我见过许多不同的写寻址运算符 (&) 和间接运算符 (*) 的方法
如果我没记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 写成 & var，两者之间有一个空格，会发生什么情况？我见过的常见语法:char* line = var;、char * line = var; 和 char *line = var;。
首先，int *pVar = var;是不正确的;这不存储 var 的地址，但它存储地址“5”，这将导致编译错误:
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;也会导致编译错误，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，要回答您的问题，在引用 ( &) 运算符和指针 (*) 运算符之间添加空格对编译器绝对没有影响。唯一有区别的是当你想分开 2 个标记时；喜欢const和 string例如。运行以下代码只是为了夸大您所要求的示例:
cout << & var << endl;
cout << * & var << endl;
cout << *pVar << endl;
产生与没有这么多空格的代码相同的结果:
0x7ffe243c3404
5
5