我可以为数组的所有成员添加一个值吗

Question

STL 中是否有一种算法可以一次向数组的所有成员添加相同的值？

例如：

KnightMoves moveKnight(int currentPossition_x, int currentPossition_y)
{
    array<int , 8> possibleMoves_x = { -2 , -2 , -1 , -1 ,  1 , 1 , 2 ,  2 };
    array<int , 8> possibleMoves_y = { -1 ,  1 , -2 ,  2 , -2 , 2 , -1 , 1 };

    for (auto it = possibleMoves_x.begin(); it != possibleMoves_x.end(); it++)
    {
        array <int, 8> newTempKnightPoss_x = currentPossition_x + possibleMoves_x;

        array <int, 8> newTempKnightPoss_y = currentPossition_y + possibleMoves_x;
    }

}

我可以做这样的事情，但我希望有更好的解决方案

KnightMoves moveKnight(int currentPossition_x, int currentPossition_y)
{
   array<int , 8> possibleMoves_x = { -2 , -2 , -1 , -1 ,  1 , 1 , 2 ,  2 };
   array<int , 8> possibleMoves_y = { -1 ,  1 , -2 ,  2 , -2 , 2 , -1 , 1 };

   for (auto it = possibleMoves_x.begin(); it != possibleMoves_x.end(); it++)
   {
       *it = *it  +currentPossition_x;

   }
   for (auto it = possibleMoves_y.begin(); it != possibleMoves_y.end(); it++)
   {
       *it = *it + currentPossition_y;

   }
}

axpepted 的结果是 2 个数组，其中每个元素是元素加上一个常量值；

Answer 1

如果你有 C++11，你可以使用基于范围的 for循环：

KnightMoves moveKnight(int currentPossition_x, int currentPossition_y){
    array<int , 8> possibleMoves_x = { -2 , -2 , -1 , -1 ,  1 , 1 , 2 ,  2 };
    array<int , 8> possibleMoves_y = { -1 ,  1 , -2 ,  2 , -2 , 2 , -1 , 1 };

    for(auto& i : possibleMoves_x){ i += currentPossition_x; }
    for(auto& i : possibleMoves_y){ i += currentPossition_y; }
}

在 C++11 之前，您可以使用std::for_each ：

struct adder{
    adder(int val): v{val}{}
    void operator()(int& n) { n += v; }
    int v;
};

KnightMoves moveKnight(int currentPossition_x, int currentPossition_y){
    array<int , 8> possibleMoves_x = { -2 , -2 , -1 , -1 ,  1 , 1 , 2 ,  2 };
    array<int , 8> possibleMoves_y = { -1 ,  1 , -2 ,  2 , -2 , 2 , -1 , 1 };

    std::for_each(possibleMoves_x.begin(), possibleMoves_x.end(), 
                  adder(currentPossition_x));
    std::for_each(possibleMoves_y.begin(), possibleMoves_y.end(),
                  adder(currentPossition_x));
}