c++ - Matlab vs C++找到等于的Matrix的第一个元素

Question

为什么此算法在Matlab中不起作用，而在C++中却起作用？

 int a[3][4] = {  
   {0, 1, 1, 1} ,   /*  initializers for row indexed by 0 */
   {1, 0, 0, 0} ,   /*  initializers for row indexed by 1 */
   {0, 0, 0, 1}   /*  initializers for row indexed by 2 */
};

 int x,y;   
        for(int i=1;i<3;i++)
            for(int j=1;j<4;j++)
            {
                if (a[i][j]==1) // if element is equal to 1, get his position and stop.
                {
                    x=i-1;
                    y=j-1;
                   break; 
                }
            }
        cout<<x<<" "<<y; // 1 2

如果我想在Matlab中做这样的事情，输出是不一样的(甚至不接近)。实际上，它根本不起作用。

Matlab代码:

for i=1:m
    for j=1:n
        if(a(i,j)==1)
            k=i-1;
            l=j-1;
            break;
        end
    end
end

我该怎么办？

Answer 1

并不是说break被忽略了，它只是跳出了内循环，而只是在外循环中继续。恐怕没有“两次破门”的指示或类似的东西。一种解决方案是让一个标志在外循环中也中断:

int x, y;
bool stop = false;
for (int i = 0; i < 3; i++)
{
    for (int j = 0; j < 4; j++)
    {
        if (a[i][j] == 1) // if element is equal to 1, get his position and stop.
        {
            x = i - 1;
            y = j - 1;
            stop = true;
            break;
        }
    }
    if (stop)
        break;
}

这是极少数情况下的一种，您可以放置​​ goto来减少麻烦:

int x, y;
for (int i = 0; i < 3; i++)
{
    for (int j = 0; j < 4; j++)
    {
        if (a[i][j] == 1) // if element is equal to 1, get his position and stop.
        {
            x = i - 1;
            y = j - 1;
            goto finishNestedLoop; // used as a "break twice"
        }
    }
}
finishNestedLoop:;

其他解决方案包括将嵌套循环放入其自己的函数中，并改为使用 return:

#include <iostream>
#include <tuple>
std::tuple<int, int> getXY() {
    int a[3][4] = {
       {0, 1, 1, 1} ,   /*  initializers for row indexed by 0 */
       {1, 0, 0, 0} ,   /*  initializers for row indexed by 1 */
       {0, 0, 0, 1}   /*  initializers for row indexed by 2 */
    };
    for (int i = 0; i < 3; i++)
    {
        for (int j = 0; j < 4; j++)
        {
            if (a[i][j] == 1) // if element is equal to 1, get his position and stop.
                return std::tuple<int, int>{i-1, j-1};
        }
    }
}

int main() {
    auto[x, y] = getXY();
    std::cout << x << " " << y;
}

另外，您确定有关 i - 1吗？要从0索引到1索引，您需要 i + 1。