What does an empty parameter list mean?

Question

The book that i am currently reading states that when you declare a function in c that accepts no arguments, but you do not use the keyword void "the function call can pass any arguments it wants". so i attempted this.

int number();

int main(void)
{
    int x =1;
    printf("%d",number(x));
}

int number()
{
    return x;
}

but it did not compile??? how does this work?

Answer 1

This is an obsolescent feature¹ from before C was standardized, decades ago.

Never use it.

In ancient C, a decade before you were born, you could declare a function with no parameters. When you called it, the compiler would promote the parameters to default types and pass them to the function. Essentially, the compiler figured out the parameter declarations from the arguments in the call.

The function should still be defined with parameters, and they should match the way the function is called.

---

¹ “Feature” is the wrong word. It was the way things were done at the time, since better ways were not yet widely developed. In a new programming language, this characteristic would be considered a deficiency.

Answer 2

In support of Eric Postpischil's answer I would like to quote from the C11 standard (6.11 Future language directions)

6.11.6 Function declarators

The use of function declarators with empty parentheses (not prototype-format parameter type declarators) is an obsolescent feature.

and

6.11.7 Function definitions

The use of function definitions with separate parameter identifier and declaration lists (not prototype-format parameter type and identifier declarators) is an obsolescent feature.

Emphasis is mine :)

Answer 3

The issue here is that your function number has no knowledge of x.
That is, when you have the function number return x, it doesn't have any x in scope to return, and so this is a compile error. If, instead, you wrote:

int number() {
    return 5;
}

it would compile fine.

Answer 4

Well, if you want pass something to it without limit , you may need a a variable argument list.Here is an example:

  #include <stdio.h>
  #include <stdarg.h>
  int number(int , ... );

  int main(void)
  {
       int x =1;
       printf("%d",number(1,x));
  }
  int number(int n, ... )
  {
          va_list         ap;
          va_start(ap,n);
          int x = va_arg(ap,int);
          va_end(ap);
          return x;
  }

Or , if you just want to pass x but not use it.

 #include <stdio.h>
  int number();

  int main(void)
  {
      int x =1;
      printf("%d",number(x));
  }

  int number()
  {
     int x = 1;
     return x;
  }

It can compile and work. When you declare int number (); in C , you mean this function can be given unspecial type . however , you can't use it.

Answer 5

Well I edited the whole answer after finding the answer.

What you want to do is number(void) not main(void). That's what you wanted and that will print any variable as a integer. If you pass the char "F" int number(void) will return the number 70, the int form of "F". In code:

int number(void e);

void main(){
    char C = 'F';
    printf("%d",number(C));
};

int number(void e){
    return e;
};

Note:

You must always pass an argument if you want a variable that's out of scope.