Function Pointer basic question

Question

I have a basic question on function pointer. In the below code snippet, how do I read this " *(FARPROC*)&pfn ="?

IFastString *CallCreateFastString(const char *psz) {

static IFastString * (*pfn)(const char *) = 0;

if (!pfn) {

const TCHAR szDll[] = _TEXT("FastString.DLL");

const char szFn[] = "CreateFastString";

HINSTANCE h = LoadLibrary(szDll);

if (h)

*(FARPROC*)&pfn = GetProcAddress(h, szFn);

}

return pfn ? pfn(psz) : 0;

}

Have a look at the documentation for GetProcAddress- you'll see that the case above is convoluted...msdn.microsoft.com/en-us/library/ms683212(v=vs.85).aspx — Nim
– Nim, Commented Aug 8, 2011 at 14:23

Kerrek SB · Accepted Answer · 2011-08-08 14:14:46Z

2

This isn't really about function pointers, but about casting in general.

Suppose pfn is of type T. Then &pfn is of type T*. This gets cast to FARPROC* by the cast expression (the stuff in the parentheses). Finally, this gets dereferenced, yielding a FARPROC&.

All in all this just means you're treating pfn as if it were of type FARPROC and assign a value to it.

Here's a generic example:

S make_an_S();

T x;
T * const px = &x;
S * const py = (S*)px;
*py = make_an_S();  // same as *(S*)&x = make_an_S();

answered Aug 8, 2011 at 14:14

Kerrek SB

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iammilind · Accepted Answer · 2011-08-08 14:14:47Z

0

*(FARPROC*)&pfn = GetProcAddress(h, szFn);

is equivalent to,

(FARPROC)pfn = GetProcAddress(h, szFn);

So, pfn is a function pointer which is type casted to FARPROC to store the address received from GetProcAddress(h, szFn).

[Note: I am not sure, if this kind of typecasting is deprecated in C++.]

answered Aug 8, 2011 at 14:14

iammilind

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5 Comments

Alexandre C. Over a year ago

It is not equivalent. You need a lvalue (eg. a reference type) in the left hand side, hence the indirection.

iammilind Over a year ago

@Alexandre, couldn't get your point. Isn't (FARPROC)pfn an lvalue ?

Alexandre C. Over a year ago

no it isn't. When you write (FARPROC)pfn, you cast pfn's value and it becomes a rvalue.

iammilind Over a year ago

@Alexandre, if you are talking about strict aliasing rules, then may be you are right. However, I have seen such code in my old DSP project, which was written in C. That's why I mentioned about deprecation.

Alexandre C. Over a year ago

If you have int x; and then you do (unsigned)x = 1 then this won't do the trick. You need *(unsigned*)&x = 1 to do type punning. Here you have the same problem: you need the indirection to build a lvalue. Note that in C++ you'd use static_cast here instead.

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