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: 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
// 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.
// 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. 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, and 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, and 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;
}
