这意味着const int *& var？

本文介绍了这意味着const int *& var？的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述

我看到有人在使用

void methodA(const int*& var);

一个答案，但不明白该参数的含义。

in one answer, but couldn't understand what the argument means.

AFAIK：

• const int var =>不能更改的const值

• const int var => const value which can't be changed

const int * var =>指向int，即const，即* var不能更改，但var可以更改

const int* var => pointer to int which is const i.e *var can't be changed but var can be changed

const int& var =>对const int的引用，即var的值无法更改

const int& var => reference to const int i.e value of var can't be changed

const int *&是什么？ var 的平均值，并且是 const int& * var 也可以吗？

What does const int*& var mean, and is const int& *var also possible?

您能否也举一些例子，例如可以做什么和不能做什么？ ？

Can you please give some example as well, like what can be done and what can't be done with it?

更新：

我不确定我的想法是否正确，但我开始考虑参考作为作为参数传递的变量的别名，因此 const int * p; methodA（p）=>在这里，我们将p作为const int *传递，但是直到我们看到方法A的定义，我们才知道这是按值传递还是传递值，

I am not sure if I am thinking right way, but I began to think reference as alias of the variable that was pass as argument, so const int * p; methodA(p) => here we are passing p as const int * but we dont know if this is pass by value or what until we see the definition of method A,

因此，如果methodA像这样 methodA（const int *& p2）==>这里p2是p的另一个名称，即p和p2从现在起在上是相同的，如果methodA（const int * p2）==>此处p2作为值传递，即p2仅在此方法本地，

so if methodA is like this methodA(const int * & p2) ==> here p2 is another name to p, i.e. p and p2 are same from now on and if methodA(const int* p2) ==> here p2 is passed as value i.e p2 is just local to this method,

如果我想错了方法，请纠正我？如果是，我可能需要进一步研究？你能指点一些不错的参考吗？

please correct me if I am thinking wrong way ? If yes, I might need to study some more about this ? Can you please point some nice references ?

更新2 如果像我这样的初学者想了解更多关于这件事的信息，可以使用 c + + decl / cdecl程序，我刚刚从此处

UPDATE 2 If some beginner like me want to know more about this thing, you can use c++decl / cdecl program, which I just discovered to very useful from here

$ c++decl Type `help' or `?' for help c++decl> explain const int&* p declare p as pointer to reference to const int c++decl> explain const int*& p declare p as reference to pointer to const int

的指针的引用，但是正如这里每个人所指出的，第一个示例在C ++中不是合法的。

But as every one here pointed, first example isnt legal in C++.

谢谢

推荐答案

还有另一篇文章有​​些相关，实际上是此处。我的答案给出了一种一般的算法来解决这些问题。

There is another post somewhat related, actually, here. My answer gives a sorta of general algorithm to figuring these things out.

此： const int& * var 没有意义，因为您没有指向引用的指针。

This: const int& *var has no meaning, because you cannot have a pointer to reference.

如果const和指针妨碍了您的操作，请记住您可以键入以下内容：

If the const's and pointers are getting in the way, remember you can typedef these things:

typedef int* IntPointer; typedef const IntPointer ConstIntPointer; void foo(ConstIntPointer&); // pass by reference void bar(const ConstIntPointer&); // pass by const reference void baz(ConstIntPointer); // pass by value

可能会更易于阅读。

如果您需要有关C ++的更多帮助，请阅读此内容。更具体地说，引用。

If you need more help on C++, read this. More specifically, references.

参考因为变量 not 不占用空间：

References as variables do not take space:

int i; // takes sizeof(int) int*pi = &i; // takes sizeof(int*) int& ri = i; // takes no space. // any operations done to ri // are simply done to i

作为参数的引用使用指针来达到最终效果：

References as parameters use pointers to achieve the end effect:

void foo(int& i) { i = 12; } void foo_transformed(int *i) { *i = 12; } int main() { int i; foo(i); // same as: foo_transformed(&i); // to the compiler (only sort of) }

因此，它实际上是在传递地址 i 在堆栈上，因此在堆栈上占用 sizeof（int *）空间。但不要开始考虑将引用作为指针。它们不相同。

So it's actually passing the address of i on the stack, so takes sizeof(int*) space on the stack. But don't start thinking about references as pointers. They are not the same.