How do I declare the main() entry point of a program without specifying all arguments in C++?

Question

In C, I can do this to have an unspecified number of arguments in a function:

#include <elf.h>
#include <stddef.h>
#include <stdlib.h>

extern char **__environ;

int __libc_start_main
(
int (*main)(),
int argc,
char **argv
)
{
    int ret;
    Elf32_auxv_t *auxv;
    size_t aux[38];

    /* ... */

    exit(main(argc, argv, __environ, aux));
}

However, when doing this in C++, the compiler emits this error:

test.c: In function ‘int __libc_start_main(int (*)(), int, char**)’:
test.c:21:45: error: too many arguments to function
         exit(main(argc, argv, __environ, aux));
                                             ^

How do I do this in C++?

I understand that the C/C++ standards don't allow this, but I'm currently writing an implementation of the standard C library.

Comments are not for extended discussion; this conversation has been [moved to chat](https://chat.stackoverflow.com/rooms/195384/discussion-on-question-by-jl2210-how-do-i-declare-the-main-entry-point-of-a-pr). — Samuel Liew, Jun 22 '19 at 01:25

score 3 · Accepted Answer · answered Jun 21 '19 at 13:17

3

The short answer is: You don't.

In C++ all functions have a prototype; there is no such thing as an "unspecified number of arguments".

If you want to call main as main(argc, argv, __environ, aux), you need to declare it as int (*main)(int, char **, char **, void *) or similar.

answered Jun 21 '19 at 13:17

melpomene

84,125
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I was wondering if one could have default values along with the function pointer definition like `int (*main)(int = 0, char ** = NULL, char ** = NULL, void * = NULL)`? – πάντα ῥεῖ Jun 21 '19 at 13:20
@πάνταῥεῖ That seems like a nifty feature, so I doubt C++ supports it. – melpomene Jun 21 '19 at 13:21
I also doubt so, just too lazy to tryout myself now ;-). Anyways, compilers seem to handle `main()` in a special way. – πάντα ῥεῖ Jun 21 '19 at 13:22
@πάνταῥεῖ g++ errors out with `error: default arguments are only permitted for function parameters`. – melpomene Jun 21 '19 at 13:23
1

@πάνταῥεῖ [main is indeed very special](https://stackoverflow.com/a/4207223/332733) – Mgetz Jun 21 '19 at 13:26
@Mgetz The `main` in question is a local variable. – S.S. Anne Jun 21 '19 at 13:27
@JL2210 hence why I didn't close vote this as a duplicate – Mgetz Jun 21 '19 at 13:28

score 2 · Answer 2 · answered Jun 21 '19 at 12:47

2

Try either:

void foo(...);

or

template <typename ... ARGS> void foo(ARGS && ... args) { ... body }

First option is the same as void foo() (little known C language fact). First option requires some sort of additional argument (for example printf(char *, ...), where first argument allows function to detect, how to parse following arguments).

Second option requires you to commit to a function body somewhere in a header.

answered Jun 21 '19 at 12:47

Radosław Cybulski

2,952
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5

`void foo(...)` is not legal C (per the grammar in C 2018 6.7.6, the *parameter-type-list* must have at least one parameter), and, if it were, it would not be equivalent to `void foo()`, due to rules about compatible types, at least. – Eric Postpischil Jun 21 '19 at 13:06

How do I declare the main() entry point of a program without specifying all arguments in C++?

2 Answers2