C - 打印行和列

Question

请向我解释如何让程序在每次通过循环增加 col.

时重置 col

我想打印 X 数量的 rows 和 cols 由 LENGTH 控制。

代码显然是一团糟，我不得不展开循环。我的问题是实际上如何正确循环？？？

实际上看起来很简单，但我真的无法理解它。目前，我想要做的就是将输出输出为如下内容；

|----------------------------------------------------------------
| row: 0, col: 0| row: 0, col: 1| row: 0, col: 2| row: 0, col: 3|
|----------------------------------------------------------------
| row: 1, col: 0| row: 1, col: 1| row: 1, col: 2| row: 1, col: 3|
|----------------------------------------------------------------
| row: 2, col: 0| row: 2, col: 1| row: 2, col: 2| row: 2, col: 3|
|----------------------------------------------------------------
| row: 3, col: 0| row: 3, col: 1| row: 3, col: 2| row: 3, col: 3|
|----------------------------------------------------------------

到目前为止它打印出来“一切正常”（或多或少），但我写的太愚蠢了，显然...... 是的，抱歉像个隐士一样编码，并感谢大家花时间！我绝对确定您知道如何做得更好，并且也许可以解释我在这里缺少的内容。

干杯！

float LENGTH    = 0.0f;
int row = 0;
int col =0;

int main(int argc, char** argv)
{
    LENGTH = 8.0f;
    
    
    row = 0;
    col =0;
    
    for (;row < LENGTH; row++)
    {
        for (;col <= LENGTH; col++)
        {
            for (;row < LENGTH; row++)
            {
                for (;col <= LENGTH; col++)
                {
                    printf("| r:%4i , c:%4i ", row, col);
                }
            }
            printf("\n");
            row = 1;
            col = 0;
            for (;row < LENGTH; row++)
            {
                for (;col <= LENGTH; col++)
                {
                    printf("| r:%4i , c:%4i ", row, col);
                }
            }
            printf("\n");
            row = 2;
            col = 0;
            for (;row < LENGTH; row++)
            {
                for (;col <= LENGTH; col++)
                {
                    printf("| r:%4i , c:%4i ", row, col);
                }
            }
            printf("\n");
            row = 3;
            col = 0;
            for (;row < LENGTH; row++)
            {
                for (;col <= LENGTH; col++)
                {
                    printf("| r:%4i , c:%4i ", row, col);
                }
            }
            printf("\n");
            row = 4;
            col = 0;
            for (;row < GRID_LENGTH; row++)
            {
                for (;col <= LENGTH; col++)
                {
                    printf("| r:%4i , c:%4i ", row, col);
                }
            }
            printf("\n");
            row = 5;
            col = 0;
            for (;row < LENGTH; row++)
            {
                for (;col <= LENGTH; col++)
                {
                    printf("| r:%4i , c:%4i ", row, col);
                }
            }
            printf("\n");
            row = 6;
            col = 0;
            for (;row < LENGTH; row++)
            {
                for (;col <= LENGTH; col++)
                {
                    printf("| r:%4i , c:%4i ", row, col);
                }
            }
            printf("\n");
            row = 7;
            col = 0;
            for (;row < LENGTH; row++)
            {
                for (;col <= LENGTH; col++)
                {
                    printf("| r:%4i , c:%4i ", row, col);
                }
            }
            printf("\n");
            row = 8;
            col = 0;
            for (;row <= LENGTH; row++)
            {
                for (;col <= LENGTH; col++)
                {
                    printf("| r:%4i , c:%4i ", row, col);
                }
            }
            printf("\n");
        }
    }
    
    printf("\n");
    return 0;
}

c89 标签在那里，因为程序需要运行在该配置中。

原因是，我的计划是动态分配一些向量。为此，我不想使用可变长度数组。

假设我们想要创建一个网格，如棋盘或棋盘或类似性质的东西。网格应该以世界为中心（x=0，y=0，z=0）。所以，如果网格只有 LENGTH = 1 那么我们仍然需要 4 个向量来绘制 4 条线，至少要得到一个四边形... 但是，如果函数的用户使用

调用它

LENGTH = 1 ,

然后我们需要给每个长度加1 (length +1) 这样当用户使用 LENGTH = 1 调用它时我们实际上从中得到了至少 2x2 个向量。

在下面的代码中，我不确定 vector 是否应该命名为 vector_array，因为我分配了我想要的所有向量。

size_t count = (((length +1) * (length +1)));
float *vector;
vector = NULL;
vector = calloc(count, sizeof(float) * 3);

我的问题是，如何分配向量数组，以及我们从循环中得到的位置，以便 row 和 col 都像下面的 row = -length 一样初始化和 col = -length

请忽略并原谅我，我交替使用了 LENGTH 和 length.. 它们的意思完全相同，但应该使用 LENGTH 将其初始化为 1也许...

