在for_each上使用仿函数

Question

为什么for_each对仿函数的调用sum::total最终不会更新？

struct sum
{
    sum():total(0){};
    int total;

    void operator()(int element) 
    { 
       total+=element; 
    }
};

int main()
{
    sum s;

    int arr[] = {0, 1, 2, 3, 4, 5};
    std::for_each(arr, arr+6, s);
    cout << s.total << endl; // prints total = 0;
}

Answer 1

for_each通过值获取仿函数 - 因此它被复制.您可以使用例如使用指向外部int的指针初始化的仿函数.

struct sum
{
    sum(int * t):total(t){};
    int * total;

    void operator()(int element)
    {
       *total+=element;
    }
};

int main()
{
    int total = 0;
    sum s(&total);

    int arr[] = {0, 1, 2, 3, 4, 5};
    std::for_each(arr, arr+6, s);
    cout << total << endl; // prints total = 15;
}

或者您可以使用返回值 for_each

struct sum
{
    sum():total(0){};
    int total;

    void operator()(int element) 
    { 
       total+=element; 
    }
};

int main()
{
    sum s;

    int arr[] = {0, 1, 2, 3, 4, 5};
    s = std::for_each(arr, arr+6, s);
    cout << s.total << endl; // prints total = 15;
}