c++ - 操作数 "="和 "!="的 vector 错误迭代器

Question

using namespace std;
    vector<IDrawable*>::const_iterator itDrawable;
    for(itDrawable= scene.getDrawables().begin(); itDrawable!=scene.getDrawables().end();itDrawable++){
        IDrawable *drawable =(*itDrawable);
        drawable->draw();
    }

此代码向我传递了错误:

Description Resource Path Location Type no match for 'operator!=' (operand types are 'std::vector<IDrawable*>::const_iterator {aka __gnu_cxx::__normal_iterator<IDrawable* const*, std::vector<IDrawable*> >}' and 'std::vector<const
IDrawable*>::const_iterator {aka __gnu_cxx::__normal_iterator<const
IDrawable* const*, std::vector<const IDrawable*> >}')

和

Description Resource Path Location Type no match for 'operator=' (operand types are 'std::vector<IDrawable*>::const_iterator {aka __gnu_cxx::__normal_iterator<IDrawable* const*, std::vector<IDrawable*> >}' and 'std::vector<const
IDrawable*>::const_iterator {aka __gnu_cxx::__normal_iterator<const
IDrawable* const*, std::vector<const IDrawable*> >}')

我已经查过这些，我应该与 const_iterator 有什么关系？然而我的scene.getDrawables()看起来像:

const std::vector<const IDrawable*>& getDrawables() const {
        return drawables;
    }

所以迭代器应该是一个 const_iterator 对吗？我不知道必须改变什么...

Answer 1

你的

const std::vector<const IDrawable*>& getDrawables() const

返回对 const IDrawable* vector 的常量引用指针。然而你的

vector<IDrawable*>::const_iterator itDrawable; 

声明一个 const_iterator到不同类型的 vector ( IDrawable* ，而不是 const IDrawable* )。将定义更改为 vector<const IDrawable*>::const_iterator itDrawable;或者简单地使用 auto在你的 for 中声明迭代器循环，

for(auto itDrawable= scene.getDrawables().cbegin(); ...)