看到“明顯調用的表達式前的括號必須具有(指針)函數類型”這句時我才發現我的語文水平有多爛,怎麼看都看不懂,折騰了半天才知道是哪裡出了問題。
舉個簡單的例子
class CTest { void (CTest::*m_pFun)(); void CallFun() { (this->*m_pFun)(); //OK,對象指針和函數名一定要用括號括起來,函數名前面要加上*號 this->*m_pFun(); //error (this->m_pFun)(); //error } //本文鏈接http://www.cnblogs.com/vcpp123/p/5902839.html };
詳細說明請參閱MSDN,鏈接:https://msdn.microsoft.com/query/dev14.query?appId=Dev14IDEF1&l=ZH-CN&k=k(C2064)&rd=true
term does not evaluate to a function taking N arguments
A call is made to a function through an expression.The expression does not evaluate to a pointer to a function that takes the specified number of arguments.
In this example, the code attempts to call non-functions as functions.The following sample generates C2064:
// C2064.cpp int i, j; char* p; void func() { j = i(); // C2064, i is not a function p(); // C2064, p doesn't point to a function }
You must call pointers to non-static member functions from the context of an object instance.The following sample generates C2064, and shows how to fix it:
// C2064b.cpp struct C { void func1() {} void func2() {} }; typedef void (C::*pFunc)(); int main() { C c; pFunc funcArray[2] = { &C::func1, &C::func2 }; (funcArray[0])(); // C2064 (c.*funcArray[0])(); // OK - function called in instance context }
Within a class, member function pointers must also indicate the calling object context.The following sample generates C2064 and shows how to fix it:
// C2064d.cpp // Compile by using: cl /c /W4 C2064d.cpp struct C { typedef void (C::*pFunc)(); pFunc funcArray[2]; void func1() {} void func2() {} C() { funcArray[0] = &C::func1; funcArray[1] = &C::func2; } void func3() { (funcArray[0])(); // C2064 (this->*funcArray[0])(); // OK - called in this instance context } };