Comment by cryptonector

Comment by cryptonector 6 months ago

> test.cpp:70:14: error: assigning to 'void ' from 'func0Ptr' (aka 'void ()(void *)') converts between void pointer and function pointer 70 | res.fn = (func0Ptr)fn;

This warning is stupid. It's part of the "we reserve the right to change the size of function pointers some day so that we can haz closures, so you can't assume that function pointers and data pointers are the same size m'kay?" silliness. And it is silly: because the C and C++ committees will never be able to change the size of function pointers, not backwards-compatibly. It's not that I don't wish they could. It's that they can't.

akdev1l 6 months ago

It’s not a warning, it’s a compile time error and I am not even using -Wall -Werror

I also believe there are platforms where a function pointer and a data pointer are not the same but idk about such esoteric platforms first hand (seems Itanium had that: https://stackoverflow.com/questions/36645660/why-cant-i-cast...)

Though my point was only that this code will not compile as is with whatever clang Apple ships*

I am not really sure how to get it to compile tbqh

Some further research ( https://www.kdab.com/how-to-cast-a-function-pointer-to-a-voi...) suggest it should be done like so:

> auto fptr = &f; void a = reinterpret_cast<void &>(fptr);

edit: I tried with GCC 15 and that compiled successfully

Reply View 4 replies

gpderetta 6 months ago

FWIW, POSIX practically requires void otr and function otr inter-convertibility hence the support from GCC.

Reply View | 2 replies
- wahern 6 months ago
  
  Not practically, but literally:
  Note that conversion from a void * pointer to a function pointer as in: fptr = (int (*)(int))dlsym(handle, "my_function"); is not defined by the ISO C standard. This standard requires this conversion to work correctly on conforming implementations.
  
  Reply View | 1 reply
  
  gpderetta 6 months ago
  
  I remembered the dlsym requirement, I wasn't sure it was de-jure.
  
  Reply View | 0 replies
comex 6 months ago
It should just be
res.fn = (void *)fn;
`res.fn` is of type `void *`, so that's what the code should be casting to. Casting to `func0Ptr` there seems to just be a mistake. Some compilers may allow the resulting function pointer to then implicitly convert to `void *`, but it's not valid in standard C++, hence the error.
Separately from that, if you enable -Wpedantic, you can get a warning for conversions between function and data pointers even if they do use an explicit cast, but that's not the default.
Reply View | 0 replies