变量声明的函数

Question

I'm reading the C Programming Language (chapter 5), and I'm confused by this example:

int n, array[SIZE], getint(int *);

Why is this function call in here like that? Is this just some tricky example and invalid code?

Answer 1

It's not calling the function; it's declaring its prototype. It's equivalent to:

int n;
int array[SIZE];
int getint(int*);

Answer 2

Since the statement began with a type specifier, namely int, then it suggests declaration. Thus what follows is a bunch of comma separated list of identifiers.

n being a single int variable.

array being an array of int.

getint being a function that returns an int and has one parameter that is an int pointer. It is unnamed and that is not important because this is a function declaration/prototype.