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明显调用的表达式前的括号必须具有(指针)函数类型 编译器错误 C2064
看到“明显调用的表达式前的括号必须具有(指针)函数类型”这句时我才发现我的语文水平有多烂，怎么看都看不懂，折腾了半天才知道是哪里出了问题。
举个简单的例子
class CTest{ void (CTest::*m_pFun)(); void CallFun() { (this->*m_pFun)(); //OK，对象指针和函数名一定要用括号括起来，函数名前面要加上*号 this->*m_pFun(); //error (this->m_pFun)(); //error } //本文链接http://www.cnblogs.com/vcpp123/p/5902839.html};
详细说明请参阅MSDN，链接：https://msdn.microsoft.com/query/dev14.query?appId=Dev14IDEF1&l=ZH-CN&k=k(C2064)&rd=true
编译器错误 C2064
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.cppint 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.cppstruct 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.cppstruct 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 }};
明显调用的表达式前的括号必须具有(指针)函数类型 编译器错误 C2064