Why is the compiler showing "Call to function with no prototype"?

Question

I am trying to print a string using DOS video memory but when I call function print_Id (); and (*old)(); it shows these warnings and does not run the code but without these functions every thing works fine.

Call to function print_id with no prototype

Call to function with no prototype

I wrote this function at the top but all in vain.

Please review my code below; I am using BORLANDC compiler.

#include<stdio.h>
#include<BIOS.H>
#include<DOS.H>
#include<conio.h>
        int j;

void interrupt (*old)();
void interrupt print_name();
void interrupt print_Id();
char st[80] ={"Bilal Maqsood$"};
char id[20]={"BC110403231$"};
char far *scr=(char far* ) 0xb8000f3C;

int main( )
{
    clrscr();

    old=getvect(0x08);
    setvect(0x08,print_name); //corrected
    return 0;
}


void interrupt print_name(){

  int i=0;
  int j=0;

while(st[i]!='$'){
    *(scr+j)=st[i];
      *(scr+j+1)=0x72;
    i++;
    j+=2;
 }
print_Id ();
}

void interrupt print_Id ( )
{
    int i=0;
  int j=0;

while(id[i]!='$'){
*(scr+j)=id[i];
    *(scr+j+1)=0x17;
  i++;
  j+=2;
 }

(*old)();
}

I did change it to `print_id` on both places but still it show both of above errors — Bilal Maqsood, Nov 22 '15 at 14:42
Then it's not about this code portion maybe. And, is old an external function? Where was it implemented? — Shakil Ahamed, Nov 22 '15 at 14:46
Its my own (user define function) in the same code not in any library — Bilal Maqsood, Nov 22 '15 at 14:48

Jonathan Leffler · Accepted Answer · 2016-07-22T14:04:53.327

You should be getting a compiler warning for the three declarations:

void interrupt (*old)();
void interrupt print_name();
void interrupt print_Id();

These declare one function pointer and two functions. None of these defines a prototype in C. The latter two declare the existence of a function and that the return types are void (and interrupt is a Borland-specific or DOS-specific noise word). But the empty parentheses mean "the argument list is undefined, except that it is not a variadic function — so there is no ... ellipsis in the actual argument list".

To make them into prototypes, you need to specify explicitly that the functions take no arguments:

void interrupt (*old)(void);
void interrupt print_name(void);
void interrupt print_Id(void);

This is different from C++, where the empty parentheses mean 'no arguments'. The difference was necessary when C was standardized because prototypes had not been a part of C before the C89/C90 standard, and all existing code had to use the empty parentheses to mean "the function exists but you don't know anything about the function arguments" and the standard would have failed had it broken all the existing C code.

The reason you are only getting the warning for the two calls is precisely because the warning is generated when the function name or function pointer is used to call the function; not when the function name is used as a function pointer. All three need to be fixed.

Incidentally, I prefer to have the declarations and definitions consistent, so I always use the explicit void argument list in the function definition too; it ensures that the function definition provides a prototype if it stands alone, too.

score 3 · Answer 2 · answered Nov 22 '15 at 14:39

3

The declaration void interrupt print_Id(); is different from print_id() because the case of i differs.

For that reason, if you call print_id() without declaration or definition of it before the call, the compiler will emit the warning.

(Although I see no print_id, including calling, in your posted code...)

answered Nov 22 '15 at 14:39

MikeCAT

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If i forget about `print_id` whats wrong with `(*old)();` this function ??? show that error on this function as well – Bilal Maqsood Nov 22 '15 at 14:44

Why is the compiler showing "Call to function with no prototype"?

2 Answers2