rSub(10,5)
通过变换bind 函数包装器 中placeholders::_1, placeholders::_2
,可以实现10-5&5-10
double Plus(int a, int b, double rate)
{
return (a + b) * rate;
}
int main()
{
function<double(int, int)> Plus1 = bind(Plus, placeholders::_1, placeholders::_2, 4.0);
function<double(int, int)> Plus2 = bind(Plus, placeholders::_1, placeholders::_2, 4.2);
function<double(int, int)> Plus3 = bind(Plus, placeholders::_1, placeholders::_2, 4.4);
cout << Plus1(5, 3) << endl;
cout << Plus2(5, 3) << endl;
cout << Plus3(5, 3) << endl;
return 0;
}
double Plus(int a, double rate,int b)
{
return (a + b) * rate;
}
int main()
{
function<double(int, int)> Plus1 = bind(Plus, placeholders::_1, 4.0 placeholders::_2);
function<double(int, int)> Plus2 = bind(Plus, placeholders::_1, 4.2,placeholders::_2);
function<double(int, int)> Plus3 = bind(Plus, placeholders::_1,4.4 ,placeholders::_2);
cout << Plus1(5, 3) << endl;
cout << Plus2(5, 3) << endl;
cout << Plus3(5, 3) << endl;
return 0;
}
主要方法分为下面三种:
&SubType::sub
&SubType::sub
,法一:先实例化出一个类SubType st;
,取其地址&st
&SubType::sub
,法二:直接传入一个匿名对象SubType()
class SubType
{
public:
static int sub(int a, int b)
{
return a - b;
}
int ssub(int a, int b, int rate)
{
return (a - b) * rate;
}
};
int main()
{
//对于静态成员函数
function<double(int, int)> Sub1 = bind(&SubType::sub, placeholders::_1, placeholders::_2);
cout << Sub1(1, 2) << endl;
//对于非静态成员函数,法一
SubType st;
function<double(int, int)> Sub2 = bind(&SubType::ssub, &st, placeholders::_1, placeholders::_2, 3);
cout << Sub2(1, 2) << endl;
//对于非静态成员函数,法二
function<double(int, int)> Sub3 = bind(&SubType::ssub, SubType(), placeholders::_1, placeholders::_2, 3);
cout << Sub3(1, 2) << endl;
return 0;
}