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Typedef function pointer?
TE0300_Open = (_TE0300_Open)GetProcAddress(hInstLibrary, "TE0300_Open"); typedef int (WINAPI *_TE0300_Open)(unsigned int* PHandle, int CardNo); Can someone explain me what this piece of code does? I know that typedef is used to assign alternative names to existing types but I don't understand this case at all.
typedef int (WINAPI *_TE0300_Open)(unsigned int* PHandle, int CardNo);
This line typedefs a function pointer to a WINAPI calling convention function returning an int, and taking an unsigned int * and an int. The function pointer type is given the alias _TE0300_Open.
Consider the following example:
typedef void (*func)(); void foo (func f) //notice we have a nice type name here { cout << "Calling function..."; f(); } void bar(){} int main() { foo (bar); } I believe C++11 added support for less icky syntax when using function pointers as well:
using func = void (*)(); As for your GetProcAddress call, this loads a function from a library. You assign it to a function pointer, and you can use that function pointer as you would the original function.
In your example, you can now call TE0300_Open as you would normally call _TE0300_Open. It should also be noted that _TE0300_Open is a name that is reserved for the implementation.
typedef usage.It's declaring a typedef, _TE0300_Open, for a function-pointer.
