What does getint (int *) mean?

Question

getint (int *);   

I am not really sure what does getint (int *) mean? Can someone explain this?

Answer 1

This is a function prototype declaring the function getint() . The function takes as pointer to an int as a parameter.

When prototyping a function it is not necessary to specify the parameters' names.

The prototype is missing a return type, which for C then defaults to int . Omiting a return type however is violating the recent C standard, so code doing so could be considered invalid.

An equivalent to

getint(int *);

though would be

int getint(int * pi); 

Answer 2

TL;DR getint (int *); is a forward declaration for a function, in a very bad programming style. Without any explicit return type, it will default to int .

The recommended way to write is is to specify the return type explicitly, like

• if int , like

   int getint(int *); //yes, omitting the identifier name is correct, see note below 

  or,

   int getint(int * outVar); //we can have the name, if we want. 

• if void , like

   void getint(int *); 

  or,

   void getint(int * outVar); 

or any return type you want.

---

NOTE:

Just for further reference, from C11 , chapter §6.7.6.3, Function declarators , (emphasis mine)

If, in the declaration 'T D1' , D1 has the form

 D( parameter-type-list ) 

...

A parameter type list specifies the types of, and may declare identifiers for, the parameters of the function.

So, the identifier name is optional.

Answer 3

The code is not valid C. If it compiles, consider upgrading to a compiler which is not older than 15 years.

History lesson:

In older, obsolete versions of the C standard you were allowed to omit the return type when writing a function declaration, in which case getint(int *); would mean the same thing as int getint(int *); , because if you specified no return type it would default to int . It was however bad practice to do so even back in 1990.