How to initialize a char array using a char pointer in C

1) pointer string2 == pointer string1

change in value of either will change the other

From poster poida

char string1[] = "hahahahaha";
char* string2 = string1;

2) Make a Copy

char string1[] = "hahahahaha";
char string2[11]; /* allocate sufficient memory plus null character */
strcpy(string2, string1);

change in value of one of them will not change the other

What you write like this:

char str[] =  "hello";

... actually becomes this:

char str[] =  {'h', 'e', 'l', 'l', 'o'};

Here we are implicitly invoking something called the initializer.
Initializer is responsible for making the character array, in the above scenario.
Initializer does this, behind the scene:

char str[5];

str[0] = 'h';
str[1] = 'e';
str[2] = 'l';
str[3] = 'l';
str[4] = 'o';

C is a very low level language. Your statement:

char str[] = another_str;

doesn't make sense to C. It is not possible to assign an entire array, to another in C. You have to copy letter by letter, either manually or using the strcpy() function. In the above statement, the initializer does not know the length of the another_str array variable. If you hard code the string instead of putting another_str, then it will work.

Some other languages might allow to do such things... but you can't expect a manual car to switch gears automatically. You are in charge of it.