C++中的生成器 - 无效使用非静态数据成员

Question

我有点理解这一点,至少是生成器的功能(我在Python中使用过它们).我理解switch语句及其内容是如何形成的.但是,我收到了这些错误.

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_'

这是代码.如果您有更好的方法来解决这个问题,请务必分享.

#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;
}

Answer 1

这是你追求的吗？

我不太确定你想要什么generator回来,但请按你的意愿修改.

这个想法是生成器对象保持它自己的状态,当你调用它的方法时,它会返回下一个值.完全取决于您定义的状态和要返回的下一个值.

在operator()可以接受的参数,如operator()(bool b),或operator()(char * c)任何你想要的价值,以及回报...

#include <iostream>
#include <vector>
#include <string>
using namespace std;

struct generator
{
    generator() : currentCh(0), currentW(0), str(0), words()
    {   // do whatever initialization you need
        words.push_back("Foobar");
        words.push_back("Barfoo");
        str = &words[currentW];
    }

    char operator()()
    { // use whatever logic you need
        if (currentCh >= str->size())
        {
            if (++currentW >= words.size()) currentW = 0;
            str = &words[currentW];
            currentCh = 0;
        }
        return str->at(currentCh++);
    }

    unsigned int currentCh;
    unsigned int currentW;
    string * str;
    vector<string> words;
};

您可以随意更新内部状态,例如添加:

char operator()(unsigned int index)
{
    currentCh = index;
    return this->operator()();
}

然后在您的代码中,您可以执行以下操作:

generator g;
g();  // get first character
g(2); // resets the index in this case ...
g();  // get next character (the logic is a bit off but it depends what you need)

用法:

int main()
{
    generator g;
    for (unsigned int i = 30; --i; ) cout << g() << endl;
}

输出:

F
o
o
b
a
r
B
a
r
f
o
o
F
o
o
b
a
r
B
a
r
f
o
o
F
o
o
b
a