Question

今天我正在通过R.Sedgewick的C算法中的Algorithms进行快速排序算法。

我理解算法的工作原理很好。编码部分让我感到困惑，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个元素然后按回车键，则会发生以下情况：

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

process returned 139(0x8b)

这是调试器返回的内容：

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的。左右指针没有任何内容。如果数组占用n个内存位置，则左指针应指向第0个位置，右指针指向n - 1位置。我没有在if条件中包含quicksort的递归函数，从而犯了一个愚蠢的错误。因此头痛。正确的计划是：

/* 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;
}

Answer 1

请在gdb之类的调试器中运行。这将显示您的segfault正在发生的确切行。如果你google＆＃34; gdb cheatsheet＆＃34;入门很容易。请记住使用-g标志进行编译。＆＃34;

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

Answer 2

l和r分别代表“左”和“右”。

分段错误正在发生，因为您正在传递l = 10，因此while (a[++i] < v);会中断。

<强> [编辑]

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

这两个循环也存在问题：您需要测试i和j是否超出范围。

Answer 3

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!!

具有分段错误的快速排序程序

3 个答案: