定义使用函数指针

1 定义一个函数指针

且不论语法，有两种不同形式的指针函数: 一个是指向普通的C函数的指针和C++的静态成员函数，另外一个是指向C++的非静态成员函数的指针。这两者的基本区别是所有指向非静态成员函数的指针都 需要这个隐含定义：指向本类的一个This指针。注意：这两种函数指针彼此不兼容。
既然一个函数指针实际上和一个变量没有什么区别，定义它的时候也就没有什么特殊。下面的例子中我们定义3个函数指针，名字是pt2Function, pt2Member 和 pt2ConstMember.
它们指向的函数，输入一个 float和两个char 类型的变量并返回一个 int 类型的变量. 对于C++程序例子，我们的指针指向的的函数，是一个叫做 TMyClass 的 类的非静态成员函数。

// 1 define a function pointer and initialize to NULL

int (*pt2Function)(float, char, char) = NULL; //
C

int (TMyClass::*pt2Member)(float, char, char) = NULL; //
C++

int (TMyClass::*pt2ConstMember)(float, char, char) const =
NULL; // C++

2 函数调用规范
通常你不需要考虑一个函数的调用规范：编译器缺省的使用__cdecl，如果你不特别指出用哪个的话。调用规范告诉编译器如何传 递参数以及如何产生函数名。一些其他的调用规范可以举例为：__stdcall, __pascal和__fastcall.
注意：使用不同调用规范的函数指针彼此不兼容。

// 2 define the calling convention

void __cdecl DoIt(float a, char b, char c); //
Borland and Microsoft

void DoIt(float a, char b, char c) __attribute__((cdecl)); //
GNU GCC

3 为函数指针分派一个地址
将函数的地址分派给函数指针很容易，在函数名字之前冠以取址符&就可以了。

// 3 assign an address to the function pointer

// Note: Although you may ommit the address operator on most compilers

// you should always use the correct way in order to write portable code.

// C

int DoIt (float a, char b, char c){ printf("DoIt\n"); return a+b+c;
}

int DoMore(float a, char b, char c)const{ printf("DoMore\n"); return a-b+c;
}

pt2Function = DoIt; // short form

pt2Function = &DoMore; // correct assignment using address operator

// C++

class TMyClass

{

public:

int DoIt(float a, char b, char c){ cout << "TMyClass::DoIt"<<
endl; return a+b+c;};

int DoMore(float a, char b, char c) const

{ cout << "TMyClass::DoMore" << endl; return a-b+c; };

/* more of TMyClass */

};

pt2ConstMember = &TMyClass::DoMore; // correct assignment using address operator

pt2Member = &TMyClass::DoIt; // note: <pt2Member> may also legally point to &DoMore

4 比较函数指针
你可以和通常方式一样使用比较符号(==, !=)。下面的例子中我们检查变量 pt2Function 和 pt2Member ，它们分别指向的函数 DoIt 和 TMyClass::DoMore.
如果指向正确就会输出代表争取的字符串。

// 4 comparing function pointers

// C

if(pt2Function >0){ // check if initialized

if(pt2Function == &DoIt)

printf("Pointer points to DoIt\n"); }

else

printf("Pointer not initialized!!\n");

// C++

if(pt2ConstMember == &TMyClass::DoMore)

cout << "Pointer points to TMyClass::DoMore" << endl;

5 使用函数指针调用函数
在C语言中我们可以通过 * 符号来调用函数指针，也可以不使用函数名而代以函数指针的名字。C++使用两个符号 * 和 ->* 来调用类的非静态函数指针。如果呼叫在其他成员函数中发生，那么就得加上this指针。

// 5 calling a function using a function pointer

int result1 = pt2Function (12, 'a', 'b'); //
C short way

int result2 = (*pt2Function) (12, 'a', 'b'); //
C

TMyClass instance1;

int result3 = (instance1.*pt2Member)(12, 'a', 'b'); //
C++

int result4 = (*this.*pt2Member)(12, 'a', 'b'); //
C++ if this-pointer can be used

TMyClass* instance2 = new TMyClass;

int result4 = (instance2->*pt2Member)(12, 'a', 'b'); //
C++, instance2 is a pointer

delete instance2;

6 如何像传递一个参数一样传递函数指针
你可以在调用一个函数的时候把函数指针当作参数来传递。在回调函数中尤其要使用到这个技术。下边这个例子演示了如何把指针传递给一个函数，这个 函数使用一个 float 和两个 char 类型的参数并有一个int类型的返回值。

//------------------------------------------------------------------------------------

// 6 How to Pass a Function Pointer

// <pt2Func> is a pointer to a function which returns an int and takes a float and two char

void PassPtr(int (*pt2Func)(float, char, char))

{

int result = (*pt2Func)(12, 'a', 'b'); //
call using function pointer

cout << result << endl;

}

// execute example code - 'DoIt' is a suitable function like defined above in 2.1-4

void Pass_A_Function_Pointer()

{

cout << endl << "Executing 'Pass_A_Function_Pointer'" << endl;

