叉/联接计算

Question

我有叉/联接计算的示例。有人能否简要介绍一下它在这里的工作原理？

def quicksortForkJoin(numbers) { 
    withPool { 
     runForkJoin(0, numbers) {index, list -> 
      def groups = list.groupBy {it <=> list[list.size().intdiv(2)]} 
      if ((list.size() < 2) || (groups.size() == 1)) { 
       return [index: index, list: list.clone()] 
      } 
      (-1..1).each { forkOffChild(it, groups[it] ?: [])} 
      return [index: index, list: childrenResults.sort {it.index}.sum {it.list}] 
     }.list 
    } 
} 

Answer 1

还好吗？

def quicksortForkJoin(numbers) { 

    // Create a pool of workers the default size 
    withPool { 

     // Run a fork with index 0 and the numbers 
     runForkJoin(0, numbers) {index, list ->   // [1] 

      // Split numbers into 3 groups: 
      // -1: those less than the "middle" number 
      //  0: those equal to the "middle" number 
      //  1: those greater than the "middle" number 
      def groups = list.groupBy {it <=> list[list.size().intdiv(2)]} 

      // If there are less than 2 numbers to sort, or all numbers are equal 
      if ((list.size() < 2) || (groups.size() == 1)) { 

       // return the index and a clone of the current list 
       return [index: index, list: list.clone()] 
      } 

      // Otherwise, fork off a child process for each of the 
      // groups above (less than, equal and greater than) 
      // forkOffChild will not block, and will effectively go back 
      // to line [1] with the new index and list 
      (-1..1).each { forkOffChild(it, groups[it] ?: [])} 

      // Block waiting for all 3 children to finish, then sort the 
      // results so the indexes are [ -1, 0, 1 ] and then join the 
      // lists of numbers back together 
      return [ index: index, 
        list: childrenResults.sort {it.index}.sum {it.list}] 

      // when done, return the `list` property from the final map 
     }.list 
    } 
}