Question

我在互联网上阅读了不同的内容并感到困惑，因为每个网站都说不同的东西。

谈到C.

我阅读了*引用运算符和&解除引用运算符;或者引用意味着使指针指向变量，并且解除引用正在访问指针指向的变量的值。所以我很困惑。

我可以获得关于“引用和取消引用”的简单但彻底的解释吗？

Answer 1

引用表示获取现有变量的地址（使用＆amp;）来设置指针变量。为了有效，必须将指针设置为与指针相同类型的变量的地址，而不是星号：

int  c1;
int* p1;
c1 = 5;
p1 = &c1;
//p1 references c1

解除引用指针意味着使用*运算符（星号字符）从指针指向的内存地址中检索值：注意：存储在指针地址中的值必须是相同类型的值作为指针＆＃34;指向＆＃34;的变量类型。 to，但有无保证除非指针设置正确，否则就是这种情况。指针指向的变量类型是最外面的星号。

int n1;
n1 = *p1;

无效解除引用可能会导致崩溃，也可能不会导致崩溃：

取消引用未初始化的指针可能会导致崩溃
使用无效的类型转换取消引用可能会导致崩溃。
取消引用指向动态分配并随后取消分配的变量的指针可能会导致崩溃
取消引用指向已超出范围的变量的指针也会导致崩溃。

无效引用更可能导致编译器错误而不是崩溃，但依赖编译器并不是一个好主意。

参考文献：

http://www.codingunit.com/cplusplus-tutorial-pointers-reference-and-dereference-operators

& is the reference operator and can be read as “address of”.
* is the dereference operator and can be read as “value pointed by”.

http://www.cplusplus.com/doc/tutorial/pointers/

& is the reference operator    
* is the dereference operator

http://en.wikipedia.org/wiki/Dereference_operator

The dereference operator * is also called the indirection operator.

Answer 2

我总是听到他们使用的是相反的意思：

&是引用运算符 - 它为您提供了某个对象的引用（指针）
*是取消引用运算符 - 它接受引用（指针）并返回引用的对象;

Answer 3

首先，您需要向后设置：&是引用，*是取消引用。

引用变量意味着访问变量的内存地址：

int i = 5;
int * p;
p = &i; //&i returns the memory address of the variable i.

取消引用变量意味着访问存储在存储器地址中的变量：

int i = 5;
int * p;
p = &i;
*p = 7; //*p returns the variable stored at the memory address stored in p, which is i.
//i is now 7

Answer 4

找到以下解释：

int main()
{
    int a = 10;// say address of 'a' is 2000;
    int *p = &a; //it means 'p' is pointing[referencing] to 'a'. i.e p->2000
    int c = *p; //*p means dereferncing. it will give the content of the address pointed by 'p'. in this case 'p' is pointing to 2000[address of 'a' variable], content of 2000 is 10. so *p will give 10. 
}

结论：

& [地址运营商]用于引用。
* [星号运算符]用于取消引用。

Answer 5

引用

&是参考运算符。它会将内存地址引用到指针变量。

实施例

int *p;
int a=5;
p=&a; // Here Pointer variable p refers to the address of integer variable a.

解引用

指针变量使用

取消引用运算符*来直接访问变量的值而不是其内存地址。

实施例

int *p;
int a=5;
p=&a;
int value=*p; // Value variable will get the value of variable a that pointer variable p pointing to.

Answer 6

*所在的上下文有时会混淆含义。

  // when declaring a function
int function(int*); // This function is being declared as a function that takes in an 'address' that holds a number (so int*), it's asking for a 'reference', interchangeably called 'address'. When I 'call'(use) this function later, I better give it a variable-address! So instead of var, or q, or w, or p, I give it the address of var so &var, or &q, or &w, or &p.   

//even though the symbol ' * ' is typically used to mean 'dereferenced variable'(meaning: to use the value at the address of a variable)--despite it's common use, in this case, the symbol means a 'reference', again, in THIS context. (context here being the declaration of a 'prototype'.) 


