why func() and func(void) are different

Question

I have a function, can write in 2 ways.

void function(void) {
        // operations....
}

and

void function() {
       // operations......
}

Both functions are of same prototype. Why we have to mention void as argument at function definition?

@Pazis: Some? All compilers that strictly adhere to the C standard should. — icktoofay, Oct 06 '13 at 05:39
@JEM: `void func()` is *not* a prototype at all. So your assertion that "both functions are of same prototype" is simply false. — AnT stands with Russia, Oct 06 '13 at 06:01

score 10 · Accepted Answer · answered Oct 06 '13 at 05:39

No, both have different prototypes.

Compile the below programs you will understand.

void function1(void)
{
   printf("In function1\n");
}

void function2()
{
   printf("In function2\n");
}

int main()
{
   function1();
   function2(100); //Won't produce any error
   return 0;
}

Program 2:

 #include <stdio.h>
 void function1(void)
 {
    printf("In function1\n");
 }

 void function2()
 {
    printf("In function2\n");
 }

int main()
{
    function1(100);   //produces an error
    function2();
    return 0;
}

score 3 · Answer 2 · answered Oct 06 '13 at 05:42

The correct way to say "no parameters" in C and C++ is

void function(void);

But when we write

void function();

It means a little different way in C and C++! It means "could take any number of parameters of unknown types", and in C++ it means the same as function(void).

why func() and func(void) are different

2 Answers2