C ++中的指针变量和引用变量之间有什么区别？ [英] What are the differences between a pointer variable and a reference variable in C++?

问题描述

我知道引用是语法糖，所以代码更容易阅读和写。

但是有什么区别？

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来自以下答案和链接的摘要：

1. - 分配任意次数，而绑定后无法重新安装引用。


2. 指针无法指向（ NULL ），而引用始终引用一个对象。




4. 没有引用算术（但是您可以获取引用指向的对象的地址，并且可以使用引用指向的对象的地址） & obj + 5 ）。

澄清一个误解：

C ++标准非常小心，以避免编译器必须
实现引用，每个C ++编译器实现
引用作为指针。也就是说，声明如：

  int& ri = i; 
  

如果未完全优化 分配相同数量的存储
作为指针，并将地址
放入该存储。

因此，指针和引用都占用相同的内存量。

作为一般规则，

• 在函数参数和返回类型中使用引用来定义有用的和自文档化的接口。


• 使用指针实现算法和数据结构。

• 我最喜欢的 C ++ FQA lite 。


• 参考与指针。


• 参考简介。


• References and const 。

解决方案

1. p>可以重新分配指针：

     int x = 5; 
    int y = 6; 
    int * p; 
    p =& x; 
    p =& y; 
    * p = 10; 
    assert（x == 5）; 
    assert（y == 10）; 
    

   引用不能，且必须在初始化时分配：

     int x = 5; 
    int y = 6; 
    int& r = x;指针在堆栈上有自己的内存地址和大小（x86上为4个字节）。   ，而引用共享相同的内存地址（与原始变量），但也占用了堆栈上的一些空间。由于引用与原始变量本身具有相同的地址，因此可以将引用视为同一变量的另一个名称。注意：指针指向的可以是堆栈或堆。同上参考。我在这个语句中的声明不是指针必须指向堆栈。指针只是一个保存内存地址的变量。这个变量在堆栈上。因为引用在堆栈上有自己的空间，并且因为地址与引用的变量相同。有关堆栈与堆的详细信息。这意味着有一个引用的真实地址，编译器不会告诉你。 
     int x = 0; 
    int& r = x; 
    int * p =& x; 
    int * p2 =& r; 
    assert（p == p2）; 
     
    




2. 您可以有指向指针的指针，提供额外的间接层。而引用仅提供一级间接。

     int x = 0; 
    int y = 0; 
    int * p =& x; 
    int * q =& y; 
    int ** pp =& p; 
    pp =& q; // * pp = q 
    ** pp = 4; 
    assert（y == 4）; 
    assert（x == 0）; 
     




3. 指针可以直接赋值为NULL，而引用不能。如果你尝试足够努力，你知道如何，你可以使引用的地址为NULL。同样，如果你努力尝试，你可以引用指针，然后该引用可以包含NULL。

     int * p = NULL; 
    int& r = NULL; < - 编译错误
     




4. 指针可以迭代数组， code> ++ 转到指针指向的下一个项目，并且 + 4 转到第5个元素。




5. 指针需要用 * 访问它指向的内存位置，而引用可以直接使用。指向类/ struct的指针使用 - > 来访问它的成员，而引用使用。 p>




6. 指针是一个保存内存地址的变量。无论如何实现引用，引用都具有与它引用的项目相同的内存地址。




7. 引用不能填充到数组中， be（由用户@litb提及）




8. Const引用可以绑定到临时表。指针不能（不是没有一些间接）：

     const int& x = int // legal C ++ 
    int * y =& int（12）; //非法取消引用临时。 
     

   这使得 const& 参数列表等。




I know references are syntactic sugar, so code is easier to read and write.

But what are the differences?

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Summary from answers and links below:

1. A pointer can be re-assigned any number of times while a reference can not be re-seated after binding.
2. Pointers can point nowhere (NULL), whereas reference always refer to an object.
3. You can't take the address of a reference like you can with pointers.
4. There's no "reference arithmetics" (but you can take the address of an object pointed by a reference and do pointer arithmetics on it as in &obj + 5).

To clarify a misconception:

The C++ standard is very careful to avoid dictating how a compiler must implement references, but every C++ compiler implements references as pointers. That is, a declaration such as:

int &ri = i;

if it's not optimized away entirely, allocates the same amount of storage as a pointer, and places the address of i into that storage.

So, a pointer and a reference both occupy the same amount of memory.

As a general rule,

• Use references in function parameters and return types to define useful and self-documenting interfaces.
• Use pointers to implement algorithms and data structures.

Interesting read:

• My alltime favorite C++ FQA lite.
• References vs. Pointers.
• An Introduction to References.
• References and const.

解决方案

1. A pointer can be re-assigned:

   int x = 5;
   int y = 6;
   int *p;
   p =  &x;
   p = &y;
   *p = 10;
   assert(x == 5);
   assert(y == 10);
   

   A reference cannot, and must be assigned at initialization:

   int x = 5;
   int y = 6;
   int &r = x;
   

2. A pointer has its own memory address and size on the stack (4 bytes on x86), whereas a reference shares the same memory address (with the original variable) but also takes up some space on the stack. Since a reference has the same address as the original variable itself, it is safe to think of a reference as another name for the same variable. Note: What a pointer points to can be on the stack or heap. Ditto a reference. My claim in this statement is not that a pointer must point to the stack. A pointer is just a variable that holds a memory address. This variable is on the stack. Since a reference has its own space on the stack, and since the address is the same as the variable it references. More on stack vs heap. This implies that there is a real address of a reference that the compiler will not tell you.

   int x = 0;
   int &r = x;
   int *p = &x;
   int *p2 = &r;
   assert(p == p2);
   

3. You can have pointers to pointers to pointers offering extra levels of indirection. Whereas references only offer one level of indirection.

   int x = 0;
   int y = 0;
   int *p = &x;
   int *q = &y;
   int **pp = &p;
   pp = &q;//*pp = q
   **pp = 4;
   assert(y == 4);
   assert(x == 0);
   

4. Pointer can be assigned NULL directly, whereas reference cannot. If you try hard enough, and you know how, you can make the address of a reference NULL. Likewise, if you try hard enough you can have a reference to a pointer, and then that reference can contain NULL.

   int *p = NULL;
   int &r = NULL; <--- compiling error
   

5. Pointers can iterate over an array, you can use ++ to go to the next item that a pointer is pointing to, and + 4 to go to the 5th element. This is no matter what size the object is that the pointer points to.

6. A pointer needs to be dereferenced with * to access the memory location it points to, whereas a reference can be used directly. A pointer to a class/struct uses -> to access it's members whereas a reference uses a ..

7. A pointer is a variable that holds a memory address. Regardless of how a reference is implemented, a reference has the same memory address as the item it references.

8. References cannot be stuffed into an array, whereas pointers can be (Mentioned by user @litb)

9. Const references can be bound to temporaries. Pointers cannot (not without some indirection):

   const int &x = int(12); //legal C++
   int *y = &int(12); //illegal to dereference a temporary.
   

   This makes const& safer for use in argument lists and so forth.

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