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!

Yes, it is the same. If there are less number of initializers than the elements in the array, then the remaining elements will be initialized as if the objects having static storage duration, (i.e., with 0).

So,

char buf[5]={0,};

is equivalent to

 char buf[5]={0,0,0,0,0};

Related Reading : From the C11 standard document, chapter 6.7.9, initalization,

If there are fewer initializers in a brace-enclosed list than there are elements or members of an aggregate, or fewer characters in a string literal used to initialize an array of known size than there are elements in the array, the remainder of the aggregate shall be initialized implicitly the same as objects that have static storage duration.