下面的一些程序要求输入以EOF终止。如果你的操作系统很难或根本无法使用重定向，请使用一些其他的测试来终止输入，如读到&字符时停止。
1.设计一个程序，统计在读到文件结尾之前读取的字符数。
include<stdio.h>
int main(void)
{
int i = 0;
while (getchar() != EOF)
i++;
printf("字符数量为%d\n", i);
return 0;
}2.编写一个程序，在遇到 EOF 之前，把输入作为字符流读取。程序要打印每个输入的字符及其相应的ASCII十进制值。注意，在ASCII序列中，空格字符前面的字符都是非打印字符，要特殊处理这些字符。如果非打印字符是换行符或制表符，则分别打印\n或\t。否则，使用控制字符表示法。例如，ASCII的1是Ctrl+A，可显示为^A。注意，A的ASCII值是Ctrl+A的值加上64。其他非打印字符也有类似的关系。除每次遇到换行符打印新的一行之外，每行打印10对值。（注意：不同的操作系统其控制字符可能不同。）
#include<stdio.h>
int main(void)
{
int ch;
int i = 0;
while ((ch = getchar()) != EOF)
{
if (ch > ' ')
printf("%c\t", ch);
else if (ch == '\n')
printf("\\n\t");
else if (ch == '\t')
printf("\\t\t");
else
printf("^%c\t", ch + 64);
printf("%d\t", ch);
i++;
if (ch == '\n' || i == 10)
{
printf("\n");
i = 0;
}
}
return 0;
}3.编写一个程序，在遇到 EOF 之前，把输入作为字符流读取。该程序要报告输入中的大写字母和小写字母的个数。假设大小写字母数值是连续的。或者使用ctype.h库中合适的分类函数更方便。
#include<stdio.h>
int main(void)
{
char ch;
int uppercase = 0;
int lowercase = 0;
while ((ch = getchar()) != EOF)
{
if (ch >= 'A' && ch <= 'Z')
uppercase++;
if (ch >= 'a' && ch <= 'z')
lowercase++;
}
printf("uppercase letters 是%d，lowercase letters 是%d\n", uppercase, lowercase);
return 0;
}4.编写一个程序，在遇到EOF之前，把输入作为字符流读取。该程序要报告平均每个单词的字母数。不要把空白统计为单词的字母。实际上，标点符号也不应该统计，但是现在暂时不同考虑这么多（如果你比较在意这点，考虑使用ctype.h系列中的ispunct()函数）。
#include<stdio.h>
int main(void)
{
char ch;
int word = 0;
int character = 0;
int space = 1;
printf("输入一些单词\n");
while ((ch = getchar()) != EOF)
{
if ((ch >= 'a' && ch <= 'z') || (ch >= 'A' && ch <= 'Z'))
{
character++;
if (space)
{
word++;
space = 0;
}
}
else
space = 1;
}
printf("the average characters of %d words is %d.\n", word, character / word);
return 0;
}5.修改程序清单8.4的猜数字程序，使用更智能的猜测策略。例如，程序最初猜50，询问用户是猜大了、猜小了还是猜对了。如果猜小了，那么下一次猜测的值应是50和100中值，也就是75。如果这次猜大了，那么下一次猜测的值应是50和75的中值，等等。使用二分查找（binary search）策略，如果用户没有欺骗程序，那么程序很快就会猜到正确的答案。
#include<stdio.h>
int main(void)
{
char ch;
int low = 1;
int high = 100;
int guess = (low + high) / 2;
printf("输入一个在1~100范围内的整数，我会尝试猜一下。\n"
"如果我猜对了，请回我一个\"y\""
"如果我猜高回我一个\"y\""
"如果我猜低回我一个\"l\"");
printf("嗯~你输入的数字是 %d？\n", guess);
while ((ch = getchar()) != 'y')
{
if (ch == 'h')
{
high = guess;
guess = (low + high) / 2;
}
else if (ch == 'l')
{
low = guess;
guess = (low + high) / 2;
}
else if (ch == '\n')
continue;
else
{
printf("请输入 y， h 或 l");
continue;
}
printf("输入一个在1~100范围内的整数，我会尝试猜一下。\n"
"如果我猜对了，请回我一个\"y\""
"如果我猜高回我一个\"y\""
"如果我猜低回我一个\"l\"");
printf("嗯~你输入的数字是 %d？\n", guess);
}
printf("我知道我能做到！\n");
return 0;
}6.修改程序清单8.8中的get_first()函数，让该函数返回读取的第1个非空白字符，并在一个简单的程序中测试。
#include<stdio.h>
char get_first(void);
int main(void)
{
char ch;
printf("输入一些词以测试\n");
ch = get_first();
printf("%c\n", ch);
return 0;
}
char get_first(void)
{
char ch;
ch = getchar();
while (cj == ' ' || ch == '\t' || ch == '\n')
ch = getchar();
while (getchar() != '\n')
continue;
return ch;
}7.修改第7章的编程练习8，用字符代替数字标记菜单的选项。用q代替5作为结束输入的标记。
#include<stdio.h>
#define TIME 40.0
#define PAY1 8.75
#define PAY2 9.33
#define PAY3 10.00
#define PAY4 11.20
#define ADD 1.5
#define RATE1 0.15
#define RATE2 0.2
#define RATE3 0.25
#define BREAK1 300
#define BREAK2 450
#define TAX1 (BREAK1 * RATE1)
#define TAX2 TAX1+(RATE2 * (BREAK2 - BREAK1))
int main()
{
float time, wage, tax, income, pay;
char ch;
printf("'*' * 30\n");
printf("enter the character corresponding to the desired pay rate or action:\n");
printf("a. $8.75/hr b. $9.33/hr\n");
printf("c. $10.00/hr d. $11.20/hr\n");
printf("q. quit\n");
printf("'*' * 30\n");
while ((ch = getchar()) != 'q')
{
if (ch < 'a' || ch > 'd')
{
printf("enter the correct option\n");
printf("'*' * 30\n");
printf("enter the character corresponding to the desired pay rate or action:\n");
