TableOfContents

实验基本要求

• 用标准C++完成程序
• 按照传统C语言风格进行命名和排版
• 在程序必要的地方进行注释

集成开发环境指导

• 参见：["C++集成开发环境"]

实验内容

1. 05瓯信算

先打开实验报告模板，查看需要完成哪些题目，然后打开判题系统地址，在判题系统上做对相应题目，最后将正确的程序复制到实验报告中，发到邮箱。

判题系统地址：[http://wzuacm.3322.org/]

完成如下实验报告：

• attachment:面向对象实验1.doc
• attachment:面向对象实验2.doc
• attachment:面向对象实验3.doc

在实验报告文件名后添加你的学号姓名，比如“面向对象实验1-0530215288-张三.doc”，然后email到： [email protected]

===实验3===

1. 拷贝5.4节中的程序：

      1    #define ALLOCSIZE 10000 /* size of available space */
      2 
      3    static char allocbuf[ALLOCSIZE]; /* storage for alloc */
      4    static char *allocp = allocbuf;  /* next free position */
      5 
      6    char *alloc(int n)    /* return pointer to n characters */
      7    {
      8        if (allocbuf + ALLOCSIZE - allocp >= n) {  /* it fits */
      9            allocp += n;
     10            return allocp - n; /* old p */
     11        } else      /* not enough room */
     12            return 0;
     13    }
     14 
     15    void afree(char *p)  /* free storage pointed to by p */
     16    {
     17        if (p >= allocbuf && p < allocbuf + ALLOCSIZE)
     18            allocp = p;
     19    }
   

2. 1.9节getline

      1   /* getline:  read a line into s, return length  */
      2    int getline(char s[],int lim)
      3    {
      4        int c, i;
      5 
      6        for (i=0; i < lim-1 && (c=getchar())!=EOF && c!='\n'; ++i)
      7            s[i] = c;
      8        if (c == '\n') {
      9            s[i] = c;
     10            ++i;
     11        }
     12        s[i] = '\0';
     13        return i;
     14    }
   

3. 5.6节

      #include <stdio.h>
      #include <string.h>
   
      #define MAXLINES 5000     /* max #lines to be sorted */
   
      char *lineptr[MAXLINES];  /* pointers to text lines */
   
      int readlines(char *lineptr[], int nlines);
      void writelines(char *lineptr[], int nlines);
   
      void qsort(char *lineptr[], int left, int right);
   
      /* sort input lines */
      main()
      {
          int nlines;     /* number of input lines read */
   
          if ((nlines = readlines(lineptr, MAXLINES)) >= 0) {
              qsort(lineptr, 0, nlines-1);
              writelines(lineptr, nlines);
              return 0;
          } else {
              printf("error: input too big to sort\n");
              return 1;
          }
      }
   
      #define MAXLEN 1000  /* max length of any input line */
      int getline(char *, int);
      char *alloc(int);
   
      /* readlines:  read input lines */
      int readlines(char *lineptr[], int maxlines)
      {
          int len, nlines;
          char *p, line[MAXLEN];
   
          nlines = 0;
          while ((len = getline(line, MAXLEN)) > 0)
              if (nlines >= maxlines || (p = alloc(len)) == NULL)
                  return -1;
              else {
                  line[len-1] = '\0';  /* delete newline */
                  strcpy(p, line);
                  lineptr[nlines++] = p;
              }
          return nlines;
      }
   
      /* writelines:  write output lines */
      void writelines(char *lineptr[], int nlines)
      {
          int i;
   
          for (i = 0; i < nlines; i++)
              printf("%s\n", lineptr[i]);
      }

   及qsort和swap

   void qsort(char *v[], int left, int right)
   {
       int i, last;
       void swap(char *v[], int i, int j);
   
       if (left >= right)  /* do nothing if array contains */
           return;         /* fewer than two elements */
       swap(v, left, (left + right)/2);
       last = left;
       for (i = left+1; i <= right; i++)
           if (strcmp(v[i], v[left]) < 0)
               swap(v, ++last, i);
       swap(v, left, last);
       qsort(v, left, last-1);
       qsort(v, last+1, right);
   }
   /* swap:  interchange v[i] and v[j] */
   void swap(char *v[], int i, int j)
   {
       char *temp;
   
       temp = v[i];
       v[i] = v[j];
       v[j] = temp;
   }

4. 所有的qsort换一个名字，比如qsort2

2. 05计本

在判题系统[http://czk.8866.org/oj/]上完成3题，并提交实验报告：

• attachment:面向对象实验模板.doc

实验报告发送到： [email protected]