Lambda用作类成员

Question

是否有可能接受两种不同类型的lambda函数作为类成员而不知道他们的模板参数提前？

struct two_functors {
    std::function<???> a;
    std::function<???> b;
    ...
};

这样的事情是可能的:

void main(){
    vector<two_functors> many_functors;

    int a = 2;
    int b = 3;
    double c = 4.7;
    double d = 8.4;

    two_functors add_and_subtract;
    add_and_subtract.a = [a, b](int x, int y){cout << x + y << endl;};
    add_and_subtract.b = [c, d](double x, double y){cout << x - y << endl;};

    two_functors multiply_and_divide;
    multiply_and_divide.a = [c, d](double x, double y){cout << x * y << endl;};
    multiply_and_divide.b = [a, b](int x, int y){cout << x / y << endl;};

    many_functors.push_back(add_and_subtract);
    many_functors.push_back(multiply_and_divide);

    for (auto functors : many_functors){
        functors.a();
        functors.b();
    }

}

Answer 1

如果您只想two_functors在不同时间构建，但稍后按顺序一次执行它们，则可以仅使用捕获的数据。

struct two_functors
{
    function<void ()> a;
    function<void ()> b;
};

int main()
{
    vector<two_functors> many_functors;

    int a = 2;
    int b = 3;
    double c = 4.7;
    double d = 8.4;

    two_functors add_and_subtract {
        [a, b](){cout << a + b << endl;},
        [c, d](){cout << c - d << endl;}
    };

    two_functors multiply_and_divide {
        [c, d](){cout << c * d << endl;},
        [a, b](){cout << a / b << endl;}
    };

    many_functors.push_back(add_and_subtract);
    many_functors.push_back(multiply_and_divide);

    for (auto functors : many_functors){
        functors.a();
        functors.b();
    }
}