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downvote(之前发生过)之前,我搜索并发现没有任何相似之处.
我和许多其他人一样,正在尝试学习C++参考变量的用法并将它们与指针联系起来.我发现制作表格更容易,我需要知道是否需要修改它.

                   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

该表希望反映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. 表和解释是否正确？
2. 表格中是否有任何情况(除了像*和n,指针指针等派生的那些)？

Answer 1

一个功能

void foo(int& n);

不接受地址(指针),也不接受文字.

所以你不能称之为

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

要么

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

但是有一个例外,如果你有一个恒定的引用,比如

int foo(const int& n);

允许使用文字值,因为这样就无法更改引用的值.

同样地

void foo(int* n);

你必须传递指针.

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

最后一个:

void foo(int n);

举例:

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