c-“引用”和“取消引用”的含义

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内容。

谈到C。

我阅读了有关*引用运算符和&解除引用运算符的信息。 或者该引用意味着使指针指向变量,而取消引用则访问该指针指向的变量的值。 所以我很困惑。

我可以得到有关“引用和取消引用”的简单但详尽的解释吗?

Milkncookiez asked 2019-11-08T02:08:54Z
6个解决方案
80 votes

引用意味着获取现有变量的地址(使用&来设置指针变量)。为了有效,必须将指针设置为与该指针相同类型的变量的地址,且不带星号:

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

取消引用指针意味着使用*运算符(星号字符)从指针所指向的内存地址中检索值:注意:指针地址处存储的值必须是“指针”指向的变量类型的“相同类型”的值,但不能保证确实如此,除非正确设置了指针。 指针所指向的变量的类型是减去最外面的星号的类型。

int n1;
n1 = *p1;

无效的解除引用可能会或可能不会导致崩溃:

  • 取消引用未初始化的指针可能会导致崩溃
  • 使用无效类型强制转换取消引用可能会导致崩溃。
  • 取消引用动态分配并随后取消分配的变量的指针可能导致崩溃
  • 取消引用自此超出范围的变量的指针也会导致崩溃。

无效的引用比崩溃更容易导致编译器错误,但是依靠编译器来实现并不是一个好主意。

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://zh.wikipedia.org/wiki/Dereference_operator]

The dereference operator * is also called the indirection operator.
A B answered 2019-11-08T02:10:14Z
17 votes

我一直听到它们的用法相反:

  • *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)并给您返回所引用的对象;

Chris Dodd answered 2019-11-08T02:10:52Z
11 votes

首先,将它们向后:引用&,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内存地址中的变量:

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
ApproachingDarknessFish answered 2019-11-08T02:11:29Z
9 votes

找到以下说明:

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:

  1. & [地址运算符]用于引用。
  2. * [star operator]用于取消引用。
nagaradderKantesh answered 2019-11-08T02:12:13Z
3 votes

参照

*是参考运算符。 它将内存地址引用到指针变量。

例:

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

取消引用

指针变量使用取消引用运算符*直接访问变量的值而不是其内存地址。

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.
zaursh answered 2019-11-08T02:13:02Z
1 votes

*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,在声明诸如int或char之类的类型的上下文中,我们将使用解引用器'*'实际表示引用(地址),如果您看到编译器发出的错误消息说:'expecting char *”,要求输入地址。

在这种情况下,当*在类型(int,char等)之后时,编译器将期望变量的地址。 我们通过在变量之前使用引用运算符,也称为地址运算符'&'来实现这一点。 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;

可以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.

}

在每个块的内部,它复制(通过“()”内的参数)传递到的变量(如果有)。 在这些块中,对变量的更改是对该变量的副本进行的,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“引用”,我们可以使用main外部或main的子代内部的块来更改变量。

Tyler Curtis Jowers answered 2019-11-08T02:13:56Z
translate from https://stackoverflow.com:/questions/14224831/meaning-of-referencing-and-dereferencing