    size_t LENGTH = 0;
    
    
    int main(int argc, const char * argv[])
    {
        LENGTH = 8;
        
        float *vector;
        vector = NULL;
        vector = calloc(3, sizeof(vector));
        if (vector == NULL)
        {
            printf("allocation of vector failed!\n");
            exit(0);
        }
        
        float **vector_array;
        vector_array = NULL;
        vector_array = calloc( ((LENGTH+1) * (LENGTH+1)), sizeof(vector));
        if (vector_array == NULL)
        {
            printf("allocation of vector array failed!\n");
            exit(0);
        }
        
        int row = 0;
        int col = 0;
        row = (int)-LENGTH;
        col = (int)-LENGTH;
        
        while (row < (int)LENGTH)
        {
            while (col < (int)LENGTH)
            {
                vector_array[row] = row;
                vector_array[col] = col;
                col++;
            }
            row++;
        }
        
        row = (int)-LENGTH;
        col = (int)-LENGTH;
        while (row <= (int)LENGTH)
        {
            
            while (col <= (int)LENGTH)
            {
                printf("c:%d\n", vector_array[col]);
                col++;
            }
            printf("r:%d\n", vector_array[row]);
            row++;
        }
        return 0;
    }

我无法更好地解释我尝试做的事情。我是初学者，我把学习 C 作为一种爱好...

非常感谢！

Answer 1

您的循环嵌套了四层。这是不必要的。您应该只需要两层嵌套。外循环的每次迭代应该处理一行，内循环的每次迭代应该处理该行的一列。

通常，您不应在 if 语句之外修改循环计数器，因为这可能会使您的代码混乱且难以理解。只有在极少数情况下才适合（这不是其中之一）。

此外，将整数与浮点数进行比较通常不是一个好主意，因为浮点数的精度有限。因此，它们往往不会完全相等，比较就会失败。例如，如果将整数 7 与 6.99999999998 进行比较，则比较结果可能为假，尽管数字实际上相等。在比较浮点值时，通常应允许一定的容差，例如判断两个值之间的差异是否超过某个阈值。

这是循环问题的符合 C89 的解决方案：

#include <stdio.h>

#define NUM_ROWS 4
#define NUM_COLS 4

int main( void )
{
    int row, col;

    for ( row = 0; row < NUM_ROWS; row++ )
    {
        // print first line containing only a static horizontal bar
        printf( "|" );
        for ( col = 0; col < NUM_COLS; col++ )
            printf( "-----------------" );
        printf( "\n" );

        //print second line containing the variable information
        for ( col = 0; col < NUM_COLS; col++ )
            printf( "| row: %i, col: %i ", row, col );

        //print vertical bar at end of line
        printf( "|\n" );
    }

    //print static horizontal bar
    printf( "|" );
    for ( col = 0; col < NUM_COLS; col++ )
        printf( "-----------------" );
    printf( "\n" );

    return 0;
}

以上代码唯一不符合 C89 的是单行注释 (//)。如果您想要严格遵守 C89，您可能必须将它们替换为多行注释（/* 和 */）。但是，大多数 C89 编译器接受单行注释作为语言扩展。

以上代码包含一些代码重复。可以通过从外部循环内部中断来删除此代码重复，如下所示：

#include <stdio.h>

#define NUM_ROWS 4
#define NUM_COLS 4

int main( void )
{
    int row, col;

    //NOTE: Leaving out the middle expression of the "for" statement
    //      will make it always true.
    for ( row = 0; ; row++ )
    {
        // print first line containing only a static horizontal bar
        printf( "|" );
        for ( col = 0; col < NUM_COLS; col++ )
            printf( "-----------------" );
        printf( "\n" );

        //break out of loop, if past last row
        if ( row == NUM_ROWS )
            break;

        //print second line containing the variable information
        for ( col = 0; col < NUM_COLS; col++ )
            printf( "| row: %i, col: %i ", row, col );

        //print vertical bar at end of line
        printf( "|\n" );
    }

    return 0;
}

代码现在更短了（如果您忽略附加注释），但可能更难理解。

这两个程序都打印以下输出：

|--------------------------------------------------------------------
| row: 0, col: 0 | row: 0, col: 1 | row: 0, col: 2 | row: 0, col: 3 |
|--------------------------------------------------------------------
| row: 1, col: 0 | row: 1, col: 1 | row: 1, col: 2 | row: 1, col: 3 |
|--------------------------------------------------------------------
| row: 2, col: 0 | row: 2, col: 1 | row: 2, col: 2 | row: 2, col: 3 |
|--------------------------------------------------------------------
| row: 3, col: 0 | row: 3, col: 1 | row: 3, col: 2 | row: 3, col: 3 |
|--------------------------------------------------------------------

