Question

Why this code is not running? Why str1 is not assigned to str2 ?? I know i have an option of using strcpy but i wish to know the reason why this is not working??

#include<stdio.h>
int main()
{
char str1[]="hello";
char str2[10];
str2=str1;
printf("%s",str2);
return 0;
}

Whereas if I use pointers than it works like here..

#include<stdio.h>
int main()
(
char *s="good morning";
char *q;
q=s;
while(*q!='\0')
{
printf("%c",*q);
q++;
}
return 0;
}

This works.
Now the string has been copied via pointers so why such difference??

Answer 1

You need to do

strcpy (str2, str1)

you need to copy each element of the str1 to str2 character by character.

What you are doing is to attempt assign the address of the string in str1 assign to the array var str2 which is not allowed. Static arrays couldn’t be used to be assigned values.

Even if you had char *str2; and then str2 = str1 although your code would work, but in such a case the string is not copied, instead the address of the string in str1 is copied in str2 so now dereferencing str2 will point to the string

Also note that when you copy string into a char *str2 always allocate enough memory before copy.
one possibility is:

str2 = malloc (sizeof (char) * (strlen (str1) + 1));
strcpy (str2, str1);