What is the meaning of Bus: error 10 in C

For one, you can't modify string literals. It's undefined behavior.

To fix that you can make str a local array:

char str[] = "First string";

Now, you will have a second problem, is that str isn't large enough to hold str2. So you will need to increase the length of it. Otherwise, you will overrun str - which is also undefined behavior.

To get around this second problem, you either need to make str at least as long as str2. Or allocate it dynamically:

char *str2 = "Second string";
char *str = malloc(strlen(str2) + 1);  //  Allocate memory
//  Maybe check for NULL.

strcpy(str, str2);

//  Always remember to free it.
free(str);

There are other more elegant ways to do this involving VLAs (in C99) and stack allocation, but I won't go into those as their use is somewhat questionable.

As @SangeethSaravanaraj pointed out in the comments, everyone missed the #import. It should be #include:

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

There is no space allocated for the strings. use array (or) pointers with malloc() and free()

Other than that

#import <stdio.h>
#import <string.h>

should be

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

NOTE:

• anything that is malloc()ed must be free()'ed
• you need to allocate n + 1 bytes for a string which is of length n (the last byte is for \0)

Please you the following code as a reference

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

int main(int argc, char *argv[])
{
    //char *str1 = "First string";
    char *str1 = "First string is a big string";
    char *str2 = NULL;

    if ((str2 = (char *) malloc(sizeof(char) * strlen(str1) + 1)) == NULL) {
        printf("unable to allocate memory \n");
        return -1; 
    }   

    strcpy(str2, str1);

    printf("str1 : %s \n", str1);
    printf("str2 : %s \n", str2);

    free(str2);
    return 0;
}