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

下面是一个非常简单的c++示例，说明如果要使用函数将指针设置为指向对象，则需要一个指针指向指针。否则，指针将继续返回null。

(一个c++的答案，但我相信在C中也是一样的)

(同样，供参考:谷歌("pass by value c++") = "默认情况下，c++中的参数是按值传递的。当实参按值传递时，实参的值被复制到函数的形参中。”)

我们想让指针b等于字符串a。

#include <iostream>
#include <string>

void Function_1(std::string* a, std::string* b) {
  b = a;
  std::cout << (b == nullptr);  // False
}

void Function_2(std::string* a, std::string** b) {
  *b = a;
  std::cout << (b == nullptr);  // False
}

int main() {
  std::string a("Hello!");
  std::string* b(nullptr);
  std::cout << (b == nullptr);  // True

  Function_1(&a, b);
  std::cout << (b == nullptr);  // True

  Function_2(&a, &b);
  std::cout << (b == nullptr);  // False
}

// Output: 10100

在Function_1(&a, b);这条线上会发生什么?

The "value" of &main::a (an address) is copied into the parameter std::string* Function_1::a. Therefore Function_1::a is a pointer to (i.e. the memory address of) the string main::a. The "value" of main::b (an address in memory) is copied into the parameter std::string* Function_1::b. Therefore there are now 2 of these addresses in memory, both null pointers. At the line b = a;, the local variable Function_1::b is then changed to equal Function_1::a (= &main::a), but the variable main::b is unchanged. After the call to Function_1, main::b is still a null pointer.

在函数_2(&a， &b);这一行发生了什么?

The treatment of the a variable is the same: within the function, Function_2::a is the address of the string main::a. But the variable b is now being passed as a pointer to a pointer. The "value" of &main::b (the address of the pointer main::b) is copied into std::string** Function_2::b. Therefore within Function_2, dereferencing this as *Function_2::b will access and modify main::b . So the line *b = a; is actually setting main::b (an address) equal to Function_2::a (= address of main::a) which is what we want.

如果你想用一个函数来修改一个东西，无论是一个对象还是一个地址(指针)，你必须传递一个指向那个东西的指针。您实际传入的内容不能被修改(在调用范围内)，因为创建了本地副本。

(一个例外是如果形参是一个引用，例如std::string& a.但通常这些是const。一般来说，如果你调用f(x)，如果x是一个对象，你应该能够假设f不会修改x。但如果x是一个指针，那么你应该假设f可能修改x指向的对象。)

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);

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。

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!!!. 

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。