Cannot convert parameter 2 from 'long (unsigned long)'
to 'long (__cdecl *)(unsigned long)'What should I do?"
Here is the code for one way to solve this problem.
//in the header class CKernel: { long (*lpFunc)(DWORD); long OLESendTC( DWORD dwInfo ); } //in the cpp File BOOL CKernel::Init() { lpFunc = OLESendTC; return TRUE; } Class methods all have one hidden argument, a pointer to the class object the method is called on. C++ uses this pointer to find the location of any class data that the method may reference. If you try and use a standard function pointer to call a class method, C++ has no way to pass this hidden argument and a conflict results.
To help deal with this and to improve type safety, C++ added three new operators, ::*, .*, and ->*, to allow for safe pointers to members. These pointers to members can point to either member functions or variables.
class CTest { public: BOOL Init(); long OLESendTC(DWORD dwInfo); }; long (CTest::*lpFunc)(DWORD dwInfo) = &CTest::OLESendTC; int main() { CTest test; (test.*lpFunc)(0); return 0; } long CTest::OLESendTC(DWORD dwInfo) { cout << "IN OLESENDTC\n"; return 0; } This example shows one use of pointers to members. The code uses the ::* operator to declare lpFunc as a pointer to a CTest member function. Note that, rather than assign a value to this pointer at run time, the pointer is simply initialized in the declaration.
In the main function, this example uses the .* operator to call the method pointed to by lpFunc. If test were a pointer here, you would use the ->* operator instead.
C++ has a lot of things going on under the hood such as hidden arguments to methods. Pointers to members allow you to safely declare a pointer to a class method and call methods through that pointer.
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