator
  int i = 0;
  char *origin_s2 = s2;
  while(s1[i] != '\0')
    *s2++ = s1[i++];
  *s2 = '\0';
  s2 = origin_s2;
  printf("%s\n", s2);
  return 0;
}

Output:

Youssef

PS: It is highly recommended to check if the dynamic allocation of the memory was successful (check if return value of malloc() is not NULL).