const char *在地图中找不到 [英] const char* not found in map find

问题描述

我有一个小程序，我试图在地图中搜索一个特定的字符串，我得到正确的结果，如果我传递静态const字符串us找到，但它不工作，如果我复制一个字符串到指针。
我不知怎么想，它试图比较传递的字符串的地址。我计划摆脱std :: string从由于某些原因。

I have a small program, I am trying to search for a particular string in the map, I get correct result if I pass static const string "us" to find but it doesn't work if I copy a string to pointer. I somehow think that it is trying to compare the address of the string that is passed. I am planning to get rid of std::string from due to some reason.

using namespace std;

static struct MarketLang {
    const char* market;
    const char* lang;
} market_lang[] = {
    {"us",  "all"},
    {"el",  "en,fr,it,de,es"},
    {"xx",  "na"},
};

class MarketLangMap {
    map<const char*, MarketLang *> table;
    public:
    MarketLangMap() {
        int TOTAL_MARKET_INFO = sizeof(market_lang)/sizeof(MarketLang);
        for (int i = 0; i < TOTAL_MARKET_INFO; i++) {
            table[market_lang[i].market] = market_lang+ i;
        }
    }

    MarketLang *operator[](const char* s) {
        if (table.find(s) != table.end()) {
            return table.find(s)->second;
        } else {
            return table.find("xx")->second;
        }
    }
};


int
main()
{

   MarketLangMap *m = new MarketLangMap();
   const char* r = "us";
   char* p = new char(10);
   strcpy(p, "us");
   std::cout<<(*m)["us"]->lang <<std::endl;``
   std::cout<<(*m)[r]->lang <<std::endl;
   std::cout<<(*m)[p]->lang <<std::endl;

}

预期输出：
all
all
all

Expected output: all all all

在此输入代码

实际输出：
all $ b b all
na

Actual output: all all na

推荐答案

std :: map 使用由其内部比较对象指示的严格弱排序标准，其默认为 std :: less 。

std::map is using strict weak ordering criterion indicated by its internal comparison object, which is defaulted to the std::less.

std :: less 不会处理 char * 参数作为一个字符串，它只会看到它们作为指针，它只是检查一个指针是否小于另一个。

std::less will not treat char* arguments as a string, it will just see them as a pointers, and it will just check if one pointer is less than the other.

但是，你可以猜猜你应该使用 std :: string 避免混合C和C ++），创建一个新类来比较 char * ，将它们视为字符串：

However, you can (but I guess you should use std::string to avoid mixing C and C++), create new class for comparing char* arguments, treating them as strings:

#include <cstring>

struct ConstCharStarComparator
{
  bool operator()(const char *s1, const char *s2) const
  {
    return strcmp(s1, s2) < 0;
  }
};

map<const char*, MarketLang *, ConstCharStarComparator> table;

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