1. The first declaration:

    char buf[10] = "";
   

   is equivalent to

    char buf[10] = {0, 0, 0, 0, 0, 0, 0, 0, 0, 0};
   

2. The second declaration:

    char buf[10] = " ";
   

   is equivalent to

    char buf[10] = {' ', 0, 0, 0, 0, 0, 0, 0, 0, 0};
   

3. The third declaration:

    char buf[10] = "a";
   

   is equivalent to

    char buf[10] = {'a', 0, 0, 0, 0, 0, 0, 0, 0, 0};
   

As you can see, no random content: if there are fewer initializers, the remaining of the array is initialized with 0. This the case even if the array is declared inside a function.

It's anyways bad practice to initialie a char array with a string literal.

The author of that comment never really justifies it, and I find the statement puzzling.

In C (and you've tagged this as C), that's pretty much the only way to initialize an array of char with a string value (initialization is different from assignment). You can write either

char string[] = "october";

or

char string[8] = "october";

or

char string[MAX_MONTH_LENGTH] = "october";

In the first case, the size of the array is taken from the size of the initializer. String literals are stored as arrays of char with a terminating 0 byte, so the size of the array is 8 ('o', 'c', 't', 'o', 'b', 'e', 'r', 0). In the second two cases, the size of the array is specified as part of the declaration (8 and MAX_MONTH_LENGTH, whatever that happens to be).

What you cannot do is write something like

char string[];
string = "october";

or

char string[8];
string = "october";

etc. In the first case, the declaration of string is incomplete because no array size has been specified and there's no initializer to take the size from. In both cases, the = won't work because a) an array expression such as string may not be the target of an assignment and b) the = operator isn't defined to copy the contents of one array to another anyway.

By that same token, you can't write

char string[] = foo;

where foo is another array of char. This form of initialization will only work with string literals.

EDIT

I should amend this to say that you can also initialize arrays to hold a string with an array-style initializer, like

char string[] = {'o', 'c', 't', 'o', 'b', 'e', 'r', 0};

or

char string[] = {111, 99, 116, 111, 98, 101, 114, 0}; // assumes ASCII

but it's easier on the eyes to use string literals.

EDIT²

In order to assign the contents of an array outside of a declaration, you would need to use either strcpy/strncpy (for 0-terminated strings) or memcpy (for any other type of array):

if (sizeof string > strlen("october"))
  strcpy(string, "october");

or

strncpy(string, "october", sizeof string); // only copies as many characters as will
                                           // fit in the target buffer; 0 terminator
                                           // may not be copied, but the buffer is
                                           // uselessly completely zeroed if the
                                           // string is shorter!

You can't - in C. In C initializing of global and local static variables are designed such that the compiler can put the values statically into the executable. It can't handle non-constant expressions as initializers. And only in C99, you can use non-constant expression in aggregate initializers - not so in C89!

In your case, since your array is an array containing characters, each element has to be an arithmetic constant expression. Look what it says about those

An arithmetic constant expression shall have arithmetic type and shall only have operands that are integer constants, floating constants, enumeration constants, character constants, and sizeof expressions.

Surely this is not satisfied by your initializer, which uses an operand of pointer type. Surely, the other way is to initialize your array using a string literal, as it explains too

All the expressions in an initializer for an object that has static storage duration shall be constant expressions or string literals.

All quotes are taken out of the C99 TC3 committee draft. So to conclude, what you want to do - using non-constant expression - can't be done with C. You have several options:

• Write your stuff multiple times - one time reversed, and the other time not reversed.
• Change the language - C++ can do that all.
• If you really want to do that stuff, use an array of char const* instead

Here is what i mean by the last option

char const c[] = "ABCD";
char const *f[] = { &c[0], &c[1], &c[2], &c[3] };
char const *g[] = { &c[3], &c[2], &c[1], &c[0] };

That works fine, as an address constant expression is used to initialize the pointers

An address constant is a null pointer, a pointer to an lvalue designating an object of static storage duration, or a pointer to a function designator; it shall be created explicitly using the unary & operator or an integer constant cast to pointer type, or implicitly by the use of an expression of array or function type. The array-subscript [] and member-access . and -> operators, the address & and indirection * unary operators, and pointer casts may be used in the creation of an address constant, but the value of an object shall not be accessed by use of these operators.

You may have luck tweaking your compiler options - another quote:

An implementation may accept other forms of constant expressions.