C ++中的生成器 - 无效使用非静态数据成员 [英] Generators in C++ -- invalid use of nonstatic data member

问题描述

我理解这一点，至少是generator的函数（我在Python中使用它们）。我理解switch语句及其内容是如何形成的。但是，我得到这些错误。

I sort of understand this, at least the function of generators (I've used them in Python). I understand how the switch statement and its content is formed. However, I get these errors.

test.cpp: In constructor 'Foo::descent::descent(int)':
test.cpp:46: error: invalid use of nonstatic data member 'Foo::index_'
test.cpp: In member function 'bool Foo::descent::operator()(std::string&)':
test.cpp:50: error: invalid use of nonstatic data member 'Foo::bars_'
test.cpp:50: error: invalid use of nonstatic data member 'Foo::index_'
test.cpp:51: error: invalid use of nonstatic data member 'Foo::index_'
test.cpp:51: error: invalid use of nonstatic data member 'Foo::bars_'
test.cpp:52: error: invalid use of nonstatic data member 'Foo::index_'

。如果你有一个更好的方式来处理这一点，一定要分享，请。

Here's the code. If you have a better way of dealing with this, by all means share please.

#include <math.h>
#include <string>
#include <vector>
#include <iostream>

#ifndef __generator_h__
#define __generator_h__

// generator/continuation for C++
// author: Andrew Fedoniouk @ terrainformatica.com
// idea borrowed from: "coroutines in C" Simon Tatham,
//                     http://www.chiark.greenend.org.uk/~sgtatham/coroutines.html

struct _generator
{
  int _line;
  _generator():_line(0) {}
};

#define $generator(NAME) struct NAME : public _generator

#define $emit(T) bool operator()(T& _rv) { \
                    switch(_line) { case 0:;

#define $stop  } _line = 0; return false; }

#define $yield(V)     \
        do {\
            _line=__LINE__;\
            _rv = (V); return true; case __LINE__:;\
        } while (0)
#endif

class Foo {
    int index_;
    std::vector<std::string> bars_;
    public:
    Foo() {
        index_ = 0;
        bars_.push_back("Foobar");
        bars_.push_back("Barfoo");
    }
    $generator(descent){
        int j;
        descent(int j) {
            index_+=j;
        }
        $emit(std::string)
            while(true) {
                $yield(bars_[index_++]);
                if(index_ >= bars_.size())
                    index_ = 0;
            }
        $stop;
    };
    //descent bar;
    void InitGenerator() { index_ = 0; }
};

using namespace std;

int main()
{
  //Foo::descent gen(1);
  //for(int n; gen(n);) // "get next" generator invocation
  //  cout << n << endl;
  return 0;
}

推荐答案

让我们来展开宏，看看它是怎样的：

Let's expand the macros to see how this really looks:

class Foo {
    int index_;
    std::vector<std::string> bars_;
    public:
    Foo() {
        index_ = 0;
        bars_.push_back("Foobar");
        bars_.push_back("Barfoo");
    }
    struct descent: public _generator {
        int j;
        descent(int j) {
            index_+=j;
        }
        bool operator()(std::string& _rv) {
         switch(_line) { case 0:;
            while(true) {
                do {
                    _line=__LINE__;
                    _rv = (bars_[index_++]); return true; case __LINE__:;
                } while (0);
                if(index_ >= bars_.size())
                    index_ = 0;
            }
         } _line = 0; return false; }
    };
    //descent bar;
    void InitGenerator() { index_ = 0; }
};

正如你所看到的，我们声明一个内部结构 Foo :: descent 。然而，与一些其他语言不同，C ++中的内部类不会自动具有指向其外部类的实例的指针。您需要添加到 descent a Foo * / code>构造函数，并使用 Foo * 引用 index _ 和 bars_ - 或将必要的成员移入下降。

As you can see, we declare an inner structure Foo::descent. However, unlike in some other languages, inner classes in C++ do not automatically have a pointer to an instance of their outer class. You need to either add to descent a Foo * which is passed in via the descent constructor, and use that Foo * to reference index_ and bars_ - or move the necessary members right into descent.

真的不明白什么 Foo 这里是为所有...一切似乎都属于下降 。

To be honest, I don't really understand what Foo here is for at all... Everything it does seems to belong right in descent.

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