That's not quite true.
Main is not required to be defined in a freestanding environment, but in a hosted environment:
> 5.1.2.2.1 Program startup
> 1
> The function called at program startup is named
> main. The implementation declares no prototype for this function. It shall be defined with a return type of int and with no parameters:
int main(void) { /* ... */ }
int main(int argc, char *argv[]) { /* ... */ }
main can take on differing types, but it then becomes undefined-behaviour, which allows the compiler to do whatever it wants.
(5.1.2.2.1 Program startup, C11 Standard http://www.open-std.org/jtc1/sc22/wg14/www/docs/n1570.pdf)
In fact, main() is just a convention of libc. You can have C without libc. (Such as when writing a kernel!)
Now, attempting to link a standalone executable without a '_start' symbol, on the other hand...
No, it's not just a convention. 'main' is defined as the execution entrypoint in at least the C11 [0], C90 [1] standards. Both have both these forms defined:
int main(void) {}
int main(int argc, char* argv[]) {}
C without libc can still expect to have a main. One doesn't imply the other. It's just without a main, you also have to manually link to _start as well.
[0] http://www.open-std.org/jtc1/sc22/wg14/www/docs/n1570.pdf
[1] http://www.open-std.org/jtc1/sc22/wg14/www/docs/n1256.pdf
GNU ld and gold both support --entry to change the entry point. Alternatively, you can specify it in your linker script.
https://gist.github.com/prattmic/1f3618025aab52b0e90e88445b6...
$ gcc -nostdlib -ffreestanding -Wl,--entry=foobar -o main main.S
$ ./main; echo $?
42
Also, TIL according to the C standard, 'main' is not a reserved identifier! (https://stackoverflow.com/questions/34764796/why-does-declar...)
If anyone can clarify: I assumed that the gcc read literals directly into their 'target' type, but it seems like some literals (such as '-1u') are read as signed integers first then typecasted to the the target type?