C ++重载分辨率 [英] C++ overload resolution

问题描述

给定以下示例，为什么我必须显式使用语句 b-> A :: DoSomething（），而不是 b - > DoSomething（）？

编译器的重载分辨率不应该决定我正在讨论的方法吗？

我使用的是Microsoft VS 2005.（注意：在这种情况下使用virtual不起作用。）

  class A 
 {
 public：
 int DoSomething（）{return 0;}; 
}; 
 
 class B：public A 
 {
 public：
 int DoSomething（int x）{return 1;}; 
}; 
 
 int main（）
 {
 B * b = new B（）; 
 b-> A :: DoSomething（）; //为什么这个？ 
 // b-> DoSomething（）; //为什么不是这个？ （给出编译器错误。）
 delete b; 
 return 0; 
} 
  

解决方案

在同一范围内。默认情况下，编译器只考虑最小可能的名称范围，直到找到名称匹配。随后进行参数匹配。在你的情况下，这意味着编译器看到 B :: DoSomething 。

一个解决方案是从 A下拉过载 into B 的范围：

  
 public：
 using A :: DoSomething; 
 // ... 
} 
  

Given the following example, why do I have to explicitly use the statement b->A::DoSomething() rather than just b->DoSomething()?

Shouldn't the compiler's overload resolution figure out which method I'm talking about?

I'm using Microsoft VS 2005. (Note: using virtual doesn't help in this case.)

class A
{
  public:
    int DoSomething() {return 0;};
};

class B : public A
{
  public:
    int DoSomething(int x) {return 1;};
};

int main()
{
  B* b = new B();
  b->A::DoSomething();    //Why this?
  //b->DoSomething();    //Why not this? (Gives compiler error.)
  delete b;
  return 0;
}

解决方案

The two "overloads" aren't in the same scope. By default, the compiler only considers the smallest possible name scope until it finds a name match. Argument matching is done afterwards. In your case this means that the compiler sees B::DoSomething. It then tries to match the argument list, which fails.

One solution would be to pull down the overload from A into B's scope:

class B : public A {
public:
    using A::DoSomething;
    // …
}

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