PassPtr(&DoIt);

}

7 如何返回一个函数指针
看上去有点怪，但是一个函数指针可以作为一个函数的返回值。下面的例子中提供了两种把函数指针作为返回值的解决方案。这个函数输入两个float类 型的参数，返回一个float类型的值。

//------------------------------------------------------------------------------------

// 7 How to Return a Function Pointer

// 'Plus' and 'Minus' are defined above. They return a float and take two float

// Direct solution: Function takes a char and returns a pointer to a

// function which is taking two floats and returns a float. <opCode>

// specifies which function to return

float (*GetPtr1(const char opCode))(float, float)

{

if(opCode == '+')

return &Plus;

else

return &Minus; // default if invalid operator was passed

}

// Solution using a typedef: Define a pointer to a function which is taking

// two floats and returns a float

typedef float(*pt2Func)(float, float);

// Function takes a char and returns a function pointer which is defined

// with the typedef above. <opCode> specifies which function to return

pt2Func GetPtr2(const char opCode)

{

if(opCode == '+')

return &Plus;

else

return &Minus; // default if invalid operator was passed

}

// Execute example code

void Return_A_Function_Pointer()

{

cout << endl << "Executing 'Return_A_Function_Pointer'" << endl;

// define a function pointer and initialize it to NULL

float (*pt2Function)(float, float) = NULL;

pt2Function=GetPtr1('+'); // get function pointer from function 'GetPtr1'

cout << (*pt2Function)(2, 4) << endl; // call function using
the pointer

pt2Function=GetPtr2('-'); // get function pointer from function 'GetPtr2'

cout << (*pt2Function)(2, 4) << endl; // call function using
the pointer

}

8 如何使用函数指针数组
使用函数指针数组非常有意思，这个技术提供了一种从索引中选择函数的方法。实现的语法看上去很复杂，经常导致理解错误。下面的例子中我们可以看 到两种定义和使用函数指针数组的方法。前一种使用 typedef ，后一种直接定义数组。使用哪一种全凭你的兴趣。

//------------------------------------------------------------------------------------

// 8 How to Use Arrays of Function Pointers

// C ---------------------------------------------------------------------------------

// type-definition: 'pt2Function' now can be used as type

typedef int (*pt2Function)(float, char, char);

// illustrate how to work with an array of function pointers

void Array_Of_Function_Pointers()

{

printf("\nExecuting 'Array_Of_Function_Pointers'\n");

// define arrays and ini each element to NULL, <funcArr1> and <funcArr2> are arrays

// with 10 pointers to functions which return an int and take a float and two char

// first way using the typedef

pt2Function funcArr1[10] = {NULL};

// 2nd way directly defining the array

int (*funcArr2[10])(float, char, char)
= {NULL};

// assign the function's address - 'DoIt' and 'DoMore' are suitable functions

// like defined above in 1-4

funcArr1[0] = funcArr2[1] = &DoIt;

funcArr1[1] = funcArr2[0] = &DoMore;

/* more assignments */

// calling a function using an index to address the function pointer

printf("%d\n", funcArr1[1](12, 'a', 'b')); //
short form

printf("%d\n", (*funcArr1[0])(12, 'a', 'b')); //
"correct" way of calling

printf("%d\n", (*funcArr2[1])(56, 'a', 'b'));

printf("%d\n", (*funcArr2[0])(34, 'a', 'b'));

}

// C++ -------------------------------------------------------------------------------

// type-definition: 'pt2Member' now can be used as type

typedef int (TMyClass::*pt2Member)(float, char, char);

// illustrate how to work with an array of member function pointers

void Array_Of_Member_Function_Pointers()

{

cout << endl << "Executing 'Array_Of_Member_Function_Pointers'" << endl;

// define arrays and ini each element to NULL, <funcArr1> and <funcArr2> are

// arrays with 10 pointers to member functions which return an int and take

// a float and two char

// first way using the typedef

pt2Member funcArr1[10] = {NULL};

// 2nd way of directly defining the array

int (TMyClass::*funcArr2[10])(float, char, char)
= {NULL};

// assign the function's address - 'DoIt' and 'DoMore' are suitable member

// functions of class TMyClass like defined above in 2.1-4

funcArr1[0] = funcArr2nd use an array of function pointers in C and C++. The first way uses a typedef, the second way directly defines the array. It's up to you which way you prefer.

[1] = &TMyClass::DoIt;

funcArr1[1] = funcArr2[0] = &TMyClass::DoMore;

/* more assignments */

// calling a function using an index to address the member function pointer

// note: an instance of TMyClass is needed to call the member functions

TMyClass instance;

cout << (instance.*funcArr1[1])(12, 'a', 'b')
<< endl;

cout << (instance.*funcArr1[0])(12, 'a', 'b')
<< endl;

cout << (instance.*funcArr2[1])(34, 'a', 'b')
<< endl;

cout << (instance.*funcArr2[0])(89, 'a', 'b')
<< endl;

}