C Primer Plus 第6版 编程练习 chapter 8

1. 第1题

1.1 题目描述

设计一个程序，统计在读到文件结尾之前读取的字符数

1.2 编程源码

# include

int main(){
	int num = 0;
	char c;
	
	while((c = getchar())!=EOF)	++num;
	
	printf("\n共计 %d 个字符\n", num);

	return 0;
}

1.3 结果显示

输入文本内容：

abc
e
f

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2. 第2题

2.1 题目描述

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）。

2.2 编程源码

# include

int main(){
	int num = 0;
	char c;
	
	while((c = getchar())!=EOF){
		++num;
		if('\n' == c) {
			printf("\n\\n-%d\t", c);
			num = 1;
		} else if('\t' == c) printf("\\t-%d\t", c);
		else printf("%c-%d\t", c,c);
		if(num%10 == 0) putchar('\n');
	}

	return 0;
}

2.3 结果显示

输入文本内容：

abc	123
e
f

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3. 第3题

3.1 题目描述

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符个数。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。

3.2 编程源码

# include
# include

int main(){
	int num = 0;
	int n=0;
	char c;
	
	while((c = getchar())!=EOF){
		if(isupper(c)) ++n;
		if(islower(c)) ++num;
	}
	printf("大写字母的个数：%d\n", n);
	printf("小写字母的个数：%d\n", num);

	return 0;
}

3.3 结果显示

输入文本内容：

ab	cde123 445
AAA AAB C
212324
f

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4. 第4题

4.1 题目描述

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。

4.2 编程源码

# include
# include

int main(){
	int num_w = 0;
	float num_c=0;
	char c;
	int flag = 0;
	
	while((c = getchar())!=EOF){
		if(flag){
			if(isalpha(c)) ++num_c;
			else{
				++num_w;
				flag = 0;
			}
		}else{
			if(isalpha(c)){
				flag = 1;
				num_c++;
			}
		}
	}
	printf("\n该文章共计%d个单词，每个单词平局%f个字符\n", num_w,num_c/num_w);

	return 0;
}

4.3 结果显示

输入文本内容：

this is a dog.
dog is running on the grass.

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5. 第5题

5.1 题目描述

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、猜对了。如果是猜小了，那么下一次猜测的值应是50和100的中值，也就是75.如果这次猜大了，那么下一次猜测的值应该是50和75的中值，等等。使用二分查找策略，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。

5.2 编程源码

# include

int main(void){
	int guess = 1;
	int low = 1;
	int up = 100;
	
	printf("Pick an integer from 1 to 100. I will try to guess ");
	printf("it. \nRespond with a y if my guess is right and with");
	printf("\nan n if it is wrong.\n");
	printf("Uh... is your number %d?\n", guess);
	
	char c;
	printf("it is big 、small or equal？please input  b s e or q(to quit):");
	while((c=getchar()) && c!=EOF && c!='q'){
		while(getchar()!= '\n');
		if(c == 'e'){
			printf("I knew I could do it !\n");
			break;
		}
		switch(c){
			case 'b': up = guess;guess = (low+up)>>1;break;
			case 's': low = guess;guess = (low+up)>>1;break;
		}
		printf("well then, is it %d ?please input  b s e or q(to quit):\n ", guess);
	}
	printf("Bye\n");
	
	return 0;
}

5.3 结果显示

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6. 第6题

6.1 题目描述

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。

6.2 编程源码

# include

char get_choice(void);
char get_first(void);
int get_int(void);
void count(void);

int main(void){
	int choice;
	void count(void);
	
	while((choice = get_choice()) != 'q'){
		switch(choice){
			case 'a': printf("Buy low, sell high.\n");break;
			case 'b': putchar('\a');break;
			case 'c': count();break;
			default: printf("Program error!\n");break;			
		}
	}
	printf("Bye.\n");
	
	return 0;
}

void count(void){
	int n,i;
	printf("Count how far? Enter an integer:\n");
	n = get_int();
	for(i=1;i<=n;++i) printf("%d\n", i);
	while(getchar() != '\n') ;
}

char get_choice(void){
	int ch;
	printf("Enter the letter of your choice:\n");
	printf("a. advice		b.bell\n");
	printf("c. count		q.quit\n");
	ch = get_first();
	
	while((ch <'a' || ch>'c') && ch != 'q'){
		printf("Please respond with a,b,c, or q.\n");
		ch = get_first();
	}
	
	return ch;
}

char get_first(void){
	int ch;
	while((ch=getchar())!=EOF && ch == ' ');
	while(getchar() !='\n') ;
	
	return ch;
}

int get_int(void){
	int input;
	char ch;
	
	while(scanf("%d", &input)!=1){
		while((ch=getchar())!='\n')putchar(ch);
		printf(" is not an integer.\nPlease enter an ");
		printf("integer value, such as 25, -178, or3:");
	}
}

6.3 结果显示

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7. 第7题

7.1 题目描述

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

7.2 编程源码

# include

# define BASE_HOURS		40
# define RATE_OVERTIEM	1.5
# define LINE_PRE		300
# define RATE_PRE		0.15
# define LINE_XU		150
# define RATE_XU		0.2
# define RATE_LAST		0.25

void display(void){
	printf("\n**************************************************************************\n");
	printf("Enter the number 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("**************************************************************************\n");
	printf("请输出您的选项（a~d,q）:");
}

float get_a(void){
	char id ;
	float num = 0;
	
	display();
	while((id=getchar()) != '\n'){
		if(id == 'q') break;
		while(getchar() != '\n') ;
		if(id < 'a' || id > 'd') {
			printf("\n2请根据菜单提示输入。");
			display();
			continue;
		}
		switch(id){
			case 'a':num = 8.75;return(num);
			case 'b':num = 9.33;return(num);
			case 'c':num = 10;return(num);
			case 'd':num = 11.2;return(num);
			default:break;
		}
	}
	
	printf("Bye\n");
	return -1;
}

int main(){
	float num = 0;
	float base = 0;
	float salary_sum = 0;
	float salary_rate = 0;
	float salary = 0;
	float work_hours = 0;
	
	while( base = get_a()){
		if(base < 0)break;
		printf("\n请输入您一周的工作小时数：");
		scanf("%f", &num);		
		while(getchar() != '\n') ;
		
		work_hours = num>BASE_HOURS?BASE_HOURS+(num-BASE_HOURS)*RATE_OVERTIEM:num;		
		salary_sum = work_hours * base;
		
		if(salary_sum <= LINE_PRE){
			salary_rate = salary_sum * RATE_PRE;
			salary = salary_sum - salary_rate;
		}else if(salary_sum <= LINE_PRE + LINE_XU){
			salary_rate = LINE_PRE * RATE_PRE + (salary_sum - LINE_PRE) * RATE_XU;
			salary = salary_sum - salary_rate;
		}else{
			salary_rate = LINE_PRE * RATE_PRE + LINE_XU * RATE_XU + (salary_sum - LINE_XU) * RATE_LAST;
			salary = salary_sum - salary_rate;
		}	
		
		printf("工资总计：%f\n", salary_sum);
		printf("工资税金：%f\n", salary_rate);
		printf("工资到手：%f\n\n", salary);
	}	

	return 0;
}

7.3 结果显示

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8. 第8题

8.1 题目描述

编写一个程序，显示一个提供加法、减法、乘法、除法的莱单。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。该程序只接受莱单提供的迭项。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作为笫2 个数(除数)，程序应提示用户垂新输入一个新值。该程序的一个运行示例如下:
Enter the operation of your choice:
a. add b. subtract
c. 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 b. subtract
c. 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 b. subtract
c. multiply d. divide
q. quit
q
Bye

8.2 编程源码

# include

# define BASE_HOURS		40
# define RATE_OVERTIEM	1.5
# define LINE_PRE		300
# define RATE_PRE		0.15
# define LINE_XU		150
# define RATE_XU		0.2
# define RATE_LAST		0.25

void display(void){
	printf("Enter the operation of your choice: \n");
	printf("a. add				b. subtract\n");
	printf("c. multiply		d. divide\n");
	printf("q. quit\n");	
}

int main(){
	char id;
	float f = 0;
	float s = 0;
	float sum = 0;
	
	display();
	while((id=getchar()) != '\n'){
		if(id == 'q') break;
		while(getchar() != '\n') ;
		printf("Enter first number:"); 
		while(scanf("%f",&f) != 1){
			char c;
			while((c=getchar())!= '\n')putchar(c);
			printf("is not a number\n");
			printf("Please enter a number, such as 2.5, -1.78E8 , or  3: ");
		}
		while(getchar() != '\n') ;
		
		printf("Enter second number:"); 
		while(scanf("%f",&s) != 1){
			char c;
			while((c=getchar())!= '\n')putchar(c);
			printf("is not a number\n");
			printf("Please enter a number, such as 2.5, -1.78E8 , or  3: ");
		}
		while(getchar() != '\n') ;
		
		int flag = 0;
		switch(id){
			case 'a':sum = f + s;printf("%f + %f = %f \n",f,s,sum);;break;
			case 'b':sum = f-s;printf("%f - %f = %f \n",f,s,sum);break;
			case 'c':sum = f*s;printf("%f * %f = %f \n",f,s,sum);break;
			case 'd':flag = 1;break;
			default:break;
		}
		
		if(flag){
			if(s< 0.0000001 || s> -0.0000001){
				printf("Enter a number other than 0: ");
				while(scanf("%f",&s) != 1){
					char c;
					while((c=getchar())!= '\n')putchar(c);
					printf("is not a number\n");
					printf("Please enter a number, such as 2.5, -1.78E8 , or  3: ");
				}
				while(getchar() != '\n') ;
			}
			
			sum = f/s;
			printf("%f / %f = %f \n",f,s,sum);
		}
		display();		
	}
	printf("Bye\n");
	return 0;
}

8.3 结果显示

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