矢量和const [英] vector and const

问题描述

 void f(vector<const T*>& p)
 {
 }
 int main()
 { 
  vector<T*> nonConstVec;
  f(nonConstVec);
 }

<向量< T *> 不能转换为向量< const T *> 从 T * 隐式转换为 const T * 。为什么是这样？

The following does not compile.The thing is that vector<T*> can not be converted to vector <const T*> , and that seems illogically to me , because there exists implicit conversion from T* to const T*. Why is this ?

向量< const T *> 不能转换为向量< T *> ，但是这是预期的，因为const T * 不能被隐式转换为 T * 。

vector<const T*> can not be converted to vector <T*> too, but that is expected because const T* can not be converted implicitly to T*.

推荐答案

我为您的代码添加了几行。这足以说明为什么不允许这样做：

I've added a few lines to your code. That's sufficient to make it clear why this is disallowed:

void f(vector<const T*>& p)
 {
    static const T ct;
    p.push_back(&ct); // adds a const T* to nonConstVec !
 }
 int main()
 { 
  vector<T*> nonConstVec;
  f(nonConstVec);
  nonConstVec.back()->nonConstFunction();
 }

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