关于嵌套的for循环

Regarding nested for loops

本文关键字：for 循环 嵌套 于嵌套      更新时间：2023-10-16

我有一个赋值使用嵌套的for循环打印出5行3列二维数组的索引，而不使用数组。

#include <iostream>
using namespace std;
int main()
{
    int row;
    int column;
    cout << "Counting with nested loops: indices of row, column" << endl << endl;
    for (row = 1; row <= 5; row++)
    {   
        for (column = 1; column <= 3; column++)
        {
        }
        cout << row << "," << column << "t" << row << "," << column << "t" << row << "," << column << "t" << endl;
    }
    cout << "nnn" << endl;
    return 0;
}

这是我到目前为止的代码。我的目标是打印

1,1,1,2,1,3

2、1、2、2、3

等等。当我运行程序时，它打印

1,4 1,4 1,4

2,4 2,4 2,4

所以行部分是正确的。谁能帮我找出我的错误是什么?

你只需要调用一次print，但是在内部循环中像这样:

for (row = 1; row <= 5; row++){
    for (column = 1; column <= 3; column++){
        std::cout << row << "," << column << "t";
    }
    std::cout << std::endl;
}

这是你必须写的代码:

int main() {
    int row;
    int column;
    cout << "Counting with nested loops: indices of row, column" << endl << endl;
    for (row = 1; row <= 5; row++) {
        for (column = 1; column <= 3; column++) {
            cout << row << "," << column << "t";
        }
        cout << endl;
    }
    cout << "nnn" << endl;
    return 0;
}

这是因为你必须在第二个循环里面打印而不是在外面。

其他人已经回答了这个问题，但我想添加一些建议，给将来考虑的事情。

首先在你引入循环变量的地方，例如int row;，你还没有初始化它们，只需要在for循环中声明它们。然后它们将不会在for循环之外可见，这将避免在循环完成后(通过cout)打印的错误。

也值得考虑循环计数器。通常从0开始。此外，数组的下标将从0开始，一直到n-1。所以你应该考虑打印0,1，…而不是1,2，…如果你想用它来遍历数组，

#include <iostream>
using namespace std;
int main()
{
    cout << "Counting with nested loops: indices of row, column" << endl << endl;
    for (int row = 0; row < 5; row++) //<--- note start at 0 and stop BEFORE n
    {   
        for (int column = 0; column < 3; column++) //<--- ditto
        {
            cout << row << "," << column << "t";
        }
        //cout << row << "," << column << "t" 
               << row << "," << column << "t"
               << row << "," << column << "t" << endl;
                 // Now compile error
                 //  And clearly was going to just give the same value 
                 //  for row and column over and over
        cout << 'n';//added to break each row in the output
    }
    cout << "nnn" << endl;
    return 0;
}
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