How to print pointer to function?

[Manfred]

On 12/17/2019 2:22 PM, Segher Boessenkool wrote:
> On Tue, Dec 17, 2019 at 01:56:48PM +0100, Vincent Lefevre wrote:
>> On 2019-12-12 11:53:05 +0100, Josef Wolf wrote:
>>> Thanks for the help, Liu!
>>>
>>> This silenced the warning.
>>
>> The goal should not be to silence warnings, but to write portable
>> code. If you find a way to silence a warning, only to silence a
>> warning, this is useless. You'd better disable warnings.
>>
>> Here your code is still non-portable, perhaps *more* non-portable,
>> as only conversions between intptr_t and void * are guaranteed to
>> work. Moreover, the intptr_t type is optional.
> 
> Any object pointer, sure (since you can convert those to pointer to
> void and back again).
> 
> Is there *any* portable way to print function pointers?  Other than
> accessing it as bytes :-)

n1570 section 6.3.2.3 p6 says "Any pointer type may be converted to an 
integer type. Except as previously specified, the result is 
implementation-defined. ..."

It says implementation-defined, not undefined, so AFAIU it is legal, 
although you can't make assumptions on the actual value of the result.

However, 7.20.1.4 specifies intptr_t and uintptr_t for pointers to void 
only, and 6.3.2.3 p1 specifies the conversion to/from pointers to void 
and pointers to "any object type" which does not include functions, AFAIU.
So, unless I am missing something, there seems indeed to be a broken 
link in the chain from pointer to function to (u)intptr_t.
But then what's the purpose of 6.3.2.3 p6 ?


> 
> 
> Segher
>