合并排序，使用文件的I / O显示分段错误

Question

The following mergesort program written in C++ works without any error when the inputs are given through the console. But when I use text files to give the inputs, it gives me a segmentation error. I have tried printing messages at various parts of the code to check where is the mistake, but it isn't printing anything except d seg-fault message. Here is the code. It works fine when I do not use files for giving the inputs/ displaying the output.

I ran this prog on a gcc compiler in a virtualbox 4.0.8(ubuntu OS) with a base memory of 1 GB. Cud the error be due to insufficient memory?

#include<iostream>
#include<fstream>
#include<sys/time.h>
using namespace std;

int array[50];
int barray[50];

void merge(int p,int q, int r)
 {
    int i,j,k,l;
    l=p;
    i=p;
    j=q+1;

    while((i<=q)&&(j<=r))
    {

     if(array[i]<=array[j])
        barray[++k]=array[i++];

     else
        barray[k++]=array[j++];

    }

    if(i>q)
    {  
       for(l=j;l<=r;l++)
       { 
         barray[k]=array[l];
         k++;
       }
    }

   else
   {
      for(l=i;l<=q;l++)
      {
        barray[k]=array[l];
        k++;
      }

   }

   for(i=p;i<=r;i++)
        array[i]=barray[i];
}

void mergesort(int p, int r)
{
   int q;

   if(p<r)
      q=(p+r)/2;

   mergesort(p,q);
   mergesort(q+1,r);
   merge(p,q,r);

}


int main()
{
  int r, p, i;

  struct timeval tv1, tv2;/* For finding the running time */
  gettimeofday(&tv1, NULL);

  ifstream fin;
  ofstream fout;

  fin.open("abc5.txt");
  fout.open("new5.txt");

  fin>>r;
  while(fin)
  {

    for(i=1;i<=r;++i)
    {
       fin>>array[i];
    }

  }

  mergesort(1,r);

  for(i=1;i<=r;++i)
  {
    fout<<array[i]<<"\n";
  }

  gettimeofday(&tv2, NULL);
  fout<<"Running Time: "<<(double) (tv2.tv_usec - tv1.tv_usec) / 1000000 + (double)    (tv2.tv_sec - tv1.tv_sec)<<" sec";


 fin.close();
 fout.close();

 return 0;

}

THE INPUT FILE 8 3 6 2 8 9 1 4 10

Answer 1

Your indexing leaves much to be desired. I will not be shy about telling you that, nor about saying simply that I would not do merge-sort this way. But it is what it is, so I'll cover it.

It is important to understand that sequence sorting using divide and conquer algorithms are all about passing some base reference within the sequence and some offsets from that reference. The algorithm should be independent of external variables (like array[] and barray[] ) and take them as parameters instead. Further, remember you're programming in C++ (or C). Part of the beauty of both languages is their native pointer-arithmetic ability. It can be wonderfully used for algorithms like merge-sort. After I demonstrate how your merge and mergesort functions should work, I'll demonstrate what I'm talking about, then I'll proffer up the most-trivial merge sort you can do in C++ if you utilize the standard libraries functionality.

---

Your Code First

First a re-roll of your function using your parameters. I've taken the liberty of renaming the parameters to they're actually meaningful to the reviewer. The algorithm should be self-explanatory and I urge you to compare it side-by-side with what you were doing.

//  low  = starting index
//  mid  = middle index
//  high = last index
void merge(int low, int mid, int high)
{
    int i=low, j=mid, k=0;

    while (i < mid && j <=high)
    {
        if (array[i] < array[j])
            barray[k++] = array[i++];
        else
            barray[k++] = array[j++];
    }

    // finish whichever segment isn't done
    while (i < mid)
        barray[k++] = array[i++];
    while (j <= high)
        barray[k++] = array[j++];

    // copy back
    for (i=0;i<k;++i)
        array[low+i] = barray[i];
}

void mergesort(int low, int high)
{
    int len = high - low;
    if (len < 2)
        return;

    int mid = low + len/2;
    mergesort(low, mid);
    mergesort(mid, high);
    merge(low, mid, high);
}

Of note to this algorithm is the initial parameters passed must be valid indexes in the sequence. In other words, if you read an array of 8-elements, the valid indexes are 0..7 and therefore you would invoke it as mergesort(0,7) .

---

A Different Approach

The traditional merge sort in C/C++ uses an array base-index and length as its only parameters. The midpoint is calculated by the sort-algorithm. Though it is not strictly needed for the merge algorithm, (it too can just use length/2), it makes for a more robust algorithm on the off-chance you ever need merging from a segment that is not split right now the middle.

Don't get hung up on the usage of std::vector, std::random_device, etc below. They're used in the main() function solely to populate an array of random numbers that is then sorted, displayed both before and after the sort operation. What I want you to get out of this is the actual algorithm itself.

#include <iostream>
#include <vector>
#include <random>

void merge(int ar[], std::size_t mid, std::size_t len)
{
    if (len < 2)
        return;

    // temporary storage. normally I use a RAII container
    //  such as std::vector<>, but I wanted to demonstrate
    //  the algorithm and indexing, not the memory management.
    //  none the less it is worth noting.
    int *tmp = new int[len];
    std::size_t i=0, j=mid, k=0;

    while (i < mid && j < len)
    {
        if (ar[i] < ar[j])
            tmp[k++] = ar[i++];
        else
            tmp[k++] = ar[j++];
    }

    // complete the unfinished segment
    while (i < mid)
        tmp[k++] = ar[i++];
    while (j < len)
        tmp[k++] = ar[j++];

    // and move back to the original array
    for (i=0; i<len; ++i)
        ar[i] = tmp[i];

    delete [] tmp;
}

void mergesort(int ar[], std::size_t len)
{
    if (len < 2)
        return;

    // note pointer arithemtic in second call.
    mergesort(ar, len/2);
    mergesort(ar+len/2, len - len/2);
    merge(ar, len/2, len);
}

int main()
{
    std::vector<int> data;
    data.reserve(20);

    std::random_device rd;
    std::default_random_engine rng(rd());
    std::uniform_int_distribution<> dist(1,50);

    // populate array
    std::generate_n(std::back_inserter(data),
                data.capacity(),
                [&](){ return dist(rng);});

    // show on-screen
    for (auto n : data)
        std::cout << n << ' ';
    std::cout << '\n';

    // mergesort
    mergesort(data.data(), data.size());

    // show on-screen
    for (auto n : data)
        std::cout << n << ' ';
    std::cout << '\n';

    return 0;
}

Output (varies)

15 10 8 38 20 21 9 43 8 22 19 45 12 16 17 36 2 32 6 37 
2 6 8 8 9 10 12 15 16 17 19 20 21 22 32 36 37 38 43 45 

---

The Part You'll Hate

After going through all this work you'll be not-so-happy to know that two algorithms that do segment merging for you are already present in the standard library, namely std::merge and std::inplace_merge . Using one of those makes implementing mergesort trivial , as you'll see below:

#include <algorithm>

void mergesort(int ar[], std::size_t len)
{
    if (len < 2)
        return;

    mergesort(ar, len/2);
    mergesort(ar+len/2, len-len/2);
    std::inplace_merge(ar, ar+len/2, ar+len);
}

Remember that if you ever need to use something besides std::sort , which honestly puts all of this to a point of near-uselessness besides fulfilling academic purpose.

Answer 2

So given your input, it looks like when p=1 and r=1 that you use an uninitialized value for q in subsequent calls to mergesort.

The error you're receiving is because of unbounded recursion in the mergesort function.