如何在 C 中采用格式化输入

Question

我必须为 c 中的一个数组输入输入。示例输入将是这样的 -

33 33 2
000000000000000000000010000000000
000000000000000000000111000000000
000000000000000000001110100000000
000000000000000000011111111000000
000000000000000000001011100000000
000000000000000000000111000000000
000000000000000000000010000000000
000000000000000000000000000000000
000000000000000000000000000000000
000000000000000000000000000000000
000000000000000000000000000000000
000000000000000000000000000000000
000000000000000000000000000000000
000000000000000000000000000000000
000000000000000000000000000000000
011111111111110000000000000000000
011011111111110000000000000000000
011111111110110000000000000000000
011111111111110000000000000000000
011111111111110000000000000000000
011111011111110000000000000000000
000000000000000000000000000000000
000000000000000000000000000000000
000000000000000000000000000000000
000000000000000000000000000000000
000000000000000000000000000000000
000000000000000000000000000000000
000000000000000000000000000000000
000000000000000000000000000000000
000000000000000000000000000000000
000000000000000000000000000000000
000000000000000000000000000000000
000000000000000000000000000000000

这里，33 和 33 是​​ m 行和 n 列。 2 是整数网格中的模式数。而其余的是矩阵。

我将如何将这个矩阵输入到 C 中的数组中。很长一段时间以来一直存在问题。这是我当前的代码。

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

int main(int argc, char *argv[]) 
{
    int **array;

    int m = atoi(argv[1]);
    int n = atoi(argv[2]);

    array = calloc(m, sizeof(int *));
    for (int i = 0; i < m; i++) {
        array[i] = calloc(n, sizeof(int));
    }

    for (int i = 0; i < m; i++) {
        for (int j = 0; j < n; j++)
        {
            /*Code to get input from*/
        }
        
    }

    for (int i = 0; i < m; i++) {
        free(array[i]);
    }

    free(array);

    return 0;
}

Answer 1

假设您正在使用类似的方式通过命令行传递数据（您的示例输入）

./program $(<data.txt)

您可以通过以下方式获取输入：

for (int i = 0; i < m; i++) {
    for (int j = 0; j < n; j++) {
        array[i][j] = argv[4 + i][j]; // Get the value as character
    }
}

然后打印

for (int i = 0; i < m; i++) {
    for (int j = 0; j < n; j++) {
        printf("%c", array[i][j]); // Print the character
    }
    printf("\n");
}

或（正如 @tshiono 在 cmets 中指出的那样）

for (int i = 0; i < m; i++) {
    for (int j = 0; j < n; j++) {
        array[i][j] = argv[4 + i][j] - '0'; // Get the value as integer
    }
}

并使用打印

for (int i = 0; i < m; i++) {
    for (int j = 0; j < n; j++) {
        printf("%d", array[i][j]); // Print the integer
    }
    printf("\n");
}

我会使用有意义的名字

int rows = atoi(argv[1]);
int cols = atoi(argv[2]);

而不是

int m = atoi(argv[1]);
int n = atoi(argv[2]);

和

for (int row = 0; row < rows; row++) {
    for (int col = 0; col < cols; col++) {

而不是

for (int i = 0; i < m; i++) {
    for (int j = 0; j < n; j++) {