a. x = (2 + 3) * 6;
b. x = (12 + 6) / 2 * 3;
c. y = x = (2 + 3) / 4;
d. y = 3 + 2 * (x = 7 / 2);
a. 30
b. 27
c. x = y = 1
d. x=3 , y=9
a. x = ( int )3.8 + 3.3;
b. x = (2 + 3) * 10.5;
c. x = 3 / 5 * 22.0;
d. x = 22.0 * 3 / 5;
a. 6(由3 + 3.3截断而来)
b. 52
c. 0(0 * 22.0)
d. 13(66.0 / 5或13.2,然后把结果赋给int类型变量)
a. 30.0 / 4.0 * 5.0;
b. 30.0 / (4.0 * 5.0);
c. 30 / 4 * 5;
d. 30 * 5 / 4;
e. 30 / 4.0 * 5;
f. 30 / 4 * 5.0;
a. 37.5(7.5 *5.0)
b. 1.5 (30.0 / 20.0)
c. 35 (7*5)
d. 37 (150 / 4)
e. 37.5(7.5*5)
f. 35.0 (7*5.0)
int main(void)
{
int i = 1,
float n;
printf("Watch out! Here come a bunch of fractions!\n");
while (i < 30)
n = 1/i;
printf(" %f", n);
printf("That's all, folks!\n");
return;
}
答:
#include
int main(void)
{
int i = 1; //第3行:末尾用分号,而不是逗号
float n;
printf("Watch out! Here come a bunch of fractions!\n");
while (i++ < 30) //第6行:while语句创建了一个无限循环。因为i的值始终为1,所以它总是小于30。推测一下,应该是想写while(i++ < 30)。
{ //这样的缩进布局不能使第7行和第8行组成一个代码块。由于没有用花括号括起来, while循环只包括第7行,所以要添加花括号。
n = 1.0/i; //第7行:因为1和i都是整数,所以当i为1时,除法的结果是1;当i为更大的数时,除法结果为0。用n = 1.0/i,i在除法运算之前会被转换为浮点数,这样就能得到非零值。
printf(" %f\n", n); //第8行:在格式化字符串中没有换行符(\n),这导致数字被打印成一行。
}
printf("That's all, folks!\n");
return 0; //第10行:应该是return 0;
}
程序清单 5.9
// min_sec.c -- 把秒数转换成分和秒
#include
#define SEC_PER_MIN 60 //1分钟 60秒
int main(void)
{
int sec, min, left;
printf("Convert seconds to minutes and seconds!\n");
printf("Enter the number of seconds (<=0 to quit):\n");
scanf("%d", &sec); // 读取秒数
while(sec > 0)
{
min = sec / SEC_PER_MIN;
left = sec % SEC_PER_MIN;
printf("%d seconds is %d minutes, %d seconds.\n", sec, min, left);
printf("Enter the number of seconds (<=0 to quit):\n");
scanf("%d", &sec);
}
printf("Done!\n");
return 0;
}
另一个版本
#include
#define S_TO_M 60
int main(void)
{
int sec, min, left;
printf("This program converts seconds to minutes and ");
printf("seconds.\n");
printf("Just enter the number of seconds.\n");
printf("Enter 0 to end the program.\n");
while (sec > 0) {
scanf("%d", &sec);
min = sec/S_TO_M;
left = sec % S_TO_M;
printf("%d sec is %d min, %d sec. \n", sec, min, left);
printf("Next input?\n");
}
printf("Bye!\n");
return 0;
}
答:这个版本最大的问题是测试条件(sec是否大于0?)和scanf()语句获取sec变量的值之间的关系。具体地说,第一次测试时,程序尚未获得sec的值,用来与0作比较的是正好在sec变量内存位置上的一个垃圾值。一个比较笨拙的方法是初始化 sec(如,初始化为 1)。这样就可通过第一次测试。不过,还有另一个问题。当最后输入0结束程序时,在循环结束之前不会检查sec,所以0也被打印了出来。因此,更好的方法是在while测试之前使用scanf()语句。可以这样修改:
scanf("%d", &sec);
while ( sec > 0 ) {
min = sec/S_TO_M;
left = sec % S_TO_M;
printf("%d sec is %d min, %d sec.\n", sec, min, left);
printf(“Next input?\n”);
scanf("%d", &sec);
}
while循环第一轮迭代使用的是scanf()在循环外面获取的值。因此,在while循环的末尾还要使用一次scanf()语句。这是处理类似问题的常用方法。
#include
#define FORMAT "%s! C is cool!\n"
int main(void)
{
int num = 10;
printf(FORMAT,FORMAT);
printf("%d\n", ++num);
printf("%d\n", num++);
printf("%d\n", num--);
printf("%d\n", num);
return 0;
}
答:
解释一下。第1个printf()语句与右面的语句相同:printf("%s! C is cool!\n","%s! C is cool!\n");
第2个printf()语句首先把num递增为11,然后打印该值。第3个printf()语句打印num的值(值为11)。第 4个printf()语句打印n当前的值(仍为12),然后将其递减为11。最后一个printf()语句打印num的当前值(值为11)。
#include
int main(void)
{
char c1, c2;
int diff;
float num;
c1 = 'S';
c2 = 'O';
diff = c1 - c2;
num = diff;
printf("%c%c%c:%d %3.2f\n", c1, c2, c1, diff, num);
return 0;
}
答:
表达式c1 -c2的值和’S’ - '0’的值相同(其对应的ASCII值是83 - 79)。
#include
#define TEN 10
int main(void)
{
int n = 0;
while (n++ < TEN)
printf("%5d", n);
printf("\n");
return 0;
}
答:
把1~10打印在一行,每个数字占5列宽度,然后开始新的一行:
#include
int main(void)
{
char ch = 'a';
while (ch <= 'g')
printf("%5c", ch++);
printf("\n");
return 0;
}
a.
