如果你想要一个字符列表(一个单词)，你可以使用char *word

如果你想要一个单词列表(一个句子)，你可以使用char **句子

如果你想要一个句子列表(独白)，你可以使用char ***monologue

如果你想要一个独白列表(传记)，你可以使用char ****传记

如果你想要一个传记列表(一个生物图书馆)，你可以使用char *****biolibrary

如果你想要一个生物库列表(a ??lol)，你可以使用char ******lol

……

是的，我知道这些可能不是最好的数据结构

一个非常非常非常无聊的lol的用法例子

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int wordsinsentence(char **x) {
    int w = 0;
    while (*x) {
        w += 1;
        x++;
    }
    return w;
}

int wordsinmono(char ***x) {
    int w = 0;
    while (*x) {
        w += wordsinsentence(*x);
        x++;
    }
    return w;
}

int wordsinbio(char ****x) {
    int w = 0;
    while (*x) {
        w += wordsinmono(*x);
        x++;
    }
    return w;
}

int wordsinlib(char *****x) {
    int w = 0;
    while (*x) {
        w += wordsinbio(*x);
        x++;
    }
    return w;
}

int wordsinlol(char ******x) {
    int w = 0;
    while (*x) {
        w += wordsinlib(*x);
        x++;
    }
    return w;
}

int main(void) {
    char *word;
    char **sentence;
    char ***monologue;
    char ****biography;
    char *****biolibrary;
    char ******lol;

    //fill data structure
    word = malloc(4 * sizeof *word); // assume it worked
    strcpy(word, "foo");

    sentence = malloc(4 * sizeof *sentence); // assume it worked
    sentence[0] = word;
    sentence[1] = word;
    sentence[2] = word;
    sentence[3] = NULL;

    monologue = malloc(4 * sizeof *monologue); // assume it worked
    monologue[0] = sentence;
    monologue[1] = sentence;
    monologue[2] = sentence;
    monologue[3] = NULL;

    biography = malloc(4 * sizeof *biography); // assume it worked
    biography[0] = monologue;
    biography[1] = monologue;
    biography[2] = monologue;
    biography[3] = NULL;

    biolibrary = malloc(4 * sizeof *biolibrary); // assume it worked
    biolibrary[0] = biography;
    biolibrary[1] = biography;
    biolibrary[2] = biography;
    biolibrary[3] = NULL;

    lol = malloc(4 * sizeof *lol); // assume it worked
    lol[0] = biolibrary;
    lol[1] = biolibrary;
    lol[2] = biolibrary;
    lol[3] = NULL;

    printf("total words in my lol: %d\n", wordsinlol(lol));

    free(lol);
    free(biolibrary);
    free(biography);
    free(monologue);
    free(sentence);
    free(word);
}

输出:

total words in my lol: 243

例如，你可能想要确保当你释放某个东西的内存时，你将指针设置为空。

void safeFree(void** memory) {
    if (*memory) {
        free(*memory);
        *memory = NULL;
    }
}

当你调用这个函数时，你会用指针的地址来调用它

void* myMemory = someCrazyFunctionThatAllocatesMemory();
safeFree(&myMemory);

现在myMemory被设置为NULL，任何重用它的尝试都将是非常明显的错误。

I have used double pointers today while I was programming something for work, so I can answer why we had to use them (it's the first time I actually had to use double pointers). We had to deal with real time encoding of frames contained in buffers which are members of some structures. In the encoder we had to use a pointer to one of those structures. The problem was that our pointer was being changed to point to other structures from another thread. In order to use the current structure in the encoder, I had to use a double pointer, in order to point to the pointer that was being modified in another thread. It wasn't obvious at first, at least for us, that we had to take this approach. A lot of address were printed in the process :)).

