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我正在编写 Coursera算法的第一个编程作业：设计和分析课程 ( https://www.coursera.org/course/algo )。它涉及使用合并排序来计算反转（http://en.wikipedia.org/wiki/Inversion_(discrete_mathematics )）。我认为这将是一个相对简单的事情，因为我之前（在学校）遇到过合并排序。

#include <iostream>
#include <fstream>

using namespace std;

int *half(int *array, int n, int start, int end)
{
    /*
     * Creates a new array which contains elements from ''array'' starting with ''start''
     * and ending with ''end - 1''.
     */

     int *new_array = new int[end-start];

     for(int i = start; i < end; i++)
     {
        new_array[i-start] = array[i];
     }

     return new_array;
}

int merge(int *array1, int n1, int *array2, int n2, int *new_array)
{
    /*
     *  Merges arrays 1 and 2 (with lengths n1 and n2) into a new_array, counting
     *  ''split inversions'' by the way.
     */
    int i = 0, j = 0, count = 0;

    for(int k = 0; k < n1 + n2; k++)
    {

         if(i >= n1)
        {
            new_array[k] = array2[j];
            j++;
            continue;
        }

         if(j >= n2)
        {

            new_array[k] = array1[i];
            i++;
            continue;
        }

        if( array1[i] <= array2[j] )
        {
            new_array[k] = array1[i];
            i++;
        }
        else
        {
            new_array[k] = array2[j];
            j++;
            count++;
        }
    }

    return count;
}

int mergesort(int *array, int n)
{

    if(n == 1) return 0; //base case

    int x, y, z;
    int odd;

    if(n%2 == 0) odd = 0;
    else odd = 1;

    int *half1 = new int [n/2];
    int *half2 = new int [n/2 + odd];

    half1 = half(array, n, 0, n/2);
    half2 = half(array, n, n/2, n);

    x = mergesort(half1, n/2);
    y = mergesort(half2, n/2 + odd);  //if n is odd, we add one
    z = merge(half1, n/2, half2, n/2 + odd, array); //we write a sorted array back in our starting array

    delete [] half1;
    delete [] half2;

    return x + y + z;
}

int main()
{
    int n;
    int *array = new int[n];

    cin >> n;

    for(int i = 0; i < n; i++)
    {
        int x;
        cin >> x;
        array[i] = x;
    }

    for(int i = 0; i < n; i++)
        cout << array[i] << " ";

    cout << endl;
    cout << "Number of inversions: " << mergesort(array, n) << endl;

    for(int i = 0; i < n; i++)
    cout << array[i] << " ";

    cout << endl;
    delete[] array;

    return 0;
}

那么，这里有什么奇怪的地方？第一件事：对我来说，它适用于某些数组，但对于其他数组会崩溃（稍后会举例）。第二件事：我将代码发送给我的朋友，他说对他来说一切正常，即使是对我来说严重崩溃的示例。

所以，例子：

For array [1 2 3 4 5 6 7] g++ produces this:

malloc.c:2451: sYSMALLOc: Assertion `(old_top == (((mbinptr) (((char *) &((av)->bins[((1) - 1) * 2])) - __builtin_offsetof (struct malloc_chunk, fd)))) && old_size == 0) || ((unsigned long) (old_size) >= (unsigned long)((((__builtin_offsetof (struct malloc_chunk, fd_nextsize))+((2 * (sizeof(size_t))) - 1)) & ~((2 * (sizeof(size_t))) - 1))) && ((old_top)->size & 0x1) && ((unsigned long)old_end & pagemask) == 0)' failed.
Aborted (core dumped)

When I ''backtrack'' it using gdb:

#0  0x00007ffff7753445 in __GI_raise (sig=<optimized out>) at ../nptl/sysdeps/unix/sysv/linux/raise.c:64
#1  0x00007ffff7756bab in __GI_abort () at abort.c:91
#2  0x00007ffff779abed in __malloc_assert (assertion=<optimized out>, file=<optimized out>, line=<optimized out>, function=<optimized out>) at malloc.c:300
#3  0x00007ffff779e0f4 in sYSMALLOc (av=0x7ffff7ad3720, nb=32) at malloc.c:2448
#4  _int_malloc (av=0x7ffff7ad3720, bytes=12) at malloc.c:3892
#5  0x00007ffff779fa45 in __GI___libc_malloc (bytes=12) at malloc.c:2924
#6  0x00007ffff7b8fded in operator new(unsigned long) () from /usr/lib/x86_64-linux-gnu/libstdc++.so.6
#7  0x00007ffff7b8ff09 in operator new[](unsigned long) () from /usr/lib/x86_64-linux-gnu/libstdc++.so.6
#8  0x0000000000400b12 in mergesort (array=0x603010, n=7) at jedan.cpp:81
#9  0x0000000000400cfe in main () at jedan.cpp:120

It does something similar (but not the same!) for the array [1 2 3 4 5 6 7 8 9 10], again connected to the new and delete[] functions. I can post it here later if somebody thinks that would be helpful, but I don't want to bloat this post too much. And it works for most of the arrays I tried (I had no problems with arrays of size <= 6, and for a pretty big number of bigger arrays).

I'm using Ubuntu 12.04, installed yesterday... pretty clean and fresh. Help?

P.S. If you find variable names a little weird, sorry... I translated them from my native language so the code could be more readable.

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3 回答 3

5 
int n;
int *array = new int[n]; // Undefined behavior

n is used uninitialized here, so you'll get a "random" length allocation.

If you're unlucky and n holds a "large" garbage value, your code could appear to work. If it holds a small value, you're likely to corrupt your heap when you fill your initial array - this will produce the type of error you're seeing.

Move the cin >> n; line before the array allocation.

Side note: I think the two allocations you're doing in mergesort are leaked (you're only deleteing the memory allocated in half, you don't actually need to allocate in mergesort itself if I read your code correctly).

于 2012-06-30T17:29:52.363 回答
1

You don't check for

n==0

in the mergesort() routine.

Then the new[] fails.

BTW, the same stuff in the half() routine:

int *new_array = new int[end-start];

You don't check for the

end == start
于 2012-06-30T17:30:51.930 回答
1

You are not specifying the quantity for the array:

int main()
{
    int n;
    int *array = new int[n];

    cin >> n;

When the program starts, you have not specified a value for n. So the size that is allocated is unknown.

I suggest moving the cin >> n; before the array allocation.

于 2012-06-30T17:31:55.813 回答