int x = 0;
while (++x < 3)
printf("%4d", x);
a. 1 2
注意,先递增x的值再比较。光标仍留在同一行。
b.
int x = 100;
while (x++ < 103)
printf("%4d\n",x);
printf("%4d\n",x);
b. 101
102
103
104
注意,这次x先比较后递增。在示例a和b中,x都是在先递增后打印。另外还要注意,虽然第2个printf()语句缩进了,但是这并不意味着它是while循环的一部分。因此,在while循环结束后,才会调用一次该printf()语句。
c.
char ch = ‘s’;
while (ch < ‘w’)
{
printf("%c", ch);
ch++;
}
printf("%c", ch);
c.stuvw
该例中,在第1次调用printf()语句后才会递增ch。
#define MESG "COMPUTER BYTES DOG"
#include
int main(void)
{
int n = 0;
while ( n < 5 )
printf("%s\n", MESG);
n++;
printf("That's all.\n");
return 0;
}
答:这个程序有点问题。while循环没有用花括号把两个缩进的语句括起来,只有printf()是循环的一部分,所以该程序一直重复打印消息COMPUTER BYTES DOG,直到强行关闭程序为止。
a. 将变量 x 的值增加10
b. 将变量 x 的值增加1
c. 将 a 与 b 之和的两倍赋给 c
d. 将 a 与 b 的两倍之和赋给 c
a. x = x + 10;
b. x++; or ++x; or x = x + 1;
c. c = 2 * (a + b);
d. c = a + 2 * b;
a. 将变量 x 的值减少1
b. 将 n 除以 k 的余数赋给 m
c. q 除以 b 减去 a,并将结果赋给 p
d. a 与 b 之和除以 c 与 d 的乘积,并将结果赋给 x
a. x–; or --x; or x = x - 1;
b. m = n % k;
c. p = q / (b - a);
d. x = (a + b) / (c * d);
#include
#define Min_Per_Hour 60
int main(void)
{
int time, hour, min;
printf("请输入您想表示的时间的分钟数:");
scanf("%d",&time);
while(time >0)
{
min = time % Min_Per_Hour;
hour = time / Min_Per_Hour;
printf("%d分钟是 %d小时 %d分钟\n", time, hour, min);
printf("请输入下一个您想表示的时间的分钟数(0退出):");
scanf("%d",&time);
}
printf("End\n");
return 0;
}
#include
int main(void)
{
int num, upper_limit;
printf("输入一个整数:");
scanf("%d",&num);
upper_limit = num+10;
while(num <= upper_limit)
{
printf("%d ",num);
num++;
}
printf("\n");
return 0;
}
#include
#define Days_Per_Week 7
int main(void)
{
int days, Day, Week;
printf("输入天数:");
scanf("%d", &days);
while(days > 0)
{
Week = days / Days_Per_Week;
Day = days % Days_Per_Week;
printf("%d days are %d weeks, %d days.\n", days, Week, Day);
printf("输入天数(非正值退出程序):");
scanf("%d", &days);
}
printf("End\n");
return 0;
}
Enter a height in centimeters: 182
182.0 cm = 5 feet, 11.7 inches
Enter a height in centimeters (<=0 to quit): 168.7
168.0 cm = 5 feet, 6.4 inches
Enter a height in centimeters (<=0 to quit): 0
bye
#include
#define CONVERT 0.3937
int main(void)
{
float height, inch;
int feet;
printf("Enter a height in centimeters:");
scanf("%f", &height);
while(height > 0)
{
feet = height / 31;
inch = CONVERT * height - feet * 12;
printf("%.1f cm = %d feet, %.1f inches\n", height, feet, inch);
printf("Enter a height in centimeters (<=0 to quit):");
scanf("%f", &height);
}
printf("bye\n");
return 0;
}
#include
int main(void)
{
int day, money, upper_limit;
day = 0;
money = 0;
printf("输入天数:");
scanf("%d", &upper_limit);
while(day++ < upper_limit)
money = money + day;
printf("money = %d\n", money);
return 0;
}
#include
int main(void)
{
int day, money, upper_limit;
day = 0;
money = 0;
printf("输入天数:");
scanf("%d", &upper_limit);
while(day++ < upper_limit)
money = money + day * day;
printf("money = %d\n", money);
return 0;
}
#include
void cube(double num);
int main(void)
{
double number;
printf("输入一个double类型的数:");
scanf("%lf", &number);
cube(number);
return 0;
}
void cube(double num)
{
double result = num * num * num;
printf("%lf 的立方值是 %lf\n", num, result);
}
This program computes moduli.
Enter an integer to serve as the