当你处理在应用程序其他地方被更改的指针时，你应该使用双指针。在处理返回和寻址给您的硬件时，您可能还会发现双指针是必须的。

有点晚了，但希望这能帮助到一些人。

在C语言中，数组总是在堆栈上分配内存，因此函数不能返回 一个(非静态)数组，因为内存分配在堆栈上 当执行到达当前块的末尾时自动释放。 当你想处理二维数组时，这真的很烦人 (即矩阵)，并实现一些可以改变和返回矩阵的函数。 要实现这一点，可以使用指针对指针来实现矩阵 动态分配内存:

/* Initializes a matrix */
double** init_matrix(int num_rows, int num_cols){
    // Allocate memory for num_rows float-pointers
    double** A = calloc(num_rows, sizeof(double*));
    // return NULL if the memory couldn't allocated
    if(A == NULL) return NULL;
    // For each double-pointer (row) allocate memory for num_cols floats
    for(int i = 0; i < num_rows; i++){
        A[i] = calloc(num_cols, sizeof(double));
        // return NULL if the memory couldn't allocated
        // and free the already allocated memory
        if(A[i] == NULL){
            for(int j = 0; j < i; j++){
                free(A[j]);
            }
            free(A);
            return NULL;
        }
    }
    return A;
} 

这里有一个例子:

double**       double*           double
             -------------       ---------------------------------------------------------
   A ------> |   A[0]    | ----> | A[0][0] | A[0][1] | A[0][2] | ........ | A[0][cols-1] |
             | --------- |       ---------------------------------------------------------
             |   A[1]    | ----> | A[1][0] | A[1][1] | A[1][2] | ........ | A[1][cols-1] |
             | --------- |       ---------------------------------------------------------
             |     .     |                                    .
             |     .     |                                    .
             |     .     |                                    .
             | --------- |       ---------------------------------------------------------
             |   A[i]    | ----> | A[i][0] | A[i][1] | A[i][2] | ........ | A[i][cols-1] |
             | --------- |       ---------------------------------------------------------
             |     .     |                                    .
             |     .     |                                    .
             |     .     |                                    .
             | --------- |       ---------------------------------------------------------
             | A[rows-1] | ----> | A[rows-1][0] | A[rows-1][1] | ... | A[rows-1][cols-1] |
             -------------       ---------------------------------------------------------

The double-pointer-to-double-pointer A points to the first element A[0] of a memory block whose elements are double-pointers itself. You can imagine these double-pointers as the rows of the matrix. That's the reason why every double-pointer allocates memory for num_cols elements of type double. Furthermore A[i] points to the i-th row, i.e. A[i] points to A[i][0] and that's just the first double-element of the memory block for the i-th row. Finally, you can access the element in the i-th row and j-th column easily with A[i][j].

下面是一个完整的例子来演示它的用法:

#include <stdio.h>
#include <stdlib.h>
#include <time.h>

/* Initializes a matrix */
double** init_matrix(int num_rows, int num_cols){
    // Allocate memory for num_rows double-pointers
    double** matrix = calloc(num_rows, sizeof(double*));
    // return NULL if the memory couldn't allocated
    if(matrix == NULL) return NULL;
    // For each double-pointer (row) allocate memory for num_cols
    // doubles
    for(int i = 0; i < num_rows; i++){
        matrix[i] = calloc(num_cols, sizeof(double));
        // return NULL if the memory couldn't allocated
        // and free the already allocated memory
        if(matrix[i] == NULL){
            for(int j = 0; j < i; j++){
                free(matrix[j]);
            }
            free(matrix);
            return NULL;
        }
    }
    return matrix;
}

/* Fills the matrix with random double-numbers between -1 and 1 */
void randn_fill_matrix(double** matrix, int rows, int cols){
    for (int i = 0; i < rows; ++i){
        for (int j = 0; j < cols; ++j){
            matrix[i][j] = (double) rand()/RAND_MAX*2.0-1.0;
        }
    }
}


/* Frees the memory allocated by the matrix */
void free_matrix(double** matrix, int rows, int cols){
    for(int i = 0; i < rows; i++){
        free(matrix[i]);
    }
    free(matrix);
}

