将文件中的值加载到2D数组指针的值中 [英] Load value from file to a value of 2D array pointer

问题描述

如何在另一个函数中将文件中的值正确加载到矩阵中?

How do I correctly load the values from file to the matrix in another function?

void someFunction(int ***matrix, n, m) {
    int c, d;
    FILE *fp = fopen("some.txt", "r"); // načtení souboru
    for (c = 0; c < m; c++) {
        for (d = 0; d < n; d++) {
            fscanf(fp, "%i", &matrix[c][d]); //4th read throws error
        }
    }
}

int main() {
    int i;
    int **matrix;
    int n = 3; // columns
    int m = 2; // rows

    first = (int **)malloc(n * sizeof(int*));
    for (i = 0; i < n; i++)
        first[i] = (int *)malloc(m * sizeof(int));

    someFunction(&matrix, n, m);

    free(matrix);
    for (i = 0; i < n; i++)
        free(first[i]);
}

fscanf(fp, "%i", &matice[c][d])对吗?

它在[1][0]处失败(内存错误)，但[0][0],[0][1],[0][2]可以正常工作.我不确定自己做错了什么.

It fails (memory error) at [1][0] but [0][0],[0][1],[0][2] works fine. I am not sure what I did wrong.

推荐答案

您有一个指针数组，每个元素都设置为指向动态分配的块.这不是2D数组，尽管您可以使用与2D数组相同的语法对整个数据结构进行索引.

You have an array of pointers, each element set to point to a dynamically-allocated block. This is not a 2D array, though you can index into the overall data structure via the same syntax that you would use for a 2D array.

您至少有两个问题:

1. 根据您的分配策略，数组的第一个索引从0到n - 1，指向指针的数组的索引从0到m - 1.在n和m不同的情况下，您可以在读取循环中反转该值，从而超出至少一个数组的边界.

1. By your allocation strategy, the first index of your array runs from 0 to n - 1, and the indexes into the pointed-to arrays run from 0 to m - 1. You reverse that in your read loop, thereby overrunning the bounds of at least one of your arrays if n and m differ.

更重要的是，您传递给fscanf()的存储指针是错误的.在someFunction()中，变量matrix的类型为int ***，并且您传递了int **的地址.因此，索引为[c] [d]的元素为(*matrix)[c][d]，该元素的地址为&(*matrix)[c][d]，它与&matrix[c][d]完全不同，甚至与matrix[c][d].

More significantly, the storage pointer you pass to fscanf() is wrong. In someFunction(), variable matrix has type int ***, and you pass an address of your int **. The element at index [c][d] is therefore (*matrix)[c][d], and the address of that element is &(*matrix)[c][d], which is not at all the same thing as &matrix[c][d], nor even the same as matrix[c][d].

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