矢量的结构 [英] Vector of Structures

问题描述

我有另一个人给我一些代码，我们有一个结构

I have some code given to me by another person in which we have a structure

struct Pair {
    string s1;
    string s2;
    bool equivalent;
  };

然后他建立这些结构的向量硬编码

Then he sets up a vector of these structs hard coded

std::vector<Pair> PairID;

  staticdata() {
      PairID={{"string","string2",true}, 
      {"string","string3",true}, 
      {"string","string4",false}, 
      {"string","string7",false}, 
      {"string3","string8",false}
    };
    }

< string2，true}，

Unfortunately my compiler is complaining on the line PairID={{"string","string2",true},

为什么？他建议使用-std = c ++ 0x编译，但我的编译器（gcc 4.2）不支持这个。有一个简单的方法来转换代码，使其工作？为什么会失败？

Why is this? He suggested to compile using -std=c++0x but my compiler (gcc 4.2) does not support this. Is there an easy way to convert the code so it works? Why is it failing??

我使用的是Mac OSX，不想更新我的编译器

I am using Mac OSX and would prefer not to update my compiler

推荐答案

您的代码不是合法的C ++。 是合法的C ++ 0x，但语言有很多变化。所以如果你想把这个代码编译成C ++代码，你需要改变它。

Your code is not legal C++. It is legal C++0x but there have been many changes to the language. So if you want to compile this code as C++ code, you'll need to change it.

PigBen的解决方案是一种方法，它的问题是临时数据建造及

PigBen's solution is one way, the problem with it being the temporary data could be constructed & destroyed many times, or live for a long time.

这是另一种方式：

    struct Pair {
        string s1;
        string s2;
        bool equivalent;
      };

    Pair make_Pair(const string& s1, const string& s2, bool equivalent)
    {
        Pair ret;
        ret.s1 = s1;
        ret.s2 = s2;
        ret.equivalent = equivalent;
        return ret;
    }

    // somewhere in the init code...
        std::vector<Pair> PairID;


   PairID.push_back(make_Pair("string","string2",true)); 
    PairID.push_back(make_Pair("string","string3",true));
    PairID.push_back(make_Pair("string","string4",false));
    PairID.push_back(make_Pair("string","string7",false));
    PairID.push_back(make_Pair("string3","string8",false));

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