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今天我正在学习 R.Sedgewick 的 Algorithms in C 中的快速排序算法。

我对算法的工作原理有了很大的了解。编码部分让我有点困惑,最后我遇到了分段错误。这是代码:

#include <stdio.h>
void quicksort(int[], int, int); // prototype

void quicksort(int a[], int l, int r) // What is the value of l. Why hasn't the author 
                                      // mentioned it. Is it the size of the array? 
{
    int i, j, v, t;
    if(r > l)
    {
        v = a[r];
        i = l - 1; // What is l here? Is it the size if yes look at first while statement
        j = r;

        for(;;)
        {

            /*The algorithm says: scan from right until an element < a[r] is found. Where 
              r is the last position in the array. But while checking in the second while
              loop elements > a[r] is searched */

            while (a[++i] < v); // How can you increment again after reaching end of arrray
                                // size if l is the size of the array
            while (a[--j] > v);
            if(i >= j) break;
            t = a[i]; a[i] = a[j]; a[j] = t;
        }
    }

    t = a[i]; a[i] = a[r]; a[r] = t;

    quicksort(a, l, i - 1);
    quicksort(a, i + 1, r);

    return;
}

int main()
{
    int i, a[10]; // assuming size is 10

    for(i = 0; i < 10; i++)
    {
        scanf("%d", &a[i]);
    }

    int l = 10; // I am passing size of the array
    int r = 9; // position of last element

    quicksort(a, l, r);
    return 0;
}

错误是这样的。假设如果我输入 10 个元素然后按 Enter,会发生以下情况:

1 4 8 2 3 6 4 7 10 9
segmentation fault.

process returned 139(0x8b)

这是调试器返回的:

Breakpoint 1, quicksort (a=0xbffff808, l=0, r=0) at quick.c:11
11      if(r > 1)
(gdb) c
Continuing.

Program received signal SIGSEGV, Segmentation fault.
0x080484fb in quicksort (a=0xbffff808, l=0, r=0) at quick.c:28
28      t = a[i]; a[i] = a[r]; a[r] = t;
(gdb) c
Continuing.

Program terminated with signal SIGSEGV, Segmentation fault.
The program no longer exists.
(gdb) c
The program is not being run.

执行上述程序的正确方法是这样。左右指针什么都没有。如果数组占用 n 个内存位置,则左指针应指向第 0 个位置,右指针应指向 n - 1 个位置。我犯了一个愚蠢的错误,没有在 if 条件中包含快速排序的递归函数。因此,所有的头痛。正确的程序是:

/* Working quicksort
 * Robert sedgewick best
 */

#include <stdio.h>

void quicksort(int[], int, int); // prototype

void quicksort(int a[], int l, int r) 
{
    int i, j, v, t;
    if(r > l)
    {
        v = a[r];
        i = l - 1;
        j = r;

        for(;;)
       {
            while (a[++i] < v); 
            while (a[--j] > v);

            if(i >= j) break;
            t = a[i]; a[i] = a[j]; a[j] = t;

        } // End for here


    t = a[i]; a[i] = a[r]; a[r] = t;

    quicksort(a, l, i - 1);
    quicksort(a, i + 1, r);

    } /* End if here. That is include the recursive
         functions inside the if condition. Then it works 
         just fine. */

    return;
}

int main()
{
    int i, a[5]; // assuming size is 10

    for(i = 0; i < 5; i++)
    {
        scanf("%d", &a[i]);
    }

    int l = 0; // I am passing size of the array
    int r = 4; // position of last element

    quicksort(a, l, r);

       int s;

    for(s = 0; s < 5; s++)
    {
        printf("%d ", a[s]);
    }
    return 0;
}
4

3 回答 3

2

请在调试器中运行,例如gdb. 这将向您显示发生段错误的确切行。如果您在 Google 上搜索“gdb 备忘单”,那么上手将非常容易。记得用-gflag编译。”

我的会话:

dan@dev1:~ $ gcc -g quick.c
dan@dev1:~ $ gdb a.out
...
(gdb) r
Starting program: /home/dan/a.out 
1 4 8 2 3 6 4 7 10 9

Program received signal SIGSEGV, Segmentation fault.
0x0000000000400572 in quicksort (a=0x7fffffffe530, l=10, r=9) at quick.c:21
21              while (a[++i] < v); // How can you increment again after reaching end of arrray
于 2013-03-14T15:43:01.753 回答
1

l分别r代表“左”和“右”。

发生分段错误是因为您正在通过l = 10,所以while (a[++i] < v);会中断。

[编辑]

while (a[++i] < v);                                
while (a[--j] > v);

这两个循环也有问题:您需要对其进行测试i并且j不会越界。

于 2013-03-14T15:39:22.453 回答
0
int a[10];
int l = 10;
int r =  9; 

quicksort(a, l, r);

called quicksort(a, l, r)
//l=10,r=9
if(r > 1) // 9 > 1 is true
{
    i = l - 1;//i = 10 - 1 = 9
    for(;;)
    {
        while (a[++i] < v);//a[++i] is a[9+1] = a[10] is out of range!!
于 2013-03-14T20:04:39.637 回答