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。
我阅读了有关*引用运算符和&取消引用运算符的信息；或者引用意味着使指针指向变量，而取消引用是访问指针指向的变量的值。所以我很困惑。
我能得到一个关于“引用和解除引用”的简单而透彻的解释吗？
问问题
131113 次
7 回答
112
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指针相同类型的变量的地址，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）从指针指向的内存地址中检索值： 注意：存储在指针地址处的值必须是与类型相同的值变量指针“指向”，但除非指针设置正确，否则不能保证会出现这种情况。指针指向的变量类型是减去最外层星号的类型。
int n1;
n1 = *p1;
无效的取消引用可能会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
- 使用无效类型转换取消引用将有可能导致崩溃。
- 取消引用指向动态分配并随后被取消分配的变量的指针可能会导致崩溃
- 取消引用指向已经超出范围的变量的指针也可能导致崩溃。
无效引用比崩溃更容易导致编译器错误，但是依赖编译器来解决这个问题并不是一个好主意。
参考：
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.
于 2013-01-08T22:15:56.457 回答
21
我总是听到它们被用在相反的意义上：
&是引用运算符 - 它为您提供某个对象的引用（指针）*是解引用操作符——它接受一个引用（指针）并返回被引用的对象；
于 2013-01-08T22:16:46.657 回答
13
首先，您将它们倒退：&是引用和*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
于 2013-01-08T22:17:25.067 回答
9
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 ：
&[地址运算符]用于引用。*[star operator] 用于取消引用。
于 2013-01-09T05:38:59.767 回答
5
* 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int 或 char 等类型的上下文中，我们将使用解引用器 ' * ' 来实际表示引用（地址），如果您看到来自编译器的错误消息说：'expecting char*' 要求地址。
在这种情况下，当 * 位于类型（int、char 等）之后时，编译器会期望变量的地址。我们通过使用引用运算符来赋予它，在变量之前也称为地址运算符'&'。更进一步，在我刚才编造的情况下，编译器期望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;
将每个变量视为具有位置（如果您熟悉数组，则为索引值）和值。可能需要一些时间来习惯认为每个变量都有两个值，一个值是它的位置，与计算机中的电力一起物理存储，另一个值代表程序员想要存储的任何数量或字母。
//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，它复制传入的变量（如果有的话）（通过“（）”中的参数）。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 之外的块或 main 的子项内部的块来更改变量。
于 2018-07-30T20:07:07.757 回答
3
参考
&是参考运算符。它将内存地址引用到指针变量。
例子：
int *p;
int a=5;
p=&a; // Here Pointer variable p refers to the address of integer variable a.
取消引用
指针变量使用解引用运算符*直接访问变量的值而不是其内存地址。
例子：
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.
于 2018-04-18T01:21:58.400 回答
0
解引用指针的引用也与指向变量的地址相同。
解释 ：-
整数变量 = 3; 诠释* 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，中性引用和取消引用。
在声明函数时，
函数的参数应该是指针，
并且在 main 方法中调用 this 函数的参数中应该使用 & 运算符。
这有点令人困惑。但请记住 int *p = &var; 与上述指针声明一样也是正确的。
于 2020-11-29T15:48:01.997 回答