通过refrence传递二维数组的函数（C程序） [英] Passing two-dimensional array to a function by refrence (C Programming)

问题描述

我学球，并得到被困了一个小时，现在，有了这个code，

I'm learning pointers, and gotten stuck for an hour now, with this code,

#include <stdio.h>

int determinant(int **mat)  /* int mat[3][3] works fine.. int *mat[3] doesn't.. neither does int *mat[] */
{
    int det;
    int a=*(*(mat+0)+0); // printf("\n%d",a);
    int b=*(*(mat+0)+1); // printf("\n%d",b);
    int c=*(*(mat+0)+2); // printf("\n%d",c);
    int d=*(*(mat+1)+0); // printf("\n%d",d);
    int e=*(*(mat+1)+1); // printf("\n%d",e);
    int f=*(*(mat+1)+2); // printf("\n%d",f);
    int g=*(*(mat+2)+0); // printf("\n%d",g);
    int h=*(*(mat+2)+1); // printf("\n%d",h);
    int i=*(*(mat+2)+2); // printf("\n%d",i);

    det = a*(e*i-h*f) - b*(d*i-g*f) + c*(d*h-e*g);
    return det;
}

int main()
{
    int mat[3][3];
    int i,j;
    printf("Enter the 3 X 3 matrix:\n\n");
    for (i=0;i<3;i++)
    {
        for (j=0;j<3;j++)
        {
            scanf("%d",*(mat+i)+j);
        }
    }
    printf("\nThe determinant of the given 3 X 3 matrix is %d",determinant(mat));
    return 0;
}

我觉得没有什么是错的函数调用。也许问题是在接受争论。 IDK，是不是垫一个指向一维数组，这将再次成为一个指向数组元素，以垫 A指针的指针？
当我在打印的地方（只是为了检查）一些文字，我发现去执行，直到在功能 INT DET 之后，并在下一步的程序崩溃。
垫[3] [3] 效果很好，但我想用一些 * 那里，因为正如我所说的，我'M'学习'。

I don't think anything is wrong with the function call. Maybe the problem is while accepting the arguments. Idk, isn't mat a pointer to an 1-dimensional array, which would again be a pointer to the array element, making mat a pointer to a pointer? When I print some text at places (just to check), i find that the execution goes till after int det in the function, and the program crashes in the next step. mat [3][3] works well, but i wanna use some * there, because as i said, i'm 'learning'..

请帮帮忙！
感谢：）

Please help! Thanks :)

推荐答案

二维数组不衰的指针的指针。您可以将它们衰变为指针，以便您的code应该像

2D array dont decay to pointer to pointer. You can decay them to pointers so your code should look like

int determinant(int *mat) {
  int det;
  int a=*((mat+0)+0); // printf("\n%d",a);
  int b=*((mat+0)+1); // printf("\n%d",b);
  int c=*((mat+0)+2); // printf("\n%d",c);
  int d=*((mat+1*3)+0); // printf("\n%d",d);
  int e=*((mat+1*3)+1); // printf("\n%d",e);
  int f=*((mat+1*3)+2); // printf("\n%d",f);
  int g=*((mat+2*3)+0); // printf("\n%d",g);
  int h=*((mat+2*3)+1); // printf("\n%d",h);
  int i=*((mat+2*3)+2); // printf("\n%d",i);

  det = a*(e*i-h*f) - b*(d*i-g*f) + c*(d*h-e*g);
  return det;
}

以上code是只是为了说明，显示2-D阵列如何衰减为1-D阵列。

The above code is just for illustration, showing how 2-D array decays to 1-D array.

当您尝试访问使用括号数组像 A [2] [1] 则编译器正在上演你。通过展开我的意思是，通过的sizeof（类型）的乘积（如上图所示乘以3）。所以，如果你要腐烂1-D，你必须自己做。

When you try to access the array using braces like a[2][1] then compiler does is unfolding for you. By unfolding I mean, the multiplication by sizeof(type) (as shown above multiply by 3). So if you decaying to 1-D you have to do it yourself.

还有一点要补充，始终尺寸的大小传给谁是有踏一维数组作为2-D的功能。像

One more thing to add, always pass the size of the dimension to the function who is has to tread the 1-D array as 2-D. like

int determinant(int *mat, int cols, rows);

编辑1：

只是，如果你想保持完整数组在函数调用补充说@JensGustedt ANS也行。

Just to add that @JensGustedt ans is also ok if you want to keep the arrays intact across function calls.

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