在c ++中的歧义功能 [英] ambigious functions in c++

问题描述

我想知道为什么这些声明不起作用（不兼容）

  void f // p1 
 void f（const int）; // p2 
 void f（int&）; // p3 
 void f（const int&）; // p4 
  

如果我理解得很好，编译器不会发现（int& &）
如果我写f（12），它将无法在两个第一个声明之间选择。
我是对吗？

原因是顶层 const 对一个值参数是不可检测的，并且在声明上是无用的。例如，您可以执行以下操作：

  void foo（int x）; //声明
 // ... 
 void foo（const int x）{
 //定义/实现
} 
  

const 这里是一个对调用者不重要的实现细节， 。该副本也是为什么它不能区别于 int 的原因，从呼叫者一侧它们是完全相同的。

注意， const int& r 没有顶层 const ，它是引用一个常量整数的引用（引用始终是常量）。对于指针，如果未声明 const ，可以更改指针，另请参见此问题 const 。

I would like to know why these declarations won't work(are not compatible)

void f(int); //p1
void f(const int);//p2
void f(int &);//p3
void f(const int &);//p4

If I understood well, the compiler won't find a difference between (int &) and (const int &) and if I write f(12) it won't be able to choose between the two first declarations.. Am I right?

解决方案

p3 and p4 are perfectly unambiguous and distinguishable, p1 and p2 are not. (And of course p1/p2 are distinguishable from p3 and p4.)

The reason is that top-level const on a value parameter is not detectable and infact useless on a declaration. You can for example do the following:

void foo(int x); // declaration
// ...
void foo(const int x){
  // definition/implementation
}

The const here is an implementation detail that's not important for the caller, since you make a copy anyways. That copy is also the reason why it's not distinguishable from just int, from the callers side it's exactly the same.

Note that const int& r does not have a top-level const, it's the reference that refers to a constant integer (references are always constant). For pointers, which may be changed if not declared const, see also this question for where to put const.

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