scanf without additional arguments in C

Question

Is it allowed to use scanf(" ") without additional arguments to ignore initial whitespaces?
I'm using getchar() to read the chars of a word, and I want to ignore the whitespaces before the word (whitespaces after are used to check the end of the word).
The code is the following, is it correct?

char *read_word() {
    int size = 2;
    int char_count = 0;
    char *s;
    char ch;

    s = mem_alloc(size);

    scanf(" ");

    while ((ch = getchar()) != EOF) {
        if (char_count >= size) {
            s = mem_realloc(s, size++);
        }

        if (ch == ' ' || ch == '\n') {
            s[char_count] = '\0';
            break;
        }

        s[char_count++] = ch;
    }

    return s;
}

Answer 1

From the definition of the scanf() function (*), emphasis mine:

The format is composed of zero or more directives: one or more white-space characters, an ordinary multibyte character (neither % nor a white-space character), or a conversion specification.

[...]

A directive composed of white-space character(s) is executed by reading input up to the first non-white-space character (which remains unread), or until no more characters can be read.

So scanf( " " ); is perfectly valid.

---

(*): ISO/IEC 9899:1999, 7.19.6.2 The fscanf function, section 3 and 5.

The other *scanf functions are defined in terms of this section.

Answer 2

To add to the other answers, all of the following are valid:

scanf(" ");      // skip over whitespace
scanf("xyz");    // skip over the longest leading substring of "xyz" if present
                 // i.e. "xyz" or "xy" or "x"
scanf(" %*s ");  // skip over the first string and whitespace around it