读书笔记：C程序设计语言，第五章：指针与数组（部分课后题解）

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2月27日终于下决心改变学习策略：

不再总结一些基础的知识
不会把书中的要点记录在博客，而是直接买书（之前都是图书馆借书），直接阅读书籍
博客主要总结难度较大的问题

这是最后一篇基础性的记录博客，但是很不完整，所以只是把题贴下来了，以后完成时会单开一篇博客。

5.2 指针与函数

Exercise 5-1. As written, getint treats a + or - not followed by a digit as a valid representation of zero. Fix it to push such a character back on the input.

答：暂时未找到合适的解法。
Exercise 5-2. Write getfloat, the floating-point analog of getint. What type does getfloat return as its function value?

答：暂时未找到合适的解法。

5.5字符指针和函数

Exercise 5-3. Write a pointer version of the function strcat that we showed in Chapter 2: strcat(s,t) copies the string t to the end of s.

答：暂时未找到合适的解法。

Exercise 5-4. Write the function strend(s,t), which returns 1 if the string t occurs at the end of the string s, and zero otherwise.

答：暂时未找到合适的解法。

Exercise 5-5. Write versions of the library functions strncpy, strncat, and strncmp, which operate on at most the first n characters of their argument strings. For example, strncpy(s,t,n) copies at most n characters of t to s. Full descriptions are in Appendix B.

答：暂时未找到合适的解法。

Exercise 5-6. Rewrite appropriate programs from earlier chapters and exercises with pointers instead of array indexing. Good possibilities include getline (Chapters 1 and 4), atoi, itoa, and their variants (Chapters 2, 3, and 4), reverse (Chapter 3), and strindex and getop (Chapter 4).
答：暂时未找到合适的解法。

5.6 指针数组以及指向指针的指针

unix有个程序sort，可以完成一个功能：根据每行字符串的第一个字符，按字典顺序排序。
本节试着写了一个例子，简化版的sort。
使用指针数组可以高效的完成这个功能，如下图所示：

由指针组成的数组，每一个指针指向一行。不用移动文本行，仅仅改变指针的指向，就完成了这个功能
移动文本行将带来复杂的储存管理和巨大的开销
在比较字符的时候我们使用了strcmp函数。
程序的划分：一个问题一个函数，主函数控制其他函数的执行。
具体的函数我倒没有觉得有什么难点，不过可以来看看指针数组的初始化，实际上也没什么区别：
```
char *lineptr[MAXLINES]
```

练习题

Exercise 5-7. Rewrite readlines to store lines in an array supplied by main, rather than calling alloc to maintain storage. How much faster is the program?答：code：

#include <stdio.h>#include <string.h>#include <stdlib.h>#include <time.h>#define TRUE     1#define FALSE    0#define MAXLINES 5000       /* maximum number of lines */#define MAXLEN   1000       /* maximum length of a line */char *lineptr[MAXLINES];char lines[MAXLINES][MAXLEN];//第一个相当于存入的一行的编号/* K&R2 p29 */int getline(char s[], int lim){  int c, i;  for (i = 0; i < lim - 1 && (c = getchar()) != EOF && c != '\n'; i++)    s[i] = c;  if (c == '\n') {    s[i++] = c;  }  s[i] = '\0';  return i;}/* K&R2 p109 */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 = malloc(len)) == NULL)//Malloc 向系统申请分配指定size个字节的内存空间      return -1;    else {      line[len - 1] = '\0';  /* delete the newline */      strcpy(p, line);//将读到的行复制到新申请的内存空间中      lineptr[nlines++] = p;    }  return nlines;}int readlines2(char lines[][MAXLEN], int maxlines){  int len, nlines;  nlines = 0;  while ((len = getline(lines[nlines], MAXLEN)) > 0)    if (nlines >= maxlines)      return -1;    else      lines[nlines++][len - 1] = '\0';   return nlines;}int main(int argc, char *argv[]){  /* read things into cache, to be fair. */  readlines2(lines, MAXLINES);  if (argc > 1 && *argv[1] == '2') {    puts("readlines2()");    readlines2(lines, MAXLINES);  } else {    puts("readlines()");    readlines(lineptr, MAXLINES);  }  return 0;}

5.7 多维数组

类似矩阵的多维数组，使用不如指针数组那么频繁
char类型在存放较小的非字符整数也是合法的

看一下二维数组的赋值，下面的二维数组表示某个月有多少天：

static char daytab[2][13] = {{0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31},{0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}};

看作矩阵的话：
```
daytab[i][j]   /* [row][col] */
```
第一个是行，第二个是列
在二位数组作为参数传递给函数的时候，在函数的声明中必须指名数组的列数，但是多少行可以不写。
```
f(int daytab[2][13]) { ... }//可以这样f(int daytab[][13]) { ... }//可以这样f(int (*daytab)[13]) { ... }//也可以这样，小心圆括号，改变优先级
```
第三种写法是因为，二维数组相当于储存了一个指向很多对象的指针，每个对象都是由13个整型元素构成的一维数组。具体说一下，就是一个数组在被调用的时候传递的就是指针。
```
int *daytab[13]//[]的优先级高，所以这样表示一个指向整型的数组指针
```
一般多维数组中，第一维可以不指定大小，其余各维都必须明确大小

练习题

Exercise 5-8. There is no error checking in day_of_year or month_day. Remedy this defect.（进行错误检查。）

答：code：

/* * A solution to exercise 5-8 in K&R2, page 112: * *There is no error checking in day_of_year or month_day. Remedy *this defect. * * The error to check for is invalid argument values. That is simple, what's * hard is deciding what to do in case of error. In the real world, I would * use the assert macro from assert.h, but in this solution I take the * approach of returning -1 instead. This is more work for the caller, of * course. * * I have selected the year 1752 as the lowest allowed year, because that * is when Great Britain switched to the Gregorian calendar, and the leap * year validation is valid only for the Gregorian calendar. * * Lars Wirzenius <liw@iki.fi> */#include <stdio.h>static char daytab[2][13] =  {{0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31},{0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31},};/* day_of_year: set day of year from month & day */int day_of_year(int year, int month, int day){int i, leap;if (year < 1752 || month < 1 || month > 12 || day < 1)return -1;leap = (year%4 == 0 && year%100 != 0) || year%400 == 0;if (day > daytab[leap][month])return -1;for (i = 1; i < month; i++)day += daytab[leap][i];return day;}/* month_day: set month, day from day of year */int month_day(int year, int yearday, int *pmonth, int *pday){int i, leap;if (year < 1752 || yearday < 1)return -1;leap = (year%4 == 0 && year%100 != 0) || year%400 == 0;if ((leap && yearday > 366) || (!leap && yearday > 365))return -1;for (i = 1; yearday > daytab[leap][i]; i++)yearday -= daytab[leap][i];*pmonth = i;*pday = yearday;return 0;}/* main: test day_of_year and month_day */int main(void){int year, month, day, yearday;for (year = 1970; year <= 2000; ++year) {for (yearday = 1; yearday < 366; ++yearday) {if (month_day(year, yearday, &month, &day) == -1) {printf("month_day failed: %d %d\n",year, yearday);} else if (day_of_year(year, month, day) != yearday) {printf("bad result: %d %d\n", year, yearday);printf("month = %d, day = %d\n", month, day);}}}return 0;}

5.8 指针数组的初始化

看下面的函数就知道怎么样对指针数组初始化：

/* month_name: return name of n-th month */char *month_name(int n){static char *name[] = {"Illegal month","January", "February", "March","April", "May", "June","July", "August", "September","October", "November", "December"};return (n < 1 || n > 12) ? name[0] : name[n];}

5.9 指针与多维数组

看例子来分析，指针数组和多维数组的区别:
```
int a[10][20];int *b[10];
```
a是真正的二维数组，分配了200个的储存空间。b只是分配了10个指针，10个指针可以指向不同的长度数组。

下图进一步说明了上面的说法：

char *name[] = { "Illegal month", "Jan", "Feb", "Mar" };

char aname[][15] = { "Illegal month", "Jan", "Feb", "Mar" };

练习题

Exercise 5-9. Rewrite the routines day_of_year and month_day with pointers instead of indexing.

答：code：

/* * A solution to exercise 5-9 in K&R2, page 114: * *Rewrite the routines day_of_year and month_day with pointers *instead of indexing. * * Lars Wirzenius <liw@iki.fi> */#include <stdio.h>static char daytab[2][13] =  {{0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31},{0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31},};/* original versions, for comparison purposes */int day_of_year(int year, int month, int day){int i, leap;leap = (year%4 == 0 && year%100 != 0) || year%400 == 0;for (i = 1; i < month; i++)day += daytab[leap][i];return day;}void month_day(int year, int yearday, int *pmonth, int *pday){int i, leap;leap = (year%4 == 0 && year%100 != 0) || year%400 == 0;for (i = 1; yearday > daytab[leap][i]; i++)yearday -= daytab[leap][i];*pmonth = i;*pday = yearday;}/* pointer versions */int day_of_year_pointer(int year, int month, int day){int i, leap;char *p;leap = (year%4 == 0 && year%100 != 0) || year%400 == 0;/* Set `p' to point at first month in the correct row. */p = &daytab[leap][1];//指向第一个月/* Move `p' along the row, to each successive month. */for (i = 1; i < month; i++) {day += *p;++p;}return day;}void month_day_pointer(int year, int yearday, int *pmonth, int *pday){int i, leap;char *p;leap = (year%4 == 0 && year%100 != 0) || year%400 == 0;p = &daytab[leap][1];//指向第一个月for (i = 1; yearday > *p; i++) {yearday -= *p;++p;}*pmonth = i;*pday = yearday;}int main(void){int year, month, day, yearday;year = 2000;month = 3;day = 1;printf("The date is: %d-%02d-%02d\n", year, month, day);printf("day_of_year: %d\n", day_of_year(year, month, day));printf("day_of_year_pointer: %d\n",day_of_year_pointer(year, month, day));yearday = 61;/* 2000-03-01 */month_day(year, yearday, &month, &day);printf("Yearday is %d\n", yearday);printf("month_day: %d %d\n", month, day);month_day_pointer(year, yearday, &month, &day);printf("month_day_pointer: %d %d\n", month, day);return 0;}

5.10 命令行参数

也就是跟着main函数的两个参数
第一个是argc，用于参数计数
第二个是argv，是参数向量，是一个字符串数组的指针，每一个字符串对应一个参数，通常用多级指针处理这些参数
c语言规定：argv[0]，是启动该程序的程序名，故argc至少为1
unix下面的回显程序：
```
echo hello, world
```
argc 为 3, argv[0], argv[1], argv[2] 分别为： "echo", "hello,", and "world"

我们在unix经常使用：

find -nx pattern find -x -n pattern
上述这样的语句，含义就是-x和-n能够对这个程序做出一些更具体的要求，这里：-n：打印行号-x：打印所有与模式不匹配的文本行pattern：所谓模式，就是查找某行里有没有某一串字符串，这个某一行字符串就是模式。下面这段代码就是讲述了如何实现这样的效果：#include <stdio.h>#include <string.h>#define MAXLINE 1000int getline(char *line, int max);/* find: print lines that match pattern from 1st arg */main(int argc, char *argv[]){char line[MAXLINE];long lineno = 0;int c, except = 0, number = 0, found = 0;while (--argc > 0 && (*++argv)[0] == '-')//先找到‘-’。另外没有{}的原因是后面直接根了一个while的程序块while (c = *++argv[0])//为什么要加[0]这是因为*argv代表字符串switch (c) {//*argv[0]代表字符串的第一个字符case 'x'://*argv[0]和**argv是等效的except = 1;//稍稍分析就可以看出这段代码既可以应对：-nx 或者 -n -x两种形式break;case 'n':number = 1;break;default:printf("find: illegal option %c\n", c);argc = 0;found = -1;break;}if (argc != 1)printf("Usage: find -x -n pattern\n");elsewhile (getline(line, MAXLINE) > 0) {lineno++;if ((strstr(line, *argv) != NULL) != except) {//如果含有该字会返回1if (number)//如果except等于1的话 1！=1 会返回0printf("%ld:", lineno);//所以就不打印了printf("%s", line);//如果不含有这个字段的分析方式类似found++;}}return found;}

练习题

Exercise 5-10. Write the program expr, which evaluates a reverse Polish expression from the command line, where each operator or operand is a separate argument. For example, expr 2 3 4 + * evaluates 2 * (3+4).答：code：

#include <ctype.h>#include <stdio.h>#include <stdlib.h>#define STACK_SIZE 1024double stack[STACK_SIZE];int stack_height = 0;void panic(const char *msg) {fprintf(stderr, "%s\n", msg);exit(EXIT_FAILURE);}void push(double value) {if (stack_height == STACK_SIZE)panic("stack is too high!");stack[stack_height] = value;++stack_height;}double pop(void) {if (stack_height == 0)panic("stack is empty!");return stack[--stack_height];}int main(int argc, char **argv) {int i;double value;for (i = 1; i < argc; ++i) {switch (argv[i][0]) {case '\0':panic("empty command line argument");break;case '0':case '1':case '2':case '3':case '4':case '5':case '6':case '7':case '8':case '9':push(atof(argv[i]));break;case '+':push(pop() + pop());break;case '-':value = pop();push(pop() - value);break;case '*':push(pop() * pop());break;case '/':value = pop();push(pop() / value);break;default:panic("unknown operator");break;}}printf("%g\n", pop());return 0;}

