如何在由指针的矩阵的子矩阵工作？ [英] How to work on a sub-matrix in a matrix by pointer?

问题描述

我有大小n矩阵。举一个例子：

I have a matrix of size n. Take an example:

我的递归函数不上躺在矩阵的边界元素的处理。现在我想调用它（递归调用）在内方阵：

My recursive function does the processing on the elements that lie in the border of the matrix. Now I want to call it (the recursive call) on the inner square matrix:

这是我的递归函数的原型：

This is the prototype of my recursive function:

void rotate(int** mat, size_t n);

我知道，一个二维数组是一个数组内的数组。我知道， *（垫+ 1）+ 1）会给这应该是我的新基的基地址的内存地址。这是我尝试：

I know that a 2D array is an array within an array. I know that *(mat+1) + 1) will give the memory address that should be the base address of my new matrix. This is what I tried:

rotate((int **)(*(mat+1) + 1), n-2)

但它不工作，我也得到段错误，当我尝试访问它 [] [] 。

推荐答案

您不能解引用垫+ 1 和reinter preT，作为一个指向整个矩阵。而要提供偏移作为参数传递给你的函数（我假设 N -by - N 方阵）：

You cannot dereference mat+1 and reinterpret that as a pointer to a whole matrix. Instead provide the offsets as arguments to your function (I assume n-by-n square matrices):

void rotate(int** mat, size_t i, size_t j, size_t n) {
    // assuming row-wise storage
    int *row0 = mat[j];     // assumes j < n
    int *row1 = mat[j + 1]; // assumes j + 1 < n
    // access row0[i..] and row1[i..]
}

如果你有持续的存储你的矩阵，你可以做以下代替：

If you had continuous storage for your matrix, you could do the following instead:

rotate(int* mat, size_t i, size_t j, size_t n) {
    int atIJ = mat[j * n + i]; // assuming row-wise storage
    // ...
}

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