字符串是使用双指针的一个很好的例子。字符串本身是一个指针，所以任何时候你需要指向一个字符串，你就需要一个双指针。

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!!!. 

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;
}

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。

Pointers to pointers also come in handy as "handles" to memory where you want to pass around a "handle" between functions to re-locatable memory. That basically means that the function can change the memory that is being pointed to by the pointer inside the handle variable, and every function or object that is using the handle will properly point to the newly relocated (or allocated) memory. Libraries like to-do this with "opaque" data-types, that is data-types were you don't have to worry about what they're doing with the memory being pointed do, you simply pass around the "handle" between the functions of the library to perform some operations on that memory ... the library functions can be allocating and de-allocating the memory under-the-hood without you having to explicitly worry about the process of memory management or where the handle is pointing.

例如:

#include <stdlib.h>

typedef unsigned char** handle_type;

//some data_structure that the library functions would work with
typedef struct 
{
    int data_a;
    int data_b;
    int data_c;
} LIB_OBJECT;

handle_type lib_create_handle()
{
    //initialize the handle with some memory that points to and array of 10 LIB_OBJECTs
    handle_type handle = malloc(sizeof(handle_type));
    *handle = malloc(sizeof(LIB_OBJECT) * 10);

    return handle;
}

void lib_func_a(handle_type handle) { /*does something with array of LIB_OBJECTs*/ }

void lib_func_b(handle_type handle)
{
    //does something that takes input LIB_OBJECTs and makes more of them, so has to
    //reallocate memory for the new objects that will be created

    //first re-allocate the memory somewhere else with more slots, but don't destroy the
    //currently allocated slots
    *handle = realloc(*handle, sizeof(LIB_OBJECT) * 20);

    //...do some operation on the new memory and return
}

void lib_func_c(handle_type handle) { /*does something else to array of LIB_OBJECTs*/ }

void lib_free_handle(handle_type handle) 
{
    free(*handle);
    free(handle); 
}


int main()
{
    //create a "handle" to some memory that the library functions can use
    handle_type my_handle = lib_create_handle();

    //do something with that memory
    lib_func_a(my_handle);

    //do something else with the handle that will make it point somewhere else
    //but that's invisible to us from the standpoint of the calling the function and
    //working with the handle
    lib_func_b(my_handle); 

    //do something with new memory chunk, but you don't have to think about the fact
    //that the memory has moved under the hood ... it's still pointed to by the "handle"
    lib_func_c(my_handle);

    //deallocate the handle
    lib_free_handle(my_handle);

    return 0;
}

希望这能有所帮助，

杰森

一个原因是你想要改变传递给函数的作为函数参数的指针的值，要做到这一点，你需要指针指向指针。

简单地说，当你想在函数调用之外保留(或保留)内存分配或分配的变化时，使用**。(因此，传递带有双指针arg的函数。)

这可能不是一个很好的例子，但会告诉你基本的用法:

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

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

int main()
{
    int *p = NULL;
    allocate(&p);
    *p = 42;
    printf("%d\n", *p);
    free(p);
}