Exercise 5-11. Modify the program entab and detab (written as exercises in Chapter 1) to accept a list of tab stops as arguments Use the default tab settings if there are no arguments.
答：没找到合适的解法。

Exercise 5-12. Extend entab and detab to accept the shorthand entab -m +n to mean tab stops every n columns, starting at column m. Choose convenient (for the user) default behavior.

答：没找到合适的解法。
Exercise 5-13. Write the program tail, which prints the last n lines of its input. By default, n is set to 10, let us say, but it can be changed by an optional argument so that
tail -n
prints the last n lines. The program should behave rationally no matter how unreasonable the input or the value of n. Write the program so it makes the best use of available storage; lines should be stored as in the sorting program of Section 5.6, not in a two-dimensional array of fixed size.

答：code：

/* K&R Exercise 5-13 *//* Steven Huang */#include <stdio.h>#include <stdlib.h>#include <string.h>#define DEFAULT_NUM_LINES      10#define MAX_LINE_LEN           1000/*   Points of interest for a novice:   1.  atoi() has a normally annoying property of not being able to       tell the caller conclusively whether the input was bad ("abc")       or it was really zero ("0"), because it returns 0 for both       cases.  Here, we exploit that property, because we only want       to accept options in the form of "-n".   2.  Try to understand how this program deals with input that       doesn't even have as many lines as the line_ptrs[] array.       That is, how does this program degenerate into just displaying       everything it read?  (Hint:  what does it mean when line_ptrs[x]       is NULL?)   3.  Using modulo arithmetic on an index to a circular array is       a common and useful technique.  Try to understand the range       of values that current_line (and j, later) will take.  In       particular, why shouldn't we just do this:       for (i = 0; i < num_lines; i++)         if (line_ptrs[i])           printf("%s", line_ptrs[i]);   4.  Why do we still use a "%s" to display what's inside line_ptrs,       rather than just:       printf(line_ptrs[i]);   5.  There is a bug in this program, where you see:       numlines = -numlines;       When will this break?*//* K&R2 p29 */int getline1(char s[], int lim){  int c, i;  for (i = 0; i < lim - 1 && (c = getchar()) != EOF && c != '\n'; i++)    s[i] = c;  if (c == '\n')    s[i++] = c;  s[i] = '\0';  return i;}/* duplicates a string */char *dupstr(const char *s){  char *p = malloc(strlen(s) + 1);  if (p)    strcpy(p, s);  return p;}int main(int argc, char *argv[]){  int num_lines = DEFAULT_NUM_LINES;  char **line_ptrs;  char buffer[MAX_LINE_LEN];  int i;  unsigned j, current_line;  if (argc > 1) {    /*       We use a little trick here.  The command line parameter should be       in the form of "-n", where n is the number of lines.  We don't       check for the "-", but just pass it to atoi() anyway, and then       check if atoi() returned us a negative number.    */    num_lines = atoi(argv[1]);    if (num_lines >= 0) {      fprintf(stderr, "Expected -n, where n is the number of lines\n");      return EXIT_FAILURE;    }    /* Now make num_lines the positive number it's supposed to be. */    num_lines = -num_lines;  }   /* First, let's get enough storage for a list of n pointers... */  line_ptrs = malloc(sizeof *line_ptrs * num_lines);  if (!line_ptrs) {    fprintf(stderr, "Out of memory.  Sorry.\n");    return EXIT_FAILURE;  }  /* and make them all point to NULL */  for (i = 0; i < num_lines; i++)    line_ptrs[i] = NULL;  /* Now start reading */  current_line = 0;  do {    getline1(buffer, sizeof buffer);    if (!feof(stdin)) {      if (line_ptrs[current_line]) {        /* there's already something here */        free(line_ptrs[current_lidne]);      }      line_ptrs[current_line] = dupstr(buffer);      if (!line_ptrs[current_line]) {        fprintf(stderr, "Out of memory.  Sorry.\n");        return EXIT_FAILURE;      }      current_line = (current_line + 1) % num_lines;    }  } while (!feof(stdin));  /* Finished reading the file, so we are ready to print the lines */  for (i = 0; i < num_lines; i++) {    j = (current_line + i) % num_lines;    if (line_ptrs[j]) {      printf("%s", line_ptrs[j]);      free(line_ptrs[j]);    }  }  return EXIT_SUCCESS;}

5.11 指向函数的指针

练习题

Exercise 5-14. Modify the sort program to handle a -r flag, which indicates sorting in reverse (decreasing) order. Be sure that -r works with -n.

/* K&R Exercise 5-14 *//* 完成者：Steven Huang *//** * 添加注释者：zy * 这段代码调用的qsort居然是库函数stdlib.h里面的 * 于是我顺带分析了一个qsort的几个参数 * 分析结果如下图。 * 此外，虽然在仅仅将函数的名字当作qsort参数传过去就可以了 * 但是书上对这段还有着比较详细的描述，因为书上的qsort函数是自己实现的 *  */#include <stdio.h>#include <string.h>#include <stdlib.h>#define TRUE 1#define FALSE 0#define MAXLINES 5000       /* maximum number of lines */char *lineptr[MAXLINES];#define MAXLEN 1000         /* maximum length of a line */int reverse = FALSE;/* K&R2 p29 */int getline1(char s[], int lim){  int c, i;  for (i = 0; i < lim - 1 && (c = getchar()) != EOF && c != '\n'; i++)    s[i] = c;  if (c == '\n') {    s[i++] = c;  }  s[i] = '\0';  return i;}/* K&R2 p109 */int readlines(char *lineptr[], int maxlines){  int len, nlines;  char *p, line[MAXLEN];  nlines = 0;  while ((len = getline1(line, MAXLEN)) > 0)    if (nlines >= maxlines || (p = malloc(len)) == NULL)      return -1;    else {      line[len - 1] = '\0';  /* delete the newline */      strcpy(p, line);      lineptr[nlines++] = p;    }  return nlines;}/* K&R2 p109 */void writelines(char *lineptr[], int nlines){  int i;  for (i = 0; i < nlines; i++)    printf("%s\n", lineptr[i]);}int pstrcmp(const void *p1, const void *p2){  char * const *s1 = reverse ? p2 : p1;  char * const *s2 = reverse ? p1 : p2;  return strcmp(*s1, *s2);}int numcmp(const void *p1, const void *p2){  char * const *s1 = reverse ? p2 : p1;  char * const *s2 = reverse ? p1 : p2;  double v1, v2;  v1 = atof(*s1);  v2 = atof(*s2);  if (v1 < v2)    return -1;  else if (v1 > v2)    return 1;  else    return 0;}int main(int argc, char *argv[]){  int nlines;  int numeric = FALSE;  int i;  for (i = 1; i < argc; i++) {    if (*argv[i] == '-') {      switch (*(argv[i] + 1)) {        case 'n':  numeric = TRUE;  break;        case 'r':  reverse = TRUE;  break;        default:          fprintf(stderr, "invalid switch '%s'\n", argv[i]);          return EXIT_FAILURE;      }    }  }  if ((nlines = readlines(lineptr, MAXLINES)) >= 0) {    qsort(lineptr, nlines, sizeof(*lineptr), numeric ? numcmp : pstrcmp);//numcmp和pstrcmp就是函数名    writelines(lineptr, nlines);    return EXIT_SUCCESS;  } else {    fputs("input too big to sort\n", stderr);    return EXIT_FAILURE;  }}

调试情况：

打印参数：

Exercise 5-15. Add the option -f to fold upper and lower case together, so that case distinctions are not made during sorting; for example, a and A compare equal.

答：

code：

该题完成前：

zy@zy:~/Documents/eclipseWorkSpace/test/src$ ./a.out adcABABacd

由于是自己花了12个小时写的，倒不是代码有多难，只是指针的使用还比较弱完成后：

/* K&R Exercise 5-15 *//* 完成者：zy * 在Exercise 5-14的基础上修改完成，修改前的作者是：Steven Huang * */#include <stdio.h>#include <string.h>#include <stdlib.h>#define TRUE 1#define FALSE 0#define MAXLINES 5000       /* maximum number of lines */char *lineptr[MAXLINES];#define MAXLEN 1000         /* maximum length of a line */int reverse = FALSE;int ignoreCase = FALSE;//表示是否忽略大小写/* K&R2 p29 */int getline1(char s[], int lim){  int c, i;  for (i = 0; i < lim - 1 && (c = getchar()) != EOF && c != '\n'; i++)    s[i] = c;  if (c == '\n') {    s[i++] = c;  }  s[i] = '\0';  return i;}/* K&R2 p109 */int readlines(char *lineptr[], int maxlines){  int len, nlines;  char *p, line[MAXLEN];  nlines = 0;  while ((len = getline1(line, MAXLEN)) > 0)    if (nlines >= maxlines || (p = malloc(len)) == NULL)      return -1;    else {      line[len - 1] = '\0';  /* delete the newline */      strcpy(p, line);      lineptr[nlines++] = p;    }  return nlines;}/* K&R2 p109 */void writelines(char *lineptr[], int nlines){  int i;  for (i = 0; i < nlines; i++)    printf("%s\n", lineptr[i]);}void swap(void *v[],int i,int j){void *temp;temp=v[i];v[i]=v[j];v[j]=temp;}/** * K&R2 P119 * 不再调用库函数：stdlib.h 里面的qsort * 自己按着书上写一个：写的原因就是因为：当我传入 A B C a b c的时候会传出来：a a b b c c * 不知道为什么库函数里面的qsort会修改了原指针的值（指向） */void myQsort(void *v[],int left,int right,int (*comp)(void *,void *)){int i,last;void swap(void *v[],int,int);if(left>=right)return;swap(v,left,(left+right)/2);last=left;for(i= left+1;i<=right;i++)if((*comp)(v[i],v[left])<0){swap(v,++last,i);}swap(v,left,last);myQsort(v,left,last-1,comp);myQsort(v,left+1,right,comp);}/** * 该函数还可以进一步改进，现在只能对第一个字母实现 不区分大小写的功能 * */int pstrcmp( char *p1,  char *p2)//这里表示传过来的是一个以p1为名的字符数组，请看下图{char  a1 = *p1;//p1：就是一个字符串，可以直接理解为一个字符数组，对其*p1是取其第一个字符char  a2 = *p2;if(ignoreCase){if(a1>64 && a1<91 ){a1+=32;}if(a2>64 && a2<91 ){a2+=32;}}if (a1==a2) return 0;else return a1-a2;}int numcmp( char *p1,  char *p2){  double v1, v2;  v1 = atof(p1);  v2 = atof(p2);  if (v1 < v2)    return -1;  else if (v1 > v2)    return 1;  else    return 0;}int main(int argc, char *argv[]){  int nlines;  int numeric = FALSE;  int i;  for (i = 1; i < argc; i++) {    if (*argv[i] == '-') {      switch (*(argv[i] + 1)) {        case 'n':  numeric = TRUE;  break;        case 'r':  reverse = TRUE;  break;        case 'f':  ignoreCase = TRUE;  break;        default:          fprintf(stderr, "invalid switch '%s'\n", argv[i]);          return EXIT_FAILURE;      }    }  }  if ((nlines = readlines(lineptr, MAXLINES)) >= 0) {    myQsort((void **)lineptr,0, nlines-1, numeric ? numcmp : pstrcmp);//numcmp和pstrcmp就是函数名    writelines(lineptr, nlines);    return EXIT_SUCCESS;  } else {    fputs("input too big to sort\n", stderr);    return EXIT_FAILURE;  }}

运行结果：

zy@zy:~/Documents/eclipseWorkSpace/test/src$ ./a.out -fABCabcAaBbCczy@zy:~/Documents/eclipseWorkSpace/test/src$

可以看出声明为一个char *p1就是一个数组的形式，下面直接打印p1的时候看出来，当*p1

Exercise 5-16. Add the -d (‘‘directory order’’) option, which makes comparisons only on letters, numbers and blanks. Make sure it works in conjunction with -f.

答：暂时未找到合适的解法。

5.12 复杂声明

Exercise 5-17. Add a field-searching capability, so sorting may bee done on fields within lines, each field sorted according to an independent set of options. (The index for this book was sorted with -df for the index category and -n for the page numbers.)

答：暂时未找到合适的解法。

Exercise 5-18. Make dcl recover from input errors.

答：

暂时未找到合适的解法。
Exercise 5-19. Modify undcl so that it does not add redundant parentheses to declarations.

答：暂时未找到合适的解法。
Exercise 5-20. Expand dcl to handle declarations with function argument types, qualifiers like const, and so on.

答：暂时未找到合适的解法。

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