How to call 'printf' using libffi?

[ShaJunxing]

在 2021/3/19 上午10:36, Kaz Kylheku (libffi) 写道:
> On 2021-03-18 19:08, ShaJunxing via Libffi-discuss wrote:
>> Yesterday after sending email I also realized maybe different
>> parameters needs different cif, and I did some experiments, it exactly
>> works. But I got two more problems. The first is, I found
>> ffi_prep_cif() works fine in this situation, so what is
>> ffi_prep_cif_var() different from?
>
> Note that what you are saying is exactly analogous to this:
>
> "When I declare printf like this:
>
>    extern int printf(const char *fmt, char *arg);
>
> everything works; I can call it as
>
>    printf("Hello, %s!\n", name);
>
> and I get the expected output.
>
> why do I have to have a correct variadic prototype by including
> <stdio.h>?
>
> It might look like it's working, but it's undefined behavior;
> it is not required to work and could prove to be nonportable.
>
> Variadic functions do not necessarily pass arguments the same way
> as ordinary functions. In order for libffi to build the correct
> parameter passing mechanism for the descriptor, it needs to be
> informed that the target function is variadic.
>
> Then, depending on the target architecture, the libffi code
> might do some things differently for calling that function.
> Those different things might depend on how many arguments
> there are and what types.
>
>> The second is, I found printf()
>> cannot properly handle float type in my machine, my code is:
>
> Note that when argument expressions of float type are passed
> as the trailing arguments of a variadic function, they are
> promoted to type double. That's a C language rule.
>
> The rule also applies to old style functions with undeclared
> parameter lists,e .g.
>
>    int old_style();
>
>    float f = 3.14;
>
>    old_style(f);  /* promoted to double! */
>
>    /* definition must be this */
>
>    int old_style(arg)
>    double arg;
>    {
>    }
>
> The %f conversion specifier in printf does not mean "float".
> It means "fixed digit format". There are two other formats:
> %e, exponential and %g, general (which chooses exponential
> or fixed digit).
>
> All of these take an argument of type double.
>
>>
>> #include <ffi.h>
>> #include <stdio.h>
>> intmain() {
>> ffi_cif cif;
>> ffi_type *atypes[2] = {&ffi_type_pointer, &ffi_type_float};
>> if(ffi_prep_cif(&cif, FFI_DEFAULT_ABI, 2, &ffi_type_sint, atypes) == 
>> FFI_OK) {
>> char*s = "hello %f\n";
>> floatf = 3.14;
>> void*avalues[2] = {&s, &f};
>> ffi_arg rvalue;
>> ffi_call(&cif, (void*)printf, &rvalue, avalues);
>> }
>> return0;
>> }
>>
>> It always outputs zero (maybe other unexpected result). I changed the
>> second argument to double, char *, int, long ... all correct, except
>> float.
>
> Not only must the FFI type be ffi_type_double, but you must make sure
> that the object which the corresponding argument points to is of
> type double. FFI doesn't perform any conversion; it has no idea:
> there is just a void * pointing at an argument object which is expected
> to match the exact type that the type descriptor states.

Thank you for your detailed answer, it's very kind of you. I'm not very 
good at C and was always thought var args are in fact macros and will be 
fixed length after compiling, now I realized it is a special mechanism 
not as simple as macros. I'll do more studying. Thank you so much.😁