printf("a. $8.75/hr b. $9.33/hr\n");
printf("c. $10.00/hr d. $11.20/hr\n");
printf("q. quit\n");
printf("'*' * 30\n");
continue;
}
switch (ch)
{
case 'a':
pay = PAY1;
break;
case 'b':
pay = PAY2;
break;
case 'c':
pay = PAY3;
break;
case 'd':
pay = PAY4;
break;
}
printf("enter the working time(hour)\n");
scanf("%f", &time);
while (getchar() != '\n')
continue;
if (time <= TIME)
wage = pay * time;
else
wage = pay * (TIME + ADD * (time - TIME));
if (wage <= BREAK1)
tax = RATE1 * wage;
else if (wage > BREAK1 && wage < BREAK2)
tax = TAX1 + RATE2 * (wage - BREAK1);
else
tax = TAX2 + RATE3 * (wage - BREAK2);
income = wage - tax;
printf("The wage is $%.2f, tax is $%.2f, net income is $%.2f\n", wage, tax, income);
printf("\nEnter the character corresponding to the desired pay rate or action:\n");
}
return 0;
}
8.编写一个程序，显示一个提供加法、减法、乘法、除法的菜单。获得用户选择的选项后，程序提示用户输入两个数字，然后执行用户刚才选择的操作。该程序只接受菜单提供的选项。程序使用float类型的变量储存用户输入的数字，如果用户输入失败，则允许再次输入。进行除法运算时，如果用户输入0作为第2个数（除数），程序应提示用户重新输入一个新值。该程序的一个运行示例如下：
Enter the operation of your choice:
a. add s. subtract
m. multiply d. divide
q. quit
a
Enter first number: 22 .4
Enter second number: one
one is not an number.
Please enter a number, such as 2.5, -1.78E8, or 3: 1
22.4 + 1 = 23.4
Enter the operation of your choice:
a. add s. subtract
m. multiply d. divide
q. quit
d
Enter first number: 18.4
Enter second number: 0
Enter a number other than 0: 0.2
18.4 / 0.2 = 92
Enter the operation of your choice:
a. add s. subtract
m. multiply d. divide
q. quit
q
Bye.
a. add s. subtract
m. multiply d. divide
q. quit
a
Enter first number: 22 .4
Enter second number: one
one is not an number.
Please enter a number, such as 2.5, -1.78E8, or 3: 1
22.4 + 1 = 23.4
Enter the operation of your choice:
a. add s. subtract
m. multiply d. divide
q. quit
d
Enter first number: 18.4
Enter second number: 0
Enter a number other than 0: 0.2
18.4 / 0.2 = 92
Enter the operation of your choice:
a. add s. subtract
m. multiply d. divide
q. quit
q
Bye.
#include<stdio.h>
float get_float(void);
void option(void);
char choice(void);
char get_first(void);
int main(void)
{
float first_number, second_number;
char ch;
option();
while ((ch = choice()) != 'q')
{
printf("Enter first number: ");
first_number = get_float();
printf("Enter second number: ");
second_number = get_float();
switch (ch)
{
case 'a':
printf("%g + %g = %g\n", first_number, second_number, first_number + second_number);
break;
case 's':
printf("%g - %g = %g\n", first_number, second_number, first_number - second_number);
break;
case 'm':
printf("%g * %g = %g\n", first_number, second_number, first_number * second_number);
break;
case 'd':
while (second_number == 0)
{
printf("Enter a number other than 0: ");
second_number = get_float();
}
printf("%g / %g = %g\n", first_number, second_number, first_number / second_number);
break;
}
option();
}
printf("Bye.\n");
return 0;
}
void option(void)
{
printf("Enter the operation of your choice:\n");
printf("a. add s. subtract\n");
printf("m. multiply d. divide\n");
printf("q. quit\n");
}
float get_float(void)
{
float number;
char ch;
while (scanf("%f", &number) != 1)
{
while ((ch = getchar()) != '\n')
putchar(ch);
printf(" is not an number.\n"
"Please enter a number, such as 2.5, -1.78E8, or 3: ");
}
return number;
}
char choice(void)
{
char ch;
ch = get_first();
while (ch != 'a' && ch != 's' && ch != 'm' && ch != 'd' && ch != 'q')
{
printf("please enter a, s, m, d or q\n");
ch = get_first();
}
return ch;
}
char get_first(void)
{
char ch;
ch = getchar();
while (ch == ' ' || ch == '\n' || ch == '\t')
ch = getchar();
while (getchar() != '\n')
continue;
return ch;
}