    //when calling a function
int main(){ 
    function(&var);  // we are giving the function a 'reference', we are giving it an 'address'
  }

因此，在声明一个类型（例如int或char）的上下文中，我们将使用 dereferencer '*'来表示引用（地址），如果您看到编译器发出的错误消息：“ expecting char *”（正在询问地址），则会使您感到困惑。

在这种情况下，当*在 type （int，char等）之后时，编译器将期望变量的地址。为此，我们在变量之前使用了引用运算符，也称为 address-of 运算符'＆'。更进一步，在上面我刚刚讲过的情况下，编译器期望该地址保存一个字符值，而不是一个数字。（类型char * ==具有字符的值的地址）

int* p;
int *a;   // both are 'pointer' declarations. We are telling the compiler that we will soon give these variables an address (with &).

int c = 10;  //declare and initialize a random variable
//assign the variable to a pointer, we do this so that we can modify the value of c from a different function regardless of the scope of that function (elaboration in a second)

p = c; //ERROR, we assigned a 'value' to this 'pointer'. We need to assign an 'address', a 'reference'.
p = &c; // instead of a value such as: 'q',5,'t', or 2.1 we gave the pointer an 'address', which we could actually print with printf(), and would be something like
//so
p = 0xab33d111; //the address of c, (not specifically this value for the address, it'll look like this though, with the 0x in the beggining, the computer treats these different from regular numbers)
*p = 10; // the value of c

a = &c; // I can still give c another pointer, even though it already has the pointer variable "p"

*a = 10;
 a = 0xab33d111;

认为每个变量都有一个位置（如果您熟悉数组，则为索引值）和一个值。可能需要一些习惯才能想到每个变量都有两个值，一个值是它的位置，用电物理存储在您的计算机中，一个值表示程序员想要存储的数量或字母。

//Why it's used
int function(b){
    b = b + 1; // we just want to add one to any variable that this function operates on.
} 

int main(){

    int c = 1;  // I want this variable to be 3.

    function(c); 
    function(c);// I call the function I made above twice, because I want c to be 3.

     // this will return c as 1. Even though I called it twice.
     // when you call a function it makes a copy of the variable.
     // so the function that I call "function", made a copy of c, and that function is only changing the "copy" of c, so it doesn't affect the original
}
  //let's redo this whole thing, and use pointers

int function(int* b){ // this time, the function is 'asking' (won't run without) for a variable that 'points' to a number-value (int). So it wants an integer pointer--an address that holds a number.
*b = *b + 1; //grab the value of the address, and add one to the value stored at that address
}

int main(){
    int c = 1; //again, I want this to be three at the end of the program
    int *p = &c; // on the left, I'm declaring a pointer, I'm telling the compiler that I'm about to have this letter point to an certain spot in my computer. Immediately after I used the assignment operator (the ' = ') to assign the address of c to this variable (pointer in this case) p. I do this using the address-of operator (referencer)' & '.
    function(p); // not *p, because that will dereference. which would give an integer, not an integer pointer ( function wants a reference to an int called int*, we aren't going to use *p because that will give the function an int instead of an address that stores an int.

    function(&c); // this is giving the same thing as above, p = the address of c, so we can pass the 'pointer' or we can pass the 'address' that the pointer(variable) is 'pointing','referencing' to. Which is &c. 0xaabbcc1122...


      //now, the function is making a copy of c's address, but it doesn't matter if it's a copy or not, because it's going to point the computer to the exact same spot (hence, The Address), and it will be changed for main's version of c as well.

}

在每个块中，它复制（通过“（）”内的参数）传递到的变量（如果有）。在这些块中，对变量的更改将在该变量的副本中进行，该变量使用相同的字母，但位于不同的地址（与原始地址不同）。通过使用原始地址的“引用”，我们可以使用main外部或main的子级内部的块来更改变量。

Answer 7

取消引用的指针的引用也与指向变量的地址相同。

说明：-

int var = 3; int * p;

p =＆var;

所以，让我们认为var的地址是：ABCDE

然后

p = ABCDE和＆* p = ABCDE;

这意味着将＆*放在一起，既不引用又不引用。

也在声明函数时，

该函数的参数应该是指针，

，并且在main方法中调用this函数时，其参数中应使用＆运算符。

这有点令人困惑。但是请记住 int * p =＆var;与上述指针声明一样正确。

“引用”和“解除引用”的含义

7 个答案:

引用

实施例

解引用

实施例