Question

我正在从一个简单的数据文件'matrix.dat'中读取数据，例如：

1 2 3
4 5 6
7 8 9

我通读文件并将数据点保存在2D数组中：

else /// Read data from file.
{
    // Check matrix is square and find number of columns/rows.
    n = CheckSquare(matrixData); /// n = dimension of square matrix
    fprintf(stdout, "n = %d\n", n); /// TESTING ///

    if(n != 0) /// Matrix is square.
    {
        int i, j; /// i = rows, j = columns.
        double MATRIX[n - 1][n - 1]; /// Declare 2D array of size n x n.

        // Loop through file "matrix.dat" and store data into 2D array "MATRIX".
        for(i = 0; i < n; i++) /// Loop rows.
        {
            for(j = 0; j < n; j++) /// Loop columns.
            {
                fscanf(matrixData, "%lf", &MATRIX[i][j]); /// Store data point as matrix element.
                fprintf(stdout, "[%d][%d] = %lf\n", i, j, MATRIX[i][j]); ///TESTING ///
            }
        }

        // Calculate the determinant of matrix "MATRIX".
        ///det = Determinant(n, MATRIX); /// COMMENT OUT ///
        ///fprintf(stdout, "Determinant = %lf\n", det); /// COMMENT OUT ///
        for(i = 0; i < n; i++) /// Loop rows.
        {
            for(j = 0; j < n; j++) /// Loop columns.
            {
                fprintf(stdout, "[%d][%d] = %lf\n", i, j, MATRIX[i][j]);
            }
        }
    }

你可以看到我打印矩阵两次用于测试目的。一旦我正在读取文件并存储数据，然后在初始for循环完成后再次。

它们应该打印出完全相同的东西，因为我没有操纵两个for循环之间的2D数组'MATRIX'。

然而，这是打印出来的（当使用上面的'matrix.dat'文件时）：

n = 3
[0][0] = 1.000000
[0][1] = 2.000000
[0][2] = 3.000000
[1][0] = 4.000000
[1][1] = 5.000000
[1][2] = 6.000000
[2][0] = 7.000000
[2][1] = 8.000000
[2][2] = 9.000000

[0][0] = 1.000000
[0][1] = 2.000000
[0][2] = 4.000000
[1][0] = 4.000000
[1][1] = 5.000000
[1][2] = 5.000000
[2][0] = 5.000000
[2][1] = 8.000000
[2][2] = 9.000000

非常感谢任何帮助。

以防万一，这是'CheckSquare'功能：

// Checks if matrix is square and returns the dimension if TRUE.
// Otherwise returns 0.
int CheckSquare(FILE* m)
{
    int dim; /// Dimension of matrix.
    int dataNum = 0; /// Holds number of data points in file "matrixData".
    double temp; /// Temporary variable to hold double data.

    // Check matrix is square and find number of columns/rows.
    while(fscanf(m, "%lf", &temp) != EOF) /// Data can still be read from file "matrixData".
    {
        dataNum++; /// Adds to number of data points.
    }
    rewind(m); /// Sets the position indicator to the beginning of the file "matrixData".

    dim = sqrt(dataNum); /// Number of columns and rows if 'dataNum' is perfect square.

    if(dim*dim == dataNum) /// Matrix is square.
    {
        return(dim);
    }

    else
    {
        return(0);
    }
}

Answer 1

如果n为3，则有效索引为0,1,2，但您声明的矩阵是这样的：

double MATRIX[n - 1][n - 1];

但应该这样说：

double MATRIX[n][n];

如果你使用这样的循环：

for(i = 0; i < n; i++) /// Loop rows.
{
    for(j = 0; j < n; j++) /// Loop columns.
    {

当n == 3时，迭代0,1和2.

2D数组丢失值而不被操纵？

1 个答案: