返回不停止功能，递归函数问题？(编程练习，动态编程，Levenshtein回溯)

Question

printOptimalAlignment函数行为不当。当函数到达位置(1，1)时，goto和return不会退出.在它应该结束的地方，没有坠毁，它似乎停留在一个任意的位置(6，6).因为由于某种原因，即使int yL，int xL的值没有增量-er，但如果它到达函数的末尾而不对if语句进行任何“命中”，那么它也会递增(但我不理解为什么它会调用自己。

完整代码：https://repl.it/@fulloutfool/Edit-Distance

void printOptimalAlignment(int** arr, string y, string x,int yL, int xL){
  int I_weight=1, D_weight=1, R_weight=1;
  bool printinfo_allot = 1,printinfo = 1; 

  if(printinfo_allot){
    cout<<"Location: "<<"("<<xL<<","<<yL<<")"<<"-------------------------------\n";
    cout<<"Same check Letters: "<<x[xL-2]<<","
      <<y[yL-2]<<"("<<(x[xL-2] == y[yL-2])<<")"<<"\n";
    cout<<"LL: "<<"("<<xL-1<<","<<yL<<")"
      <<":"<<arr[yL][xL-1]
      <<":"<<(arr[yL][xL-1]+I_weight)
      <<":"<<(arr[yL][xL])
      <<":"<<(((arr[yL][xL-1]+I_weight) == arr[yL][xL])==1)
      <<":"<<(yL>=1 && xL>=1)<<"\n";
    cout<<"xL state:"<<((&x[xL]))<<":"<<(x[xL-1])<<"\n";
    cout<<"yL state:"<<((&y[yL]))<<":"<<(y[yL-1])<<"\n";
    string tx = &x[xL];
    cout<<x.length()<<","<<(tx.length()+1)<<"\n";
  }

  string tx = &x[xL]; // slopy hotfix
  if(x.length()==(tx.length()+1)){
    cout<<"return functionality not working?-=-=-=-=-=-=-=-=\n";
    cout<<"-> Prep last, current distance = "<<arr[yL][xL] <<"\n";
    return;
    //printOptimalAlignment(arr,y,x,yL-1,xL-1);
    //cant use this goto... but where does it go?
    //goto because_Im_a_terrible_person;
    throw "how?... breaking rules... make it stop";
  }

  if(yL>=1 && xL>=1 && (x[xL-2] == y[yL-2]) == 1){
    if(printinfo){
      cout<<"-> Same (same char), current distance = "<<arr[yL][xL] <<"\n";
    }
    printOptimalAlignment(arr,y,x,yL-1,xL-1);
  }

  if(yL>=1 && xL>=1 && (arr[yL-1][xL-1] == arr[yL][xL])){
    if(printinfo){
      cout<<"-> Swap (same int), current distance = "<<arr[yL][xL] <<"\n";
      if(arr[yL-1][xL-1]==0)cout<<"---this is last---\n";
    }
    printOptimalAlignment(arr,y,x,yL-1,xL-1);
  }

  if(yL>0 && xL>0 && (arr[yL-1][xL]+D_weight == arr[yL][xL])){
    if(printinfo){
      cout<<"-> Delete, current distance = "<<arr[yL][xL]<<"\n";
    }
    printOptimalAlignment(arr,y,x,yL-1,xL);
  }

  //really weird ((yL>1 && xL>1) && (((arr[yL][xL-1]+I_weight) == arr[yL][xL])==1))
  //not true if it is?
  bool seperate = (((arr[yL][xL-1]+I_weight) == arr[yL][xL])==1);
  if(yL>=1 && xL>=1){
    if((((arr[yL][xL-1]+I_weight) == arr[yL][xL])==1) && (true)){
      if(printinfo){
        cout<<"-> Insert, current distance = "<<arr[yL][xL]<<"\n";
        cout<<"Next Location1: "<<"("<<xL-1<<","<<yL<<")"<<"\n";
      }
      printOptimalAlignment(arr,y,x,yL,xL-1);
      return;
      //how does it get here... also return gets ignored... prob another stack issue
      cout<<"insert function broke?????? @ (1,1) ???????????????\n";
      //return;

    }
  }
  return;
  cout<<"END... Hopefully.. if you see this Something went wrong\n";
  because_Im_a_terrible_person:
  cout<<"QUIT\n";
}

Answer 1

我怀疑您的问题是，您的函数调用了自己，并且您似乎没有考虑到在调用完之后接下来应该发生什么。所以你到了最后的状态，你说return不工作，但它确实.它只是返回到以前对printOptimalAlignment的调用中停止的地方，在返回到调用方之前仍然可能会做一些事情，依此类推。您有三个不同的站点，在这些站点中，递归调用printOptimalAlignment时，没有紧接着返回语句，在任何一个站点上，代码都可能会继续并触发另一个条件块。