c++ - 我正在尝试将 C++ 引用与指针相关联

Question

在说这将是一个重复的问题和否决票(就像以前发生的那样)之前，我进行了搜索，但没有发现任何相似的东西。
我和许多其他人一样，正在尝试学习 C++ 引用变量的使用并将它们与指针相关联。我发现制作表格更容易，我需要知道它是否需要修改。

                   int *n   int n    int &n    caller/local
void foo(int *n)     n       &n        &n          caller
void foo(int n)     *n        n         n           local
void foo(int &n)    *n        n         n          caller

表格要反射(reflect)所有合法传递的参数。

[1,1]: passing by reference (trivial)  
[1,2]: passing by reference  
[1,3(1)]: passing by reference, an is an address(?)  
[1,3(2)]: passing by reference, as n is used as alias(?)  
[2,1]: passing by value, as dereferencing  
[2,2]: passing by value (trivial)  
[2,3(1)]: passing by value, using value of n (where n is an alias)  
[2,3(2)]: passing by value (dereferencing n, which is an address)  
[3,1(1)]: passing by reference, as foo accepts address  
[3,1(2)]: passing by reference, reference of value at address n  
[3,2(1)]: passing by reference (trivial)  
[3,2(2)]: passing by reference, as foo accepts address or reference  
[3,3]: passing by reference (trivial, as argument matches parameter exactly)  

1. 表格和解释是否正确？
2. 表中是否遗漏了任何情况(派生的情况除外，如 *&n、指向指针的指针等)？

Answer 1

一个函数

void foo(int& n);

不接受地址(指针)，也不接受文字。

所以你不能这样调用它

int a = ...;
foo(&a);  // Trying to pass a pointer to a function not taking a pointer

或

foo(1);  // Passing R-value is not allowed, you can't have a reference to a literal value

但是有一个异常(exception)，如果你有一个常量引用，比如

int foo(const int& n);

然后允许使用文字值，因为这样引用的值就不能更改。

---

同样适用于

void foo(int* n);

您必须传递一个指针。

例如:

int a = ...;
int& ra = a;   // ra references a

foo(&a);  // OK
foo(&ra); // OK
foo(a);   // Fail, a is not a pointer
foo(ra);  // Fail, ra is not a pointer
foo(1);   // Fail, the literal 1 is not a pointer

---

最后:

void foo(int n);

举例说明:

int a = ...;
int& ra = a;   // ra references a
int* pa = &a;  // pa points to a

foo(a);   // OK, the value of a is copied
foo(ra);  // OK, the value of the referenced variable is copied
foo(*pa); // OK, dereferences the pointer, and the value is copied
foo(pa);  // Fail, passing a pointer to a function not expecting a pointer
foo(1);   // OK, the literal value 1 is copied