向量的uint8_t值问题

Question

我正在尝试创建一个带有uint8_t值的10 x 10网格。使用一些坐标，我尝试在这些索引处放置255个值。当我在这些坐标上替换255时，它工作得很好，但是当我试图打印出这些索引上的值时，它似乎不是我想要的值。

float grid_width = 10;
float grid_height = 10;
float grid_resolution = 0.05;
float rows = grid_width/grid_resolution;
float cols = grid_height/grid_resolution;

std::vector<std::vector<float>> landmarks = {{2,2},{8,8},{2,8},{8,2},{3,2},{3.5,2.1},{4.0,3.5},{4.5,2.0},{5.0,3.0}};
std::vector<std::vector<uint8_t>> grid(rows, std::vector<uint8_t>(cols, 0));

for(auto i: landmarks)
{
    for(auto m = i[0]-(grid_resolution*2) ; m <= i[0] +(grid_resolution*2) ; m=m+grid_resolution )
    {
        for(auto n = i[1]-(grid_resolution*2); n <= i[1]+(grid_resolution*2); n=n+grid_resolution)
        {
            int m_co = m/grid_resolution;
            int n_co = n/grid_resolution;
            grid[m_co][n_co] = 255;
            cout<<"m_co: "<< m_co <<" n_co:" << n_co <<endl;
            cout<<"grid[m_co][n_co]: "<< unsigned(grid[m_co][n_co])  <<endl; //check the printout over here for index values of 18,22 you will see 255 but when you try it outside the for loops those values are rest to 0
            
        }
        
    }
}

// for(auto m: grid)
// {
//     for(auto n: m)
//     {
//         cout<<unsigned(n)<<" ";
//     }
//     cout<<endl;
// }
cout<<unsigned(grid[18][18])<<endl; //expected 255 got 255
cout<<unsigned(grid[18][22])<<endl; //expected 255 got 0
cout<<unsigned(grid[22][22])<<endl; //expected 255 got 0
cout<<unsigned(grid[22][18])<<endl; //expected 255 got 0
cout<<unsigned(grid[17][17])<<endl; //expected 255 got 0
cout<<unsigned(grid[20][18])<<endl; //expected 255 got 255

因此，我使用地标向量来获取索引，并尝试在地标的索引上将网格向量中的0替换为255，但我也想在该索引周围的索引上替换0。例如，如果2,2是我的坐标，我想替换1,1和1, 2 ,3和2,1和2，以及2,3和3,1和3,2和3,3到255，但我得不到1,3和2，3,3,255我得到的是0这只是我尝试实现的一个示例

Answer 1

我找到了解决方案。旧的逻辑太复杂了，不能与不同的分辨率一起工作。所以我做了一个更简单的，它确实需要处理四舍五入的问题。首先获取每个点，并尝试用索引替换值。

for(auto i: landmarks)
{
    for(int m = (i[0]/grid_resolution)-2; m <= (i[0]/grid_resolution)+2; m++)
    {
        for(int n = (i[1]/grid_resolution)-2; n <= (i[1]/grid_resolution)+2; n++)
        {
            grid[m][n] = 255;
        }
    }
}

Answer 2

在循环中，您打印的是n/grid_resolution。当它打印22时，它实际上是21.99999809265，但是因为iostream的默认精度是6，所以它四舍五入为22。另一方面，当您将它赋给int时，它会向下舍入为21，因此您实际上将255存储在grid[18][21]中。

Answer 3

问题是m/grid_resolution的类型是float，而m_co的类型是int：

int main()
{
    float grid_width = 10;
    float grid_height = 10;
    float grid_resolution = 0.1;
    float rows = grid_width/grid_resolution;
    float cols = grid_height/grid_resolution;
    cout.precision(5);
    
    std::vector<std::vector<float>> landmarks = {{2,2},{8,8},{2,8},{8,2},{3,2},{3.5,2.1},{4.0,3.5},{4.5,2.0},{5.0,3.0}};
    std::vector<std::vector<uint8_t>> grid(rows, std::vector<uint8_t>(cols, 0));
    
    for(auto i: landmarks)
    {
        for(auto m = i[0]-(grid_resolution*2) ; m <= i[0] +(grid_resolution*2) ; m=m+grid_resolution )
        {
            for(auto n = i[1]-(grid_resolution*2); n <= i[1]+(grid_resolution*2); n=n+grid_resolution)
            {
                int m_co = m/grid_resolution;
                int n_co = n/grid_resolution;
                grid[m_co][n_co] = 255;
                // Uncomment the following two lines for showing that are different
                //if (m_co != m/grid_resolution || n_co != n/grid_resolution)
                //    cout << "different" << endl;
                cout << "grid["<<m_co<<"]["<<n_co<<"]: "<< unsigned(grid[m_co][n_co]) << endl;
            }
            
        }
    }
    
    // for(auto m: grid)
    // {
    //     for(auto n: m)
    //     {
    //         cout<<unsigned(n)<<" ";
    //     }
    //     cout<<endl;
    // }
    cout<<"grid[18][18]" << unsigned(grid[18][18])<<endl;
    cout<<"grid[18][22]" << unsigned(grid[18][22])<<endl;
    cout<<"grid[22][22]" << unsigned(grid[22][22])<<endl;
    cout<<"grid[22][18]" << unsigned(grid[22][18])<<endl;
    cout<<"grid[17][17]" << unsigned(grid[17][17])<<endl;
    cout<<"grid[20][18]" << unsigned(grid[20][18])<<endl;
    return 0;
}

如果运行该代码，将打印m_co和n_co的正确值。

诚挚的问候。