分配数组元素值时，C下标值既不是数组，也不是指针，也不是矢量

问题描述

很抱歉询问已经回答的问题，我是C的新手，不了解解决方案.这是我的功能

Sorry for asking the already answered question, I am a newbie to C and don't understand the solutions.Here is my function

int rotateArr(int *arr) {
    int D[4][4];
    int i = 0, n =0;
    for(i; i < M; i ++ ){
        for(n; n < N; n++){
            D[i][n] = arr[n][M - i + 1];
        }
    }
    return D;
}

它引发错误

在线

怎么了?我只是将一个数组元素的值设置为另一个数组元素.

What's wrong? I am just setting the value of an array element to another array element.

所传递的arr被声明为

The arr passed is declared as

int S[4][4] = { { 1, 4, 10, 3 }, { 0, 6, 3, 8 }, { 7, 10 ,8, 5 },  { 9, 5, 11, 2}  };

推荐答案

C使您可以在数组和指针上使用下标运算符[].在指针上使用此运算符时，结果类型是指针指向的类型.例如，如果将[]应用于int*，则结果将是int.

C lets you use the subscript operator [] on arrays and on pointers. When you use this operator on a pointer, the resultant type is the type to which the pointer points to. For example, if you apply [] to int*, the result would be an int.

这正是正在发生的事情:您正在传递int*，它对应于整数向量.一次在其上使用下标会使其变为int，因此您无法在其上应用第二个下标.

That is precisely what's going on: you are passing int*, which corresponds to a vector of integers. Using subscript on it once makes it int, so you cannot apply the second subscript to it.

从您的代码看来，arr应该是一个二维数组.如果将其实现为锯齿状"数组(即指针数组)，则参数类型应为int **.

It appears from your code that arr should be a 2-D array. If it is implemented as a "jagged" array (i.e. an array of pointers) then the parameter type should be int **.

此外，您似乎正在尝试返回本地数组.为了合法地执行此操作，您需要动态分配数组，并返回一个指针.但是，更好的方法是为4x4矩阵声明一个特殊的struct，并使用它包装固定大小的数组，如下所示:

Moreover, it appears that you are trying to return a local array. In order to do that legally, you need to allocate the array dynamically, and return a pointer. However, a better approach would be declaring a special struct for your 4x4 matrix, and using it to wrap your fixed-size array, like this:

// This type wraps your 4x4 matrix
typedef struct {
    int arr[4][4];
} FourByFour;
// Now rotate(m) can use FourByFour as a type
FourByFour rotate(FourByFour m) {
    FourByFour D;
    for(int i = 0; i < 4; i ++ ){
        for(int n = 0; n < 4; n++){
            D.arr[i][n] = m.arr[n][3 - i];
        }
    }
    return D;
}
// Here is a demo of your rotate(m) in action:
int main(void) {
    FourByFour S = {.arr = {
        { 1, 4, 10, 3 },
        { 0, 6, 3, 8 },
        { 7, 10 ,8, 5 },
        { 9, 5, 11, 2}
    } };
    FourByFour r = rotate(S);
    for(int i=0; i < 4; i ++ ){
        for(int n=0; n < 4; n++){
            printf("%d ", r.arr[i][n]);
        }
        printf("\n");
    }
    return 0;
}

此打印以下内容:

3 8 5 2
10 3 8 11
4 6 10 5
1 0 7 9

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