Can you output data in C++ without using any includes? - c++

Question

This is a thought experiment.
This question is intentionally trying to do something easy in a difficult way. I am aware of the many ways to do this using iostream, std.h, etc. Please do not suggest answers that circumvent the challenge.
Using C++,
Only using main() without any header files included, is it possible to write a program that creates a variable of any type, and outputs the value of the variable in any way, such as printing it in console, saving it to a file, anything like that?
The goal is to have a .main file that looks like:
int main(){
Some code;
return 0;
}
I've tried using C's printf but that requires a header.
The best I could come up with is saving the value to a specific space in memory then using another program to read that block. But that is somewhat circumventing the goal of outputting only from main.
This may be impossible.

Answer 1

If you want to have no code outside of main, in C you could do this:
int main()
{
int puts(const char *);
puts("Sup!\n");
return 0;
}
Or the same thing with printf, or any other C standard library function.
It wouldn't work in C++ though, since in C++ the puts declaration needs extern "C" to disable name mangling, which can only be used outside of functions.
There's a trick that lets you do it in C++:
int main()
{
extern int puts;
((int(*)(const char *))&puts)("Sup!\n");
return 0;
}
Variable names are not mangled, so by tricking the compiler into thinking that puts is a variable rather than a function, you can get rid of the mangling without extern C.

Answer 2

Technically, this satisfies your constraints:
int main() {
return 123;
}
Most OSs allow you to check a program's return code. So this communicates one number to the outside world. The available range of values may differ with the OS.

Answer 3

You wrote "only main", but extern "C" cannot go into a block. So, if you can live with that line outside the main function, here's the code you asked for:
extern "C" int printf(const char *fmt, ...);
int main() {
printf("Hello %s\n", "world");
return 0;
}
Proof: https://godbolt.org/z/41WhGYc79

Answer 4

Perhaps I haven't understood correctly, but if you want to avoid the includes/libraries then the short answer is no. A C++ header (apart from templates) is just a declaration of a possible link-time function availability and you will always need them.
Almost anything meaningful you can do with C++ (as in any programming language, even assembly) requires a call to the underlying OS and this requires a header and a library link.
Even something as simple as a char output requires thousands of lines to communicate with a device driver to output to screen.
Even if you don't include anything, the compiler will inject code (CRT) that calls the OS by using a library no matter what. Even a simple main() requires lots of calls to the OS, for actually there is no "main" entry point in the executable.

Answer 5

That does not exactly answer your question, but if you were doing this in C rather than C++, then for most compilers that would be a valid program (without any includes):
int main()
{
printf("hello");
return 0;
}
Although it generates a warning.

Answer 6

Combining two of the answers above I was able to use printf by replacing the parts of
extern int puts;
((int(*)(const char *))&puts)("Sup!\n");
With parts that would allow printf:
extern int printf;
((int(*)(const char *fmt, ...))&printf)("%d\n", 10);
So the full code would be:
int main(){
int num = 10;
extern int printf;
((int(*)(const char *fmt, ...))&printf)("%d\n", num);
return 0;
}
I hope this helps, this is a combination of two responses not my own.