Question

如何使用数组指针（int *）创建和打印这种二维数组：

0 1 2 3 4 5 6 7 8...
1 2 3 4 5 6 7 8 9...
2 3 4 5 6 7 8 9 10...
3 4 5 6 7 8 9 ...
4 5 6 7 8 9...
5...

我目前有这段代码：

#include <stdio.h>
#include <stdlib.h>

void createMatrix(int * matrix, int row, int column)
{
    puts("Begin creating matrix.");
    int i, j;
    for(i = 0; i < row; i++)
    {
        for(j = 0; j < column; j++)
        {
            *(matrix) = i + j;
            matrix++;
        }
    }
    puts("Success!");
}

int printMatrix(int * matrix, int row, int column)
{
    puts("Begin printing matrix!");
    int i, j, valuesPrinted;
    for(i = 0; i < row; i++)
    {
        for(j = 0; j < column; j++)
        {
            printf("%5d ", *(matrix + i * j));
            valuesPrinted++;
        }
        puts("");
    }
    puts("Printing success!");
    return valuesPrinted;
}

int main()
{
    int row = 11;
    int column = 13;
    int * matrix = malloc(sizeof(int) * row * column);

    createMatrix(matrix, row, column);
    printMatrix(matrix, row, column);
    return 0;   
}

我想要做的是：使用declare数组并赋值给指针，然后将指针传递给函数并给它如上所述的值。我认为问题在于我通过指针初始化2D数组并访问它。

结果是：

    0     0     0     0     0     0     0     0     0     0     0     0     0
    0     1     2     3     4     5     6     7     8     9    10    11    12
    0     2     4     6     8    10    12     2     4     6     8    10    12
    0     3     6     9    12     3     6     9    12     3     6     9    12
    0     4     8    12     4     8    12     4     8    12     4     8    12
    0     5    10     3     8    13     6    11     4     9    14     7    12
    0     6    12     6    12     6    12     6    12     6    12     6    12
    0     7     2     9     4    11     6    13     8    15    10    17    12
    0     8     4    12     8     4    12     8    16    12     8    16    12
    0     9     6     3    12     9     6    15    12     9    18    15    12
    0    10     8     6     4    14    12    10     8    18    16    14    12

Answer 1

您的打印代码使用错误的表达式来访问矩阵的元素。你有：

printf("%5d ", *(matrix + i * j));

你需要：

printf("%5d ", *(matrix + i * column + j));

或者：

printf("%5d ", matrix[i * column + j]);

Answer 2

int *matrix = malloc(sizeof(int) * row * column);
// could better be written as
int *matrix = malloc((row*column) * sizeof *matrix);

上述语句动态分配大小为row * column的整数数组。如果要访问此数组的元素，就像访问2D数组的元素一样，则行i和列j的元素具有索引

i * column + j

将printMatrix功能修改为 -

int printMatrix(int * matrix, int row, int column)
{
    puts("Begin printing matrix!\n");
    int i, j, valuesPrinted;
    for(i = 0; i < row; i++)
    {
        for(j = 0; j < column; j++)
        { 
            // this is equivalent and more readabable
            // printf("%5d ", matrix[i*column + j]);
            printf("%5d ", *(matrix + i*column + j));
                                      //  ^ note i is multiplied by column
            valuesPrinted++;
        }
        puts("");
    }
    puts("Printing success!\n");
    return valuesPrinted;
}

Answer 3

printf("%5d ", *(matrix + i * column + j));

而不是

printf("%5d ", *(matrix + i * j));

声明和访问2D数组

3 个答案: