C中的收益率＃

Question

這是否與c有任何等價關係？

Answer 1

否。在Windows上，您可以使用光纖實現類似的效果。

Answer 2

纖維？呵呵，這樣的：

產量返回迭代器本機C++使用光纖
http://www.codeproject.com/KB/library/fiber-based_iterator.aspx

Answer 3

雖然C沒有在關於枚舉一個集合它也有創造協同程序的能力相同的概念產量和纖維。

這裏有一些維基百科的文章可能會感興趣：

http://en.wikipedia.org/wiki/Setcontext http://en.wikipedia.org/wiki/Setjmp/longjmp

Answer 4

不可以。不過，就可以實現與使用setjmp, longjmp用C 類似的效果，但它是非常棘手。

Answer 5

產量由編譯器實現爲實現狀態機的自定義類。雖然不能輕鬆獲得語法（除非您使用之前指定的光纖方法），但您可以非常簡單地複製結果，雖然它非常繁瑣。下面是（我將在C＃中顯示，你就必須做在C適當的事情++取決於你所使用的類型）：

假設下面的代碼：

public IEnumerable<T> GetOddStrings(
    IEnumerable<IEnumerable<string>> stringCollections) 
{ 
    foreach(var stringCollection in stringCollections) 
     foreach(var str in stringCollection) 
     { 
     if(str.Length %2 != 0) yield return str; 
     if(str.Length == 42) yield break; // 42 is BAD! Stop immediately 
     } 
} 

1）攤開所有的foreach方法爲顯性枚舉來電：

public IEnumerable<T> GetOddStrings(
    IEnumerable<IEnumerable<string>> stringCollections) 
{ 
    var firstEnumerator = stringCollection.GetEnumerator(); 
    while(firstEnumerator.MoveNext()) 
    { 
    var secondEnumerator = firstEnumerator.Current.GetEnumerator(); 
    while(secondEnumerator.MoveNext()) 
    { 
     var str= secondEnumerator.Current; 
     if(str.Length %2 != 0) yield return str; 
     if(str.Length == 42) yield break; 
    } 
    } 
} 

2）將所有的局部變量的方法的頂部：

public IEnumerable<T> GetOddStrings(
    IEnumerable<IEnumerable<string>> stringCollections) 
{ 
    IEnumerator<IEnumerable<string>> firstEnumerator; 
    IEnumerator<string> secondEnumerator; 
    string str; 

    firstEnumerator = stringCollections.GetEnumerator(); 
    while(firstEnumerator.MoveNext()) 
    { 
    secondEnumerator = firstEnumerator.Current.GetEnumerator(); 
    while(secondEnumerator.MoveNext()) 
    { 
     str= secondEnumerator.Current; 
     if(str.Length %2 != 0) yield return str; 
     if(str.Length == 42) yield break; 
    } 
    } 
} 

3）使用嵌套switch語句移至循環結構。
a）更改狀態並繼續每個收益回報的循環。 b）如果條件爲 c）每個退出條件的收益率突破（下面我們反轉if）。

public IEnumerable<T> GetOddStrings(
    IEnumerable<IEnumerable<string>> stringCollections) 
{ 
    IEnumerator<IEnumerable<string>> firstEnumerator; 
    IEnumerator<string> secondEnumerator; 
    string str; 
    int state = 0; 

    while(true) 
    { 
    switch(state) 
    { 
     case 0: 
     firstEnumerator = stringCollections.GetEnumerator(); 
     // This could be "yield break" but I want to show how you 
     // could split ifs with significant code in the else 
     if(!firstEnumerator.MoveNext()) 
     { 
      state = 1; 
      continue; 
     } 

     secondEnumerator = firstEnumerator.Current; 
     if(!secondEnumerator.MoveNext) continue; 
     state = 2; 
     if(str.Length %2 != 0) yield return str; 
     continue; 

     case 1: 
     yield break; 

     case 2: 
     if(str.Length == 42) yield break; 
     state = 0; 
     continue; 
    } 
    } 
} 

4）移動到一個類，並從你的方法返回類：一 ）產量突破成爲「返回false;」 b）產量回報變成「this.Current = ??; return true;」

public IEnumerable<T> GetOddStrings(
    IEnumerable<IEnumerable<string>> stringCollections) 
{ 
    return new OddStringEnumerable(stringCollections); 
} 

private class OddStringEnumerable : IEnumerable<string> 
{ 
    IEnumerable<IEnumerable<string>> stringCollections; 
    IEnumerator<IEnumerable<string>> firstEnumerator; 
    IEnumerator<string> secondEnumerator; 
    string str; 
    int state; 

    public OddStringEnumerable(IEnumerable<IEnumerable<string>> stringCollections) 
    { 
    this.stringCollections = stringCollections; 
    } 

    public string Current { get; private set; } 

    public bool MoveNext() 
    { 
    while(true) 
    { 
     switch(state) 
     { 
     case 0: 
      firstEnumerator = this.stringCollections.GetEnumerator(); 
      if(!this.firstEnumerator.MoveNext()) 
      { 
      this.state = 1; 
      continue; 
      } 

      this.secondEnumerator = this.firstEnumerator.Current; 
      if(!secondEnumerator.MoveNext) continue; 

      this.state = 2; 
      if(str.Length %2 != 0) 
      { 
      this.Current = str; 
      return true; 
      } 
      continue; 

     case 1: 
      return false; 

     case 2: 
      if(str.Length == 42) return false; 
      this.state = 0; 
      continue; 
     } 
    } 
    } 
} 

