从std :: vector中删除项目

Question

在下面的第一个代码片段中，我试图基于提供给std :: remove函数的静态条件函数从成员函数中的向量中删除元素。然后我收到第二个片段中显示的大量模板错误。你能告诉我我错过了什么吗？

SNIPPET 1（代码）

void removeVipAddress(std::string &uuid)
{
          struct RemoveCond
          {
            static bool condition(const VipAddressEntity & o)
            {
              return o.getUUID() == uuid;
            }
          };

          std::vector<VipAddressEntity>::iterator last =
            std::remove(
                    mVipAddressList.begin(),
                    mVipAddressList.end(),
                    RemoveCond::condition);

          mVipAddressList.erase(last, mVipAddressList.end());

}

SNIPPET 2 （编辑输出）

 /usr/include/c++/4.7/bits/random.h:4845:5: note: template<class _IntType> bool      std::operator==(const std::discrete_distribution<_IntType>&, const   std::discrete_distribution<_IntType>&)
 /usr/include/c++/4.7/bits/random.h:4845:5: note:   template argument deduction/substitution failed:
 In file included from /usr/include/c++/4.7/algorithm:63:0,
             from Entity.hpp:12:
 /usr/include/c++/4.7/bits/stl_algo.h:174:4: note:   ‘ECLBCP::VipAddressEntity’ is not  derived from ‘const std::discrete_distribution<_IntType>’
 In file included from /usr/include/c++/4.7/random:50:0,
               from /usr/include/c++/4.7/bits/stl_algo.h:67,
               from /usr/include/c++/4.7/algorithm:63,
               from Entity.hpp:12:
 /usr/include/c++/4.7/bits/random.h:4613:5: note: template<class _RealType> bool  std::operator==(const std::extreme_value_distribution<_RealType>&, const  std::extreme_value_distribution<_RealType>&)
 /usr/include/c++/4.7/bits/random.h:4613:5: note:   template argument deduction/substitution failed:
 In file included from /usr/include/c++/4.7/algorithm:63:0,
             from Entity.hpp:12:
 /usr/include/c++/4.7/bits/stl_algo.h:174:4: note:   ‘ECLBCP::VipAddressEntity’ is not  derived from ‘const std::extreme_value_distribution<_RealType>’

Answer 1

我猜您正在寻找std::remove_if()，而不是std::remove()。

std::remove_if()接受谓词作为第三个参数，并删除满足该谓词的元素。

std::remove()将值作为第三个参数，并删除等于该值的元素。

修改

要使其工作，您还必须将RemoveCond定义转换为谓词对象，因为它需要状态。像这样：

void removeVipAddress(std::string &uuid)
{
      struct RemoveCond : public std::unary_function<VipAddressEntity, bool>
      {
        std::string uuid;

        RemoveCond(const std::string &uuid) : uuid(uuid) {}

        bool operator() (const VipAddressEntity & o)
        {
          return o.getUUID() == uuid;
        }
      };

      std::vector<VipAddressEntity>::iterator last =
        std::remove(
                mVipAddressList.begin(),
                mVipAddressList.end(),
                RemoveCond(uuid));

      mVipAddressList.erase(last, mVipAddressList.end());

}