/* Outputs the matrix to the console */
void print_matrix(double** matrix, int rows, int cols){
    for(int i = 0; i < rows; i++){
        for(int j = 0; j < cols; j++){
            printf(" %- f ", matrix[i][j]);
        }
        printf("\n");
    }
}


int main(){
    srand(time(NULL));
    int m = 3, n = 3;
    double** A = init_matrix(m, n);
    randn_fill_matrix(A, m, n);
    print_matrix(A, m, n);
    free_matrix(A, m, n);
    return 0;
}

假设你有一个指针。取值为地址。 但现在你想更改地址。 你可以。通过执行pointer1 = pointer2，你给了pointer1一个pointer2的地址。 但是!如果在函数中执行此操作，并且希望结果在函数完成后仍然存在，则需要做一些额外的工作。你需要一个新的pointer3来指向pointer1。将指针3传递给函数。 这里有一个例子。先看看下面的输出，以便理解。

#include <stdio.h>

int main()
{

    int c = 1;
    int d = 2;
    int e = 3;
    int * a = &c;
    int * b = &d;
    int * f = &e;
    int ** pp = &a;  // pointer to pointer 'a'

    printf("\n a's value: %x \n", a);
    printf("\n b's value: %x \n", b);
    printf("\n f's value: %x \n", f);
    printf("\n can we change a?, lets see \n");
    printf("\n a = b \n");
    a = b;
    printf("\n a's value is now: %x, same as 'b'... it seems we can, but can we do it in a function? lets see... \n", a);
    printf("\n cant_change(a, f); \n");
    cant_change(a, f);
    printf("\n a's value is now: %x, Doh! same as 'b'...  that function tricked us. \n", a);

    printf("\n NOW! lets see if a pointer to a pointer solution can help us... remember that 'pp' point to 'a' \n");
     printf("\n change(pp, f); \n");
    change(pp, f);
    printf("\n a's value is now: %x, YEAH! same as 'f'...  that function ROCKS!!!. \n", a);
    return 0;
}

void cant_change(int * x, int * z){
    x = z;
    printf("\n ----> value of 'a' is: %x inside function, same as 'f', BUT will it be the same outside of this function? lets see\n", x);
}

void change(int ** x, int * z){
    *x = z;
    printf("\n ----> value of 'a' is: %x inside function, same as 'f', BUT will it be the same outside of this function? lets see\n", *x);
}

以下是输出:(先阅读这个)

 a's value: bf94c204

 b's value: bf94c208 

 f's value: bf94c20c 

 can we change a?, lets see 

 a = b 

 a's value is now: bf94c208, same as 'b'... it seems we can, but can we do it in a function? lets see... 

 cant_change(a, f); 

 ----> value of 'a' is: bf94c20c inside function, same as 'f', BUT will it be the same outside of this function? lets see

 a's value is now: bf94c208, Doh! same as 'b'...  that function tricked us. 

 NOW! lets see if a pointer to a pointer solution can help us... remember that 'pp' point to 'a' 

 change(pp, f); 

 ----> value of 'a' is: bf94c20c inside function, same as 'f', BUT will it be the same outside of this function? lets see

 a's value is now: bf94c20c, YEAH! same as 'f'...  that function ROCKS!!!. 

添加到Asha的响应，如果你使用单个指针指向下面的例子(例如alloc1())，你将失去对函数内部分配的内存的引用。

#include <stdio.h>
#include <stdlib.h>

void alloc2(int** p) {
    *p = (int*)malloc(sizeof(int));
    **p = 10;
}

void alloc1(int* p) {
    p = (int*)malloc(sizeof(int));
    *p = 10;
}

int main(){
    int *p = NULL;
    alloc1(p);
    //printf("%d ",*p);//undefined
    alloc2(&p);
    printf("%d ",*p);//will print 10
    free(p);
    return 0;
}

发生这种情况的原因是在alloc1中，指针是按值传入的。因此，当它被重新分配给alloc1内部的malloc调用的结果时，更改不属于不同作用域中的代码。