[cfe-dev] Default argument type for function without prototype?

[Kirill Balunov via cfe-dev]

Hi,

I am a beginner to C and I previously asked this question on SO (and it was
marked as duplicate :-) ) I still do not understand the problem to the end. I
understand the need of function prototypes and the default argument
promotion machinery. But I am still confused with the obtained warning and
final result for this example:

#include <stdio.h>

void impl();

int main(void){

impl(3.0);

return 0;

}

void impl(val){

printf("%.2f", val);

}

I get the following warning: *format specifies type 'double' but the
argument has type 'int' [-Wformat].*I can not understand why it is supposed
by a compiler that *val* should be treated as *int* and not to skip the
number and types of arguments and then perform default argument promotion
at run time. Also I read that if you don't provide argument types in ANSI C:

> An ANSI C compiler will assume that you have decided to forego function
> prototyping, and it will not check arguments.

The result which is printed to the console is *0.00*, but as I understood
the caller should place double on the stack and *%f* also means double in
*printf* so it reads double off the stack, and the result should not be
affected.

So I'm still confused what is happening here. Also I can't find where it is
stated that the argument without a type should be treated by a compiler as
type *int*. I understand that the question is stupid in nature, for this I
apologize in advance, but it somehow sat down in my head and I nowhere can
I find why it happens. Thank you!

p.s.: Compiled with *clang 6.0* and *--std=c89* flag under Windows 10.
Apologies if this question is off-topic for this list, but i can't find any
other.

With kind regards,

-gdg
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