继 c++ set 重载无法正确调用,给了一个完整的 demo - V2EX
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huanyingch01
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继 c++ set 重载无法正确调用,给了一个完整的 demo

  •  
  •   huanyingch01 2017-10-30 12:26:42 +08:00 2824 次点击
    这是一个创建于 2991 天前的主题,其中的信息可能已经有所发展或是发生改变。

    // buffer

    #ifndef __Buffer_H__ #define __Buffer_H__ #include <iostream> #include <vector> #include <list> #include <deque> #include <map> #include <set> class CBuffer { public: CBuffer() { } ~CBuffer() { } template <class T> CBuffer& operator<<(T &val) { std::cout << "11111" << std::endl; return *this; } CBuffer& operator<<(std::string &val) { std::cout << "22222" << std::endl; return *this; } template <class T> CBuffer& operator<<(std::list<T> &val) { auto iter = val.begin(); while (iter != val.end()) { *this << (*iter++); } return *this; } template <class T> CBuffer& operator<<(std::deque<T> &val) { auto iter = val.begin(); while (iter != val.end()) { *this << (*iter++); } return *this; } template <class T> CBuffer& operator<<(std::vector<T> &val) { for (size_t i = 0; i < val.size(); ++i) { *this << val[i]; } return *this; } template <class T> CBuffer& operator<<(std::set<T> &val) { auto iter = val.begin(); while (iter != val.end()) { // static_cast<T> 不加上这个无法正确重载,原因未知 *this << (*iter++); } return *this; } }; #endif 

    // main

    #include <iostream> #include "Buffer.h" using namespace std; int main() { CBuffer buf; std::set<std::string> a1 = { "aaa"}; std::vector<std::string> a2 = { "aaa"}; std::list<std::string> a3 = { "aaa" }; std::deque<std::string> a4 = { "aaa" }; buf << a1 << a2 << a3 << a4; system("pause"); return 0; } 

    输出

    11111 22222 22222 22222 

    set 的<<重载调用与其他 vector,list,deque 不同。 不解!!!

    7 条回复    2017-10-30 14:10:31 +08:00
    huanyingch01
        1
    huanyingch01  
    OP
       2017-10-30 12:27:32 +08:00
    环境 vs2015
    gulucn
        2
    gulucn  
       2017-10-30 12:36:01 +08:00
    set 里面的值应该是不能变化的,因为其实调用的是
    CBuffer& operator<<(T &val)
    这个函数吧, T 为 const std::string
    wevsty
        3
    wevsty  
       2017-10-30 12:45:29 +08:00
    @gulucn 正解
    jlsk
        4
    jlsk  
       2017-10-30 12:55:54 +08:00
    2L+1
    一个 const 就会造成是不同的类型
    再加个重载就知道了
    CBuffer& operator<<(const std::string &val) {
    std::cout << "33333" << std::endl;
    return *this;
    }
    huanyingch01
        5
    huanyingch01  
    OP
       2017-10-30 12:59:32 +08:00
    @gulucn 谢谢,确实是这样。
    huanyingch01
        6
    huanyingch01  
    OP
       2017-10-30 13:05:52 +08:00
    @jlsk 是的,3q。
    gnaggnoyil
        7
    gnaggnoyil  
       2017-10-30 14:10:31 +08:00
    这算是个老的 Defect Reports 了……见 https://cplusplus.github.io/LWG/issue103

    所以现在标准里在[associative.reqmts]这一节直接有这么一句话:"For associative containers
    where the value type is the same as the key type, both iterator and const_iterator are constant iterators."
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