Question

我已经看到了许多不同的编写地址运算符（＆amp;）和间接运算符（*）的方法

如果我没弄错的话应该是这样的：

//examples
int var = 5;
int *pVar = var;

cout << var << endl; //this prints the value of var which is 5

cout << &var << endl; //this should print the memory address of var

cout << *&var << endl; //this should print the value at the memory address of var

cout << *pVar << endl; //this should print the value in whatever the pointer is pointing to

cout << &*var << endl; //I've heard that this would cancel the two out

例如，如果您将&var写为& var并且两者之间有空格，会发生什么？我见过的常见语法：char* line = var;，char * line = var;和char *line = var;。

Answer 1

首先，int *pVar = var;不正确;这不存储var的地址，但它存储地址＆＃34; 5＆＃34;，这将导致编译错误说：

main.cpp: In function 'int main()':
main.cpp:9:15: error: invalid conversion from 'int' to 'int*' [-fpermissive]
    int *pVar = var;
                ^~~

var需要在*pvar的初始化中引用：

int *pVar = &var;

其次，cout << &*var << endl;也会在编译时导致错误，因为var不是指针（int*）类型变量：

main.cpp: In function 'int main()':
  main.cpp:19:13: error: invalid type argument of unary '*' (have 'int')
     cout << &*var << endl; //I've heard that this would cancel the two out
               ^~~

现在，为了回答您的问题，在引用（&）运算符和指针（*）运算符之间添加空格将使编译器完全没有区别。它唯一有意义的是你想要分开2个代币;比如const和string。运行以下代码只是为了夸大你所问的例子：

cout <<             &             var << endl;  
cout <<                            *                    & var << endl;
cout <<                                   *pVar << endl;

产生与没有太多空格的代码相同的结果：

0x7ffe243c3404
5
5

地址（＆amp;）和间接运算符的正确格式是什么

1 个答案: