指针的指针与正常指针 [英] Pointers to pointers vs. normal pointers

问题描述

的指针的目的是为了保存特定变量的地址。然后按照code的存储结构应该是这样的：

  int类型的= 5;
为int * B =＆放大器;一个;
 

  

......内存地址......值

  一个... 0x000002 ................... 5

  b ... 0x000010 ................... 0x000002

好，好。再假设现在我要救指针* b的地址。然后，我们通常定义一个双指针，** C，如

  int类型的= 5;
为int * B =＆放大器;一个;
INT ** C =和b;
 

那么内存结构如下：

  

......内存地址......值

  一个... 0x000002 ................... 5

  b ... 0x000010 ................... 0x000002

  ç... 0x000020 ................... 0x000010

所以，** C指* b的地址。

现在我的问题是，为什么这种类型的code的，

  int类型的= 5;
为int * B =＆放大器;一个;
INT * C =和b;
 

生成一个警告？

如果指针的目的只是为了节省内存地址，我觉得不应该有层次，如果我们要保存地址指的是一个变量，指针，双指针等，这样下面的类型$的C $ C应该有效。

  int类型的= 5;
为int * B =＆放大器;一个;
INT * C =和b;
为int * D =和C;
为int * E =和D;
为int * F =急症室;
 

解决方案

在

  int类型的= 5;
为int * B =＆放大器;一个;
INT * C =和b;
 

您得到一个警告，因为和b 的类型为 INT ** ，并尝试初始化类型的变量为int * 。有这两种类型之间不存在隐式转换，导致了警告。

要带你想要的工作，如果我们试图取消引用更长的例子˚F编译器会给出我们一个 INT ，不是指针，我们可以进一步取消引用。

另外请注意，在许多系统上 INT 和为int * 是不一样的尺寸（如指针可能64位长和 INT 32位）。如果取消引用˚F，并得到一个 INT ，你失去了一半的价值，然后你甚至不能投它一个有效的指针。

The purpose of a pointer is to save the address of a specific variable. Then the memory structure of following code should look like:

int a = 5;
int *b = &a;

...... memory address ...... value
a ... 0x000002 ................... 5
b ... 0x000010 ................... 0x000002

Okay, fine. Then assume that now I want to save the address of pointer *b. Then we generally define a double pointer, **c, as

int a = 5;
int *b = &a;
int **c = &b;

Then the memory structure looks like:

...... memory address ...... value
a ... 0x000002 ................... 5
b ... 0x000010 ................... 0x000002
c ... 0x000020 ................... 0x000010

So **c refers the address of *b.

Now my question is, why does this type of code,

int a = 5;
int *b = &a;
int *c = &b;

generate a warning?

If the purpose of pointer is just to save the memory address, I think there should be no hierarchy if the address we are going to save refers to a variable, a pointer, a double pointer, etc., so the below type of code should be valid.

int a = 5;
int *b = &a;
int *c = &b;
int *d = &c;
int *e = &d;
int *f = &e;

解决方案

In

int a = 5;
int *b = &a;   
int *c = &b;

You get a warning because &b is of type int **, and you try to initialize a variable of type int *. There's no implicit conversions between those two types, leading to the warning.

To take the longer example you want to work, if we try to dereference f the compiler will give us an int, not a pointer that we can further dereference.

Also note that on many systems int and int* are not the same size (e.g. a pointer may be 64 bits long and an int 32 bits long). If you dereference f and get an int, you lose half the value, and then you can't even cast it to a valid pointer.

这篇关于指针的指针与正常指针的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！