Just tell me why this happens.

**lydiapeter** · 08-17-2005

Code:

#include<stdio.h>
#include<conio.h>
void get()
{
printf("Hello\n");
}
main()
{
clrscr();
get(10,20);
getch();
return 0;
}

When a call is made to the get fn with (any number of) auguments it works. Why is that happening? Help me out please.

**Brian** · 08-17-2005

Because you defined get with an undefined number of arguments. Try void get(void).

**lydiapeter** · 08-17-2005

The same gives errors in c++.Is that the default type of augument in c is void and c++ is not void?

**hk_mp5kpdw** · 08-17-2005

Originally Posted by lydiapeter

The same gives errors in c++.Is that the default type of augument in c is void and c++ is not void?

I think so (I could be wrong)... specifying no arguments when you create your function "get" as you have done for C means that the function accepts an undefined number of arguments (as already mentioned by Brian). In C++ however, specifying no arguments means that the function does not accept any arguments at all and you will get errors if you try to call the function with any arguments.

**orbitz** · 08-17-2005

C has declarations and prototypes. They are seperate things. C++ has declarations and prototypes too, however they are the same thing.

In C: T fn() is a declaration. It provides no information on the parameters and that basicaly means anything goes. T fn(void) in C is a declaration, and a prototype. All prototypes are declarations, not all declarations are prototypes. This is a prototype because it provides type information for the parameters (in this case no parameters).

**dwks** · 08-19-2005

With void get(), you are neglecting to provide information about the arguments, so the compiler accepts any number. void get(void) tells the compiler that this function takes no arguments.

On the same line, you should declare main like this:

Code:

int main(void) {
    /* ... */
    return 0;
}

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Just tell me why this happens.