首页 > > 程序设计 > C/C++ >

明显调用的表达式前的括号必须具有(指针)函数类…

2018-06-17 23:43:51来源：未知阅读 ()

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
	}
};

标签：

版权申明：本站文章部分自网络，如有侵权，请联系：west999com@outlook.com
特别注意：本站所有转载文章言论不代表本站观点，本站所提供的摄影照片，插画，设计作品，如需使用，请与原作者联系，版权归原作者所有