矢量大小不正确 - 我们如何在结构内初始化矢量

Question

I have below sample code:

#include <iostream>
#include <vector>

typedef struct defs
{
    std::vector<int> myvec;
} dvec;

typedef struct devdef: public dvec
{
    dvec dvec1;
    devdef()
    {
          dvec1.myvec.push_back(10);
    }
}myVec;

class A
{
public:
    A():v1()
    {
         std::cout << "my vec size is " << v1.myvec.size() << std::endl;
    }
private:
    myVec v1;
};

int main()
{
     A a1;
}

Compile and execute:

$ c++ -std=c++11 try58.cpp

$ ./a.exe
my vec size is 0

I was expecting the size of vector to be 1 - why the same is 0?

Answer 1

You have dvecs as both a parent to myVec as well as a member of myVec . In your constructor you push a value to dvec1.myvec , but in A you read v1.myvec aka the member of the parent. You simply have two instances of dvecs in your structure and you're using different ones in your functions.

Answer 2

This version, I think, does what you wanted.

#include <iostream>
#include <vector>

struct dvec
{
    std::vector<int> myvec;
}

struct myVec : public dvec
{
    myVec()
    {
          myvec.push_back(10);
    }
}

class A
{
public:
    A():v1()
    {
         std::cout << "my vec size is " << v1.myvec.size() << std::endl;
    }
private:
    myVec v1;
};

int main()
{
     A a1;
}

Some notes about what I've changed.

1. There's no need to use a typedef to create a new type. In C++ both class and struct define new types.
2. When deriving the new type myVec from dvec , it contains the vectror myvec , so there is no need to add a new one.
3. In the constructor for the derived type myVec , we can access the myvec member of the base type directly, since it is a member of this object.