使用共享库时的函数原型

Question

I'm not sure of the correct terminology to use to describe this problem, but I will try my best.

I'm writing a toy program that prints the factorial of 12 and 13 in test.c:

#include <stdio.h>

int main(void)
{
    printf("%ld\n", factorial(12));
    printf("%ld\n", factorial(13));
    return 0;
}

The factorial function is defined in a shared library with source fact.c

long factorial(int n)
{
    long r = 1;
    while (n > 1) r *= n--;
    return r;
}

I am compiling the shared library and my program with the following commands:

$ gcc -shared -fPIC -o libfact.so fact.c
$ gcc -L. -lfact test.c

I am on x86-64, so I expect that factorial(13) (13! = 6227020800) does not overflow a 64-bit long. However, I get a strange result.

$ LD_LIBRARY_PATH=. ./a.out
479001600
1932053504

Here, 1932053504 happens to be the decimal value of the lower 32 bits of the correct result. If, however, I insert the function prototype long factorial(int) at the top of test.c and recompile, I get the correct result.

$ LD_LIBRARY_PATH=. ./a.out
479001600
6227020800

This leaves me with a couple questions:

1. What is the assumed return type if I do not specify a prototype in test.c? Is this compiler dependent?
2. Without a prototype, I seem to be able to pass in any arbitrary number of arguments to factorial . Is this related to this question about C void arguments and this question about function prototypes ?
3. Why doesn't the linker throw an undefined reference error?

Answer 1

As of the 1990 version of the C standard, if you call a function with no visible declaration the compiler assumes that it returns type int . So for the calls to factorial() in your main program, the compiler will most likely interpret the long value it returns as if it were an int value. If long and int happen to have the same representation, this is likely to work. If long is wider than int , it might happen to work, or it can fail. But the behavior is undefined either way.

The 1999 version of the C standard (C99) removed the "implicit int" rule. Calling a function with no visible declaration is a constraint violation , requiring a compiler diagnostic. The compiler may then either reject the program, or continue to compile it -- but if it does, the behavior is undefined.

To correct this, you need to have a visible declaration of factorial() before you call it. The best way to do this is to create a header file, factorial.h :

factorial.h:

#ifndef FACTORIAL_H
#define FACTORIAL_H

long factorial(int n);

#endif

factorial.c:

#include <stdio.h>
#include "factorial.h"

long factorial(int n)
{
    printf("factorial(%d)", n);
    long r = 1;
    while (n > 1) r *= n--;
    printf(" --> %ld\n", r);
    return r;
}

main.c:

#include <stdio.h>
#include "factorial.h"

int main(void)
{
    printf("%ld\n", factorial(12));
    printf("%ld\n", factorial(13));
    return 0;
}

Note that this still won't work if long is only 32 bits, since 13 factorial exceeds 2 ³¹ -1. Check the value of LONG_MAX and/or sizeof (long) on your system:

printf("LONG_MAX = %ld, sizeof (long) = %d\n", LONG_MAX, (int)sizeof (long));

Consider using long long rather than long -- or, better yet, int64_t or uint64_t , defined in <stdint.h> .

(As far as I know, your use of a shared library doesn't affect any of this.)

Answer 2

• If function return type is unknown, it is considered by the compiler to be int . The cast that occurs in this case explains the wired output of your printf statement. Anyhow, prefer using long long int , as long int are defined in the standard to handle as maximal value minimum 2^31 - 1

§5.2.4.2.1

— maximum value for an object of type long int

LONG_MAX +2147483647 // 2^31− 1

• The linker does not throw any undefined reference error because it does not check in the dynamic library if the function really exists. It binds it, and will try to load it at execution time.

Answer 3

long doesn't have to be 8 bytes on 64-bit architecture, standard requires it to be not less than 4 bytes. You have an overflow.