未指定的参数个数和传递给C函数的变量的个数之间有什么区别？

Question

#include <stdio.h>
#define L 11

int oo1(){
printf("oooooo\n");
}

int oo2(int C ,...)
{
 //some code here
 printf("in oo2\n");
}
int main(void){
 oo1();
 oo1(1,4,7,"f");
 oo2(2);
 oo2(2,5,7,3,11);
 printf("%d\n",L);
}

OUTPUT :

oooooo

oooooo

in oo2

in oo2

11

This code has 2 functions oo1 and oo2 . oo2 is a function that accepts variable number of arguments . However , oo1 accepts variable or any number of arguments too as it seems . What is the difference between these two and how do they work ? Any link or reference would be helpful too if this seems too obvious . Thanks .

Answer 1

The difference is that it's unspecified what happens for oo1 .

oo1 accepts variable or any number of arguments too as it seems

The C standard does not guarantee this. In fact, all the C standard says on this regard is that a function definition with an empty argument list is compatible with a function declaration with an argument list of void (ISO/IEC 9899 6.9.1, footnote 162).

That is, the function definition

int oo1() {…}

Is compatible with the prototype declaration

int oo1(void);

In other words, the only safe thing you may do is call oo1 without arguments. But the C compiler is not obliged to diagnose violations of this. That said, a modern C compiler will warn:

warning: too many arguments in call to 'oo1'

Heed this warning. You mustn't call oo1 with any arguments.