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.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
}
};
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