请注意，如果数字大于 10，格式将不再正常工作，因为它们将需要更多 space。

关于如何计算绘制二维网格所需的线的第二个问题：

如果要绘制 10*10 瓷砖网格的线条，则需要绘制 11 条水平线和 11 条垂直线。您可以使用以下逻辑生成这些线的坐标：

#include <stdio.h>

#define PIXELS_PER_TILE 50
#define SIZE_X 10
#define SIZE_Y 10

int main( void )
{
    int i;

    printf( "Processing lines along X axis:\n" );
    for ( i = 0; i <= SIZE_Y; i++ )
    {
        printf(
            "Draw line from [%3d,%3d] to [%3d,%3d]\n",
            0      * PIXELS_PER_TILE, i * PIXELS_PER_TILE,
            SIZE_X * PIXELS_PER_TILE, i * PIXELS_PER_TILE
        );
    }

    printf( "\nProcessing lines along Y axis:\n" );
    for ( i = 0; i <= SIZE_X; i++ )
    {
        printf(
            "Draw line from [%3d,%3d] to [%3d,%3d]\n",
            i * PIXELS_PER_TILE, 0      * PIXELS_PER_TILE,
            i * PIXELS_PER_TILE, SIZE_Y * PIXELS_PER_TILE
        );
    }
}

这是该程序的输出：

Processing lines along X axis:
Draw line from [  0,  0] to [500,  0]
Draw line from [  0, 50] to [500, 50]
Draw line from [  0,100] to [500,100]
Draw line from [  0,150] to [500,150]
Draw line from [  0,200] to [500,200]
Draw line from [  0,250] to [500,250]
Draw line from [  0,300] to [500,300]
Draw line from [  0,350] to [500,350]
Draw line from [  0,400] to [500,400]
Draw line from [  0,450] to [500,450]
Draw line from [  0,500] to [500,500]

Processing lines along Y axis:
Draw line from [  0,  0] to [  0,500]
Draw line from [ 50,  0] to [ 50,500]
Draw line from [100,  0] to [100,500]
Draw line from [150,  0] to [150,500]
Draw line from [200,  0] to [200,500]
Draw line from [250,  0] to [250,500]
Draw line from [300,  0] to [300,500]
Draw line from [350,  0] to [350,500]
Draw line from [400,  0] to [400,500]
Draw line from [450,  0] to [450,500]
Draw line from [500,  0] to [500,500]

如果要绘制 10*10*10 平铺网格（即 3D 网格）的线，则需要沿 X 轴绘制 11*11 条线，11*11 行沿 Y 轴，11*11 行沿 Z 轴（共 363 行）。您可以使用以下逻辑生成这些线的坐标：

#include <stdio.h>

#define PIXELS_PER_TILE 50
#define SIZE_X 10
#define SIZE_Y 10
#define SIZE_Z 10

int main( void )
{
    int i;
    int j;

    printf( "Processing lines along X axis:\n" );
    for ( i = 0; i <= SIZE_Y; i++ )
    {
        for ( j = 0; j <= SIZE_Z; j++ )
        {
            printf(
                "Draw line from [%3d,%3d,%3d] to [%3d,%3d,%3d]\n",
                0      * PIXELS_PER_TILE, i * PIXELS_PER_TILE, j * PIXELS_PER_TILE,
                SIZE_X * PIXELS_PER_TILE, i * PIXELS_PER_TILE, j * PIXELS_PER_TILE
            );
        }
    }

    printf( "\nProcessing lines along Y axis:\n" );
    for ( i = 0; i <= SIZE_X; i++ )
    {
        for ( j = 0; j <= SIZE_Z; j++ )
        {
            printf(
                "Draw line from [%3d,%3d,%3d] to [%3d,%3d,%3d]\n",
                i * PIXELS_PER_TILE, 0      * PIXELS_PER_TILE, j * PIXELS_PER_TILE,
                i * PIXELS_PER_TILE, SIZE_Y * PIXELS_PER_TILE, j * PIXELS_PER_TILE
            );
        }
    }

    printf( "\nProcessing lines along Z axis:\n" );
    for ( i = 0; i <= SIZE_X; i++ )
    {
        for ( j = 0; j <= SIZE_Y; j++ )
        {
            printf(
                "Draw line from [%3d,%3d,%3d] to [%3d,%3d,%3d]\n",
                i * PIXELS_PER_TILE, j * PIXELS_PER_TILE, 0      * PIXELS_PER_TILE,
                i * PIXELS_PER_TILE, j * PIXELS_PER_TILE, SIZE_Z * PIXELS_PER_TILE
            );
        }
    }
}

我不会post第二个程序的输出，因为它会打印几百行输出。

C - 打印行和列

C - printing row and column

c

c89