Question

我只是在C中堆积一个数组。但是在运行时它给我分段错误（核心转储）...我不知道我在哪里尝试访问未分配的内存!!

#include<stdio.h>

int n;

int left(i)
{
    return (2*i);
}

int right(i)
{
    return (2*i + 1);
}


void min_heap(int a[],int i)
{
    int l=left(i);
    int r=right(i);
    int min;

    if((l<=n)&&(a[l]<=a[i])&&(a[l]<=a[r]))
    {
        min=a[l];
        a[i]=a[i]+a[l];
        a[l]=a[i]-a[l];
        a[i]=a[i]-a[l];
    }
    else if((r<=n)&&(a[r]<=a[i])&&(a[r]<=a[l]))
    {
        min=a[r];
        a[i]=a[i]+a[r];
        a[r]=a[i]-a[r];
        a[i]=a[i]-a[r];
    }

    min_heap(a,min);
}



int main()
{
    printf("The no is : ");
    scanf("%d",&n);
    int i,a[n+1];

    for(i=1;i<=n;i++)
    {
        scanf("%d",&a[i]);
    }

    for(i=n/2;i>=1;i--)
    {
        min_heap(a,i);
    }

    for(i=1;i<=n;i++)
    {
        printf("%d",a[i]);
    }

    return 0;
}

Answer 1

您在min_heap(a,i)时致电i == n/2。

在这种情况下，在min_heap()内，对right()的调用将生效：

(2 * (n/2) + 1)

当n为偶数时，会产生n+1的正确索引并且访问a[r]（带r == n+1）超出了您已分配的数组的末尾

我不确定这是否是你的段错误的原因;我猜可能还有其他问题。

您可能应该只使用调试器来完成运行。

Answer 2

以下是Jon Bentley撰写的More Programming Pearls代码，作为C文件中的注释编写。完整代码与您无关;它是通用接口，如bsearch()和qsort()的接口，但这是用awk编写的。

/*
** See Appendix 2 of Jon Bentley "More Programming Pearls".
** See also Column 14 of Jon Bentley "Programming Pearls, 2nd Edn".
** Note that MPP algorithms are in terms of an array indexed from 1.
** C, of course, indexes arrays from zero.
**
** 1-based identities.
** root          = 1
** value(i)      = x(i)
** leftchild(i)  = 2*i
** rightchild(i) = 2*i+1
** parent(i)     = i/2
** null(i)       = (i < 1) or (i > n)
**
** 0-based identities.
** root          = 0
** value(i)      = x(i)
** leftchild(i)  = 2*(i+1)-1   = 2*i+1
** rightchild(i) = 2*(i+1)+1-1 = leftchild(i)+1
** parent(i)     = (i+1)/2-1
** null(i)       = (i < 0) or (i >= n)  # NB: i < 0 irrelevant for unsigned numbers
*/

/*
**  function swap(i, j  t) {
**      # x[i] :=: x[j]
**      t = x[i]
**      x[i] = x[j]
**      x[j] = t
**  }
**
**  function siftup(l, u,    i, p) {
**      # pre  maxheap(l, u-1)
**      # post maxheap(l, u)
**      i = u
**      while (1) {
**          # maxheap(l, u) except between i and its parent
**          if (i <= l) break
**          p = int(i/2)        # p = parent(i)
**          if (x[p] >= x[i]) break
**          swap(p, i)
**          i = p
**      }
**  }
**
**  function siftdown(l, u,  i, c) {
**      # pre  maxheap(l+1, u)
**      # post maxheap(l,u)
**      i = l
**      while (1) {
**          # maxheap(l, u) except between i and its children
**          c = 2*i         # c = leftchild(i)
**          if (c > u) break;
**          if (c + 1 <= u && x[c+1] > x[c]) c++
**          if (x[i] >= x[c]) break
**          swap(c, i)
**          i = c
**      }
**  }
**
**  function hsort(    i) {
**      # post sorted(1, n)
**      for (i = int(n/2); i >= 1; i--)
**          siftdown(i, n)
**      for (i = n; i >= 2; i--) {
**          swap(1, i)
**          siftdown(1, i-1)
**      }
**  }
*/

在代码中，正在排序的数组是x，从1索引到N。

堆积时的分段错误

2 个答案: