Getting array declaration error - c

Question

I have this array called arr_[6], having an idea of including six strings...but when I declare this array compiler throws error.
#include <stdio.h>
#include <stdlib.h>
int main()
{
int i;
char arr_1[]= {"My_name","your Name", "His Name"};
char *arr_p;
arr_p = malloc(sizeof(char)*6);
arr_p = arr_1;
printf("%s\n",*arr_p);
system("PAUSE");
return 0;
}
Shown errors are as follows:
> main.c: In function `main': main.c:9: error: excess elements in char
> array initializer main.c:9: error: (near initialization for `arr_1')
> main.c:9: error: excess elements in char array initializer main.c:9:
> error: (near initialization for `arr_1')
>
> make.exe: *** [main.o] Error 1
Please help me!

Answer 1

#include <stdio.h>
#include <stdlib.h>
int main()
{
int i;
const char *arr_1[]= {"My_name","your Name", "His Name"}; // has to be an array of <char *>
//arr_p is not necessary
printf("%s\n",*arr_1); // will print the first string, "My_name"
printf("%s\n",arr_1[1]); // will print the second string, "your Name"
printf("%s\n",arr_1[2]); // will print the third string, "His Name"
system("PAUSE");
return 0;
}

Answer 2

I believe what you're looking for is this:
#include <stdio.h>
#include <stdlib.h>
int main()
{
int i;
char* arr_1[]= {"My_name","your Name", "His Name", NULL};
char** arr_p;
arr_p = arr_1;
i = 0;
while (arr_p[i] != NULL)
{
printf("%s\n",(arr_p[i]));
++i;
}
system("PAUSE");
return 0;
}
These are the list of changes I've made:
Use char* arr_1[] to declare an array of strings, since each string is an array of characters.
If you need a pointer to a char*, you need to declare the pointer to be of datatype char**
Used NULL as the last element in the array, so that you know when you've reached the end of the array of strings.
Use a while loop to iterate over all the strings.