c - 传递参数从整数生成指针，没有强制转换/预期的“双 *”，但参数的类型为“int”

Question

我对 C 非常陌生，并且在我的 C 程序作业中遇到错误。我正在生成一个 4x4 希尔伯特矩阵，然后通过克莱默规则找到它的行列式。但是对于函数“main”中的两个参数，我的代码中出现以下错误：

In function ‘main’:
passing argument 1 of ‘determinant’ makes pointer from integer without a cast [enabled by default]
  printf("\nThe determinant using Cramer’s Rule is:\%lf\n", determinant(n,H));
  ^
note: expected ‘double *’ but argument is of type ‘int’
double determinant(double * matrix, int n){

这是我的代码：

#include <malloc.h>
#include <stdio.h>

void Hilbert(int n, double H[n][n]) {
    int i, j;
    for (i = 0; i < n; i++) {
        for (j = 0; j < n; j++) {
            H[i][j] = (double)1.0 / ((i + 1) + (j + 1) - 1.0);
        }
    }
}

void printMatrix(int n, double matrix[n][n]) {
    int i, j;
    for (i = 0; i < n; i++) {
        for (j = 0; j < n; j++) {
            printf("%lf\t", matrix[i][j]);
        }
        printf("\n");
    }
}

double determinant(double * matrix, int n) {
    if (1 >= n) return matrix[0];
    double *subMatrix = (double*)malloc((n - 1)*(n - 1) * sizeof(double));
    double result = 0.0;
    int i, j, k;
    for (i = 0; i < n; i++) {
        for (j = 0; j < n - 1; j++) {
            for (k = 0; k < i; k++)
                subMatrix[j*(n - 1) + k] = matrix[(j + 1)*n + k];
            for (k = i + 1; k < n; k++)
                subMatrix[j*(n - 1) + (k - 1)] = matrix[(j + 1)*n + k];
        }
        if (i % 2 == 0)
            result += matrix[0 * n + i] * determinant(subMatrix, n - 1);
        else
            result -= matrix[0 * n + i] * determinant(subMatrix, n - 1);
    }
    free(subMatrix);
    return result;
}

int main() {
    int i, j;
    int n = 4;
    double H[n][n];
    Hilbert(n, H);
    printf("\nThe Hilbert matrix is:\n");
    printMatrix(n, H);
    printf("\nThe determinant using Cramer’s Rule is: \%lf\n", determinant(n, H));
}

标签： c

Answer 1

只需像这样交换调用determinant中的参数printf：

determinant(H,n)

为了抑制警告，您可能还需要参考第一行而不是整个矩阵：

determinant(H[0],n)