把函数指针铸造成另一种类型

假设我有一个函数，它接受一个 void (*)(void*)函数指针作为回调函数:

void do_stuff(void (*callback_fp)(void*), void* callback_arg);

现在，如果我有一个这样的函数:

void my_callback_function(struct my_struct* arg);

我能安全地做这件事吗？

do_stuff((void (*)(void*)) &my_callback_function, NULL);

我看过 这个问题和一些 C 标准，它们说你可以强制转换为“兼容的函数指针”，但是我找不到“兼容的函数指针”的定义。

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As C code compiles to instruction which do not care at all about pointer types, it's quite fine to use the code you mention. You'd run into problems when you'd run do_stuff with your callback function and pointer to something else then my_struct structure as argument.

I hope I can make it clearer by showing what would not work:

int my_number = 14;
do_stuff((void (*)(void*)) &my_callback_function, &my_number);
// my_callback_function will try to access int as struct my_struct
// and go nuts

or...

void another_callback_function(struct my_struct* arg, int arg2) { something }
do_stuff((void (*)(void*)) &another_callback_function, NULL);
// another_callback_function will look for non-existing second argument
// on the stack and go nuts

Basically, you can cast pointers to whatever you like, as long as the data continue to make sense at run-time.

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If you think about the way function calls work in C/C++, they push certain items on the stack, jump to the new code location, execute, then pop the stack on return. If your function pointers describe functions with the same return type and the same number/size of arguments, you should be okay.

Thus, I think you should be able to do so safely.

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最佳答案

As far as the C standard is concerned, if you cast a function pointer to a function pointer of a different type and then call that, it is undefined behavior. See Annex J.2 (informative):

The behavior is undefined in the following circumstances:

  • A pointer is used to call a function whose type is not compatible with the pointed-to type (6.3.2.3).

Section 6.3.2.3, paragraph 8 reads:

A pointer to a function of one type may be converted to a pointer to a function of another type and back again; the result shall compare equal to the original pointer. If a converted pointer is used to call a function whose type is not compatible with the pointed-to type, the behavior is undefined.

So in other words, you can cast a function pointer to a different function pointer type, cast it back again, and call it, and things will work.

The definition of compatible is somewhat complicated. It can be found in section 6.7.5.3, paragraph 15:

For two function types to be compatible, both shall specify compatible return types127.

Moreover, the parameter type lists, if both are present, shall agree in the number of parameters and in use of the ellipsis terminator; corresponding parameters shall have compatible types. If one type has a parameter type list and the other type is specified by a function declarator that is not part of a function definition and that contains an empty identifier list, the parameter list shall not have an ellipsis terminator and the type of each parameter shall be compatible with the type that results from the application of the default argument promotions. If one type has a parameter type list and the other type is specified by a function definition that contains a (possibly empty) identifier list, both shall agree in the number of parameters, and the type of each prototype parameter shall be compatible with the type that results from the application of the default argument promotions to the type of the corresponding identifier. (In the determination of type compatibility and of a composite type, each parameter declared with function or array type is taken as having the adjusted type and each parameter declared with qualified type is taken as having the unqualified version of its declared type.)

127) If both function types are ‘‘old style’’, parameter types are not compared.

The rules for determining whether two types are compatible are described in section 6.2.7, and I won't quote them here since they're rather lengthy, but you can read them on the draft of the C99 standard (PDF).

The relevant rule here is in section 6.7.5.1, paragraph 2:

For two pointer types to be compatible, both shall be identically qualified and both shall be pointers to compatible types.

Hence, since a void* is not compatible with a struct my_struct*, a function pointer of type void (*)(void*) is not compatible with a function pointer of type void (*)(struct my_struct*), so this casting of function pointers is technically undefined behavior.

In practice, though, you can safely get away with casting function pointers in some cases. In the x86 calling convention, arguments are pushed on the stack, and all pointers are the same size (4 bytes in x86 or 8 bytes in x86_64). Calling a function pointer boils down to pushing the arguments on the stack and doing an indirect jump to the function pointer target, and there's obviously no notion of types at the machine code level.

Things you definitely can't do:

  • Cast between function pointers of different calling conventions. You will mess up the stack and at best, crash, at worst, succeed silently with a huge gaping security hole. In Windows programming, you often pass function pointers around. Win32 expects all callback functions to use the stdcall calling convention (which the macros CALLBACK, PASCAL, and WINAPI all expand to). If you pass a function pointer that uses the standard C calling convention (cdecl), badness will result.
  • In C++, cast between class member function pointers and regular function pointers. This often trips up C++ newbies. Class member functions have a hidden this parameter, and if you cast a member function to a regular function, there's no this object to use, and again, much badness will result.

Another bad idea that might sometimes work but is also undefined behavior:

  • Casting between function pointers and regular pointers (e.g. casting a void (*)(void) to a void*). Function pointers aren't necessarily the same size as regular pointers, since on some architectures they might contain extra contextual information. This will probably work ok on x86, but remember that it's undefined behavior.
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You have a compatible function type if the return type and parameter types are compatible - basically (it's more complicated in reality :)). Compatibility is the same as "same type" just more lax to allow to have different types but still have some form of saying "these types are almost the same". In C89, for example, two structs were compatible if they were otherwise identical but just their name was different. C99 seem to have changed that. Quoting from the c rationale document (highly recommended reading, btw!):

Structure, union, or enumeration type declarations in two different translation units do not formally declare the same type, even if the text of these declarations come from the same include file, since the translation units are themselves disjoint. The Standard thus specifies additional compatibility rules for such types, so that if two such declarations are sufficiently similar they are compatible.

That said - yeah strictly this is undefined behavior, because your do_stuff function or someone else will call your function with a function pointer having void* as parameter, but your function has an incompatible parameter. But nevertheless, i expect all compilers to compile and run it without moaning. But you can do cleaner by having another function taking a void* (and registering that as callback function) which will just call your actual function then.

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The point really isn't whether you can. The trivial solution is

void my_callback_function(struct my_struct* arg);
void my_callback_helper(void* pv)
{
my_callback_function((struct my_struct*)pv);
}
do_stuff(&my_callback_helper);

A good compiler will only generate code for my_callback_helper if it's really needed, in which case you'd be glad it did.

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Void pointers are compatible with other types of pointer. It's the backbone of how malloc and the mem functions (memcpy, memcmp) work. Typically, in C (Rather than C++) NULL is a macro defined as ((void *)0).

Look at 6.3.2.3 (Item 1) in C99:

A pointer to void may be converted to or from a pointer to any incomplete or object type

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I asked about this exact same issue regarding some code in GLib recently. (GLib is a core library for the GNOME project and written in C.) I was told the entire slots'n'signals framework depends upon it.

Throughout the code, there are numerous instances of casting from type (1) to (2):

  1. typedef int (*CompareFunc) (const void *a, const void *b)
  2. typedef int (*CompareDataFunc) (const void *b, const void *b, void *user_data)

It is common to chain-thru with calls like this:

int stuff_equal (GStuff      *a,
GStuff      *b,
CompareFunc  compare_func)
{
return stuff_equal_with_data(a, b, (CompareDataFunc) compare_func, NULL);
}


int stuff_equal_with_data (GStuff          *a,
GStuff          *b,
CompareDataFunc  compare_func,
void            *user_data)
{
int result;
/* do some work here */
result = compare_func (data1, data2, user_data);
return result;
}

See for yourself here in g_array_sort(): http://git.gnome.org/browse/glib/tree/glib/garray.c

The answers above are detailed and likely correct -- if you sit on the standards committee. Adam and Johannes deserve credit for their well-researched responses. However, out in the wild, you will find this code works just fine. Controversial? Yes. Consider this: GLib compiles/works/tests on a large number of platforms (Linux/Solaris/Windows/OS X) with a wide variety of compilers/linkers/kernel loaders (GCC/CLang/MSVC). Standards be damned, I guess.

I spent some time thinking about these answers. Here is my conclusion:

  1. If you are writing a callback library, this might be OK. Caveat emptor -- use at your own risk.
  2. Else, don't do it.

Thinking deeper after writing this response, I would not be surprised if the code for C compilers uses this same trick. And since (most/all?) modern C compilers are bootstrapped, this would imply the trick is safe.

A more important question to research: Can someone find a platform/compiler/linker/loader where this trick does not work? Major brownie points for that one. I bet there are some embedded processors/systems that don't like it. However, for desktop computing (and probably mobile/tablet), this trick probably still works.

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Well, unless I understood the question wrong, you can just cast a function pointer this way.

void print_data(void *data)
{
// ...
}




((void (*)(char *)) &print_data)("hello");

A cleaner way would be to create a function typedef.

typedef void(*t_print_str)(char *);
((t_print_str) &print_data)("hello");

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