如何在C ++中使用嵌套向量? [英] How do I work with nested vectors in C++?

问题描述

我正在尝试为我正在编写的数独难题求解器使用整数向量的向量.

I'm trying to work with vectors of vectors of ints for a sudoku puzzle solver I'm writing.

问题1:

如果我要按索引访问我的2d向量，是否必须先用适当的大小对其进行初始化?

If I'm going to access a my 2d vector by index, do I have to initialize it with the appropriate size first?

例如:

typedef vector<vector<int> > array2d_t;

void readAPuzzle(array2d_t grid)
{
    for(int i = 0; i < 9; i++)
        for(int j = 0; j < 9; j++)
            cin >> grid[i][j];
    return;
}

int main()
{
    array2d_t grid;
    readAPuzzle(grid);
}

将出现段故障.我认为这是因为它正在尝试访问尚未初始化的网格元素?

Will seg fault. I assume this is because it is trying to access elments of grid that have not yet been initialized?

我已将网格的声明行换成:

I've swapped out grid's declaration line with:

array2d_t grid(9, vector<int>(9, 0));

这似乎摆脱了这个段错误.这是正确的处理方式吗?

And this seems to get rid of this seg fault. Is this the right way to handle it?

问题2:

为什么当我尝试从cin读取网格并打印出网格时，网格为空白?

Why is it that when I try to read into my grid from cin, and then print out the grid, the grid is blank?

我正在使用以下代码来这样做:

I'm using the following code to do so:

void printGrid(array2d_t grid)
{
    for (int i = 0; i < 9; i++)
    {
        for (int j = 0; j < 9; j++)
        {
            cout << grid[i][j] + " ";
        }
        cout << endl;
    }
}

void readAPuzzle(array2d_t grid)
{
    for(int i = 0; i < 9; i++)
        for(int j = 0; j < 9; j++)
            cin >> grid[i][j];

    return;
}

int main()
{
    array2d_t grid(9, vector<int>(9, 0));
    printGrid(grid);
    readAPuzzle(grid);
    printGrid(grid);
}

然后我尝试运行程序，如下所示:

And I attempt to run my program like:

./a.out < sudoku-test

sudoku-test是一个包含以下内容的文件:

Where sudoku-test is a file containing the following:

3 0 0 0 0 0 0 0 0  
5 8 4 0 0 2 0 3 0  
0 6 0 8 3 0 0 7 5  
0 4 1 0 0 6 0 0 0  
7 9 0 0 2 0 0 5 1  
0 0 0 9 0 0 6 8 0  
9 3 0 0 1 5 0 4 0  
0 2 0 4 0 0 5 1 8  
0 0 0 0 0 0 0 0 6  

第一次调用printGrid()会给出一个空白网格，相反，我应该看到一个0的9x9网格，因为这是我初始化它的方式.第二个调用应包含上面的网格.但是，两次都为空.

The first call to printGrid() gives a blank grid, when instead I should be seeing a 9x9 grid of 0's since that is how I initialized it. The second call should contain the grid above. However, both times it is blank.

任何人都可以对此有所了解吗?

Can anyone shed some light on this?

推荐答案

Q1:是的，这是处理它的正确方法.但是，请注意，嵌套向量是实现2D数组的一种相当低效的方法.一个向量并按 x + y * width 计算索引通常是一个更好的选择.

Q1: Yes, that is the correct way to handle it. However, notice that nested vectors are a rather inefficient way to implement a 2D array. One vector and calculating indices by x + y * width is usually a better option.

Q2A:计算 grid [i] [j] +" 不会连接两个字符串(因为左侧是int，而不是字符串)，而是将数字值添加到指针(字符串"的第一个字符的内存地址).使用 cout<<grid [i] [j]<" 代替.

Q2A: Calculating grid[i][j] + " " does not concatenate two strings (because the left hand side is int, not a string) but instead adds the numeric value to a pointer (the memory address of the first character of the string " "). Use cout << grid[i][j] << " " instead.

Q2B:您正在按值传递数组(它被复制)用于 readAPuzzle .该函数读入其本地副本，该本地副本在函数返回时被销毁.而是通过引用传递(这样避免了复制，而改用原始的):

Q2B: You are passing the array by value (it gets copied) for readAPuzzle. The the function reads into its local copy, which gets destroyed when the function returns. Pass by reference instead (this avoids making a copy and uses the original instead):

void readAPuzzle(array2d_t& grid)

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