5）根據需要進行優化。

Answer 6

Coroutines in C使用了一些預處理器hackery，但實現了相當自然的（相對於C中的其他任何東西）yield。

而你會在Python寫：

"""This is actually a built-in function. 
def range(start, stop, step): 
    i = start 
    while i < stop: 
     yield i 
     i = i + step 
""" 

if __name__ == '__main__': 
    import sys 
    start = int(sys.argv[1]) if len(sys.argv) > 2 else 0 
    stop = int(sys.argv[2]) if len(sys.argv) > 2 else int(sys.argv[1]) 
    step = int(sys.argv[3]) if len(sys.argv) > 3 else 1 
    for i in range(start, stop, step): 
     print i, 
    print 

coroutine.h，您可以用C寫：

#include <coroutine.h> 
#include <stdio.h> 

int range(int start, int stop, int step) { 
    static int i; 

    scrBegin; 
    for (i = start; i < stop; i += step) 
     scrReturn(i); 
    scrFinish(start - 1); 
} 

int main(int argc, char **argv) { 
    int start, stop, step, i; 

    start = argc > 2 ? atoi(argv[1]) : 0; 
    stop = argc > 2 ? atoi(argv[2]) : atoi(argv[1]); 
    step = argc > 3 ? atoi(argv[3]) : 1; 

    while ((i = range(start, stop, step)) >= start) 
     printf("%d ", i); 
    printf("\n"); 
} 

 
$ cc range.c 
$ ./a.out 10 
0 1 2 3 4 5 6 7 8 9 

對於更復雜的東西，並要求重入的Hamming numbers在Python ：

def hamming(): 
    yield 1 

    i2 = (2*x for x in hamming()) 
    i3 = (3*x for x in hamming()) 
    i5 = (5*x for x in hamming()) 

    m2, m3, m5 = i2.next(), i3.next(), i5.next() 

    while True: 
     if m2 < m3: 
      if m2 < m5: 
       yield m2 
       m2 = i2.next() 
      else: 
       if m2 > m5: yield m5 
       m5 = i5.next() 
     elif m2 == m3: m3 = i3.next() 
     elif m3 < m5: 
      yield m3 
      m3 = i3.next() 
     else: 
      if m3 > m5: yield m5 
      m5 = i5.next() 

if __name__ == '__main__': 
    import sys 
    it = hamming() 
    for i in range(str(sys.argv[1]) if len(sys.argv) > 1 else 25): 
     print it.next(), 
    print 

和C：

#include <coroutine.h> 
#include <stdio.h> 

int hamming(ccrContParam) { 
    ccrBeginContext; 
    ccrContext z[3]; 
    int m2, m3, m5; 
    ccrEndContext(state); 

    ccrBegin(state); 
    state->z[0] = state->z[1] = state->z[2] = 0; 
    ccrReturn(1); 

#define m2_next (2*hamming(&state->z[0])) 
#define m3_next (3*hamming(&state->z[1])) 
#define m5_next (5*hamming(&state->z[2])) 

    state->m2 = m2_next, state->m3 = m3_next, state->m5 = m5_next; 

    while (1) { 
     if (state->m2 < state->m3) { 
      if (state->m2 < state->m5) { 
       ccrReturn(state->m2); 
       state->m2 = m2_next; 
      } else { 
       if (state->m2 > state->m5) ccrReturn(state->m5); 
       state->m5 = m5_next; 
      } 
     } else if (state->m2 == state->m3) state->m3 = m3_next; 
     else if (state->m3 < state->m5) { 
      ccrReturn(state->m3); 
      state->m3 = m3_next; 
     } else { 
      if (state->m3 > state->m5) ccrReturn(state->m5); 
      state->m5 = m5_next; 
     } 
    } 
    ccrFinish(-1); 
} 

int main(int argc, char **argv) { 
    int count = argc > 1 ? atoi(argv[1]) : 25, i; 
    ccrContext z = 0; 

    for (i = 0; i < count; i++) 
     printf("%d ", hamming(&z)); 
    printf("\n"); 
} 

 
$ cc hamming.c 
$ ./a.out 
1 2 3 4 5 6 8 9 10 12 15 16 18 20 24 25 27 30 32 36 40 45 48 50 54 

Answer 7

在C＃收益簡化的集合創建IEnumberables的。

在C++中，你必須使用STL迭代器。