函数指针几乎可以有任何返回类型。考虑这个例子：
#include "stdafx.h"
#include
using namespace std;
// this defines a type called MathOp that is a function pointer returning
// an int, that takes 2 int arguments
typedef int (*MathOp)(int, int);
enum MathOpType
{
Add = 1,
Subtract = 2
};
// some generic math operation to add two numbers
int AddOperation(int a, int b)
{
return a+b;
}
// some generic math operation to subtract two numbers
int SubtractOperation(int a, int b)
{
return a-b;
}
// function to return a math operation function pointer based on some argument
MathOp GetAMathOp(MathOpType opType)
{
if (opType == MathOpType::Add)
return &AddOperation;
if (opType == MathOpType::Subtract)
return &SubtractOperation;
return NULL;
}
int _tmain(int argc, _TCHAR* argv[])
{
// declare a variable with the type MathOp, which is a function pointer to
// a function taking two int arguments, and returning an int.
MathOp op = &AddOperation;
std::cout << op(2, 3) << std::endl;
// switch the operation we want to perform by calling our one op variable
op = &SubtractOperation;
std::cout << op(2, 3) << std::endl;
// just an example of using a function that returns a function pointer
std::cout << GetAMathOp(MathOpType::Subtract)(5, 1) << std::endl;
std::getchar();
return 0;
}
上面的程序打印出 5、-1，然后是 4。函数指针可以有任何返回类型，并且非常强大。