为什么不修改关联容器的键？ [英] Why not modify key of associative container?

问题描述

我知道在关联容器中更改对象的键是一个可怕的想法，但我不知道标准在哪里禁止我这样做。请考虑：

I know that it's a terrible idea to change the key of an object in an associative container, but I wonder where exactly the standard forbids me to do so. Consider:

#include <map>
#include <memory>

struct X { int i; };

struct lt
{
  bool operator()( const std::shared_ptr< X >& lhs,
                   const std::shared_ptr< X >& rhs ) const
  {
    return lhs->i < rhs->i;
  }
};

int main()
{
  std::map< std::shared_ptr< X >, int, lt > m;
  auto x = std::make_shared< X >();
  x->i = 1;
  m.insert( std::make_pair( x, 2 ) );

  x->i = 42; // change key wrt the container!
}

我假设上述但我现在正在阅读标准一段时间，我找不到任何实际上使 是非法的。它在哪里？

I assume that the above should be illegal, but I was reading the standard for some time now and I can't find anything that actually makes it illegal. Where is it? Or is it hiding in a future defect report?

推荐答案

这会在程序中注入未定义行为，如果您以某种方式修改值

This injects Undefined Behavior in your program if you modify the values in a way that the comparison of any two keys is different after the change according to the comparator you specified.

根据C ++ 11标准的第23.2.4 / 3节（[ associative.reqmts]）：

Per Paragraph 23.2.4/3 of the C++11 Standard ([associative.reqmts]):

短语等价键是指通过比较施加的等价关系，而不是
operator == 。也就是说，如果比较
对象<$，则认为两个键 k1 和 k2 c $ c> comp ， comp（k1，k2）== false&& comp（k2，k1）== false 。 对于
同一容器中的任何两个键 k1 和 k2 ，调用<$ c $

The phrase "equivalence of keys" means the equivalence relation imposed by the comparison and not the operator== on keys. That is, two keys k1 and k2 are considered to be equivalent if for the comparison object comp, comp(k1, k2) == false && comp(k2, k1) == false. For any two keys k1 and k2 in the same container, calling comp(k1, k2) shall always return the same value.

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