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

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。

例如，如果您想随机访问不连续的数据。

p -> [p0, p1, p2, ...]  
p0 -> data1
p1 -> data2

——用C

T ** p = (T **) malloc(sizeof(T*) * n);
p[0] = (T*) malloc(sizeof(T));
p[1] = (T*) malloc(sizeof(T));

存储一个指针p，它指向一个指针数组。每个指针指向一段数据。

如果sizeof(T)很大，则可能无法分配sizeof(T) * n字节的连续块(即使用malloc)。

这里的大多数答案或多或少都与应用程序编程有关。下面是一个嵌入式系统编程的例子。例如，以下是NXP Kinetis KL13系列微控制器参考手册的摘录，此代码片段用于从固件中运行驻留在ROM中的引导加载程序:

＂ 为了获得入口点的地址，用户应用程序读取包含引导加载程序API树指针的单词，该指针位于引导加载程序向量表的0x1C偏移量处。向量表被放置在引导加载器地址范围的底部，ROM的地址范围是0x1C00_0000。因此，API树指针位于地址0x1C00_001C。

引导加载程序API树是一个包含指向其他结构的指针的结构，这些结构具有引导加载程序的函数和数据地址。引导加载程序入口点总是API树的第一个单词。 ＂

uint32_t runBootloaderAddress;
void (*runBootloader)(void * arg);
// Read the function address from the ROM API tree.
runBootloaderAddress = **(uint32_t **)(0x1c00001c);
runBootloader = (void (*)(void * arg))runBootloaderAddress;
// Start the bootloader.
runBootloader(NULL);

我今天看到了一个很好的例子，从这篇博客文章，我总结如下。

假设您有一个链表中节点的结构，可能是这样

typedef struct node
{
    struct node * next;
    ....
} node;

现在您想实现一个remove_if函数，它接受删除条件rm作为参数之一，并遍历链表:如果一个条目满足该条件(例如rm(entry)==true)，它的节点将从链表中删除。最后，remove_if返回链表的头(可能与原始头不同)。

你可以写信

for (node * prev = NULL, * curr = head; curr != NULL; )
{
    node * const next = curr->next;
    if (rm(curr))
    {
        if (prev)  // the node to be removed is not the head
            prev->next = next;
        else       // remove the head
            head = next;
        free(curr);
    }
    else
        prev = curr;
    curr = next;
}

就像你的for循环。这里的信息是，如果没有双指针，您必须维护一个prev变量来重新组织指针，并处理两种不同的情况。

但是使用双指针，你实际上可以写

// now head is a double pointer
for (node** curr = head; *curr; )
{
    node * entry = *curr;
    if (rm(entry))
    {
        *curr = entry->next;
        free(entry);
    }
    else
        curr = &entry->next;
}

你现在不需要一个prev，因为你可以直接修改什么prev->next指向。

为了使事情更清楚，让我们稍微跟随一下代码。拆卸过程中:

如果entry ==* head:它将是*head (==*curr) =* head->next - head现在指向新标题节点的指针。您可以通过直接将head的内容更改为一个新的指针来实现这一点。 如果entry != *head:类似地，*curr是prev->next所指向的，现在指向entry->next。

无论哪种情况，您都可以使用双指针以统一的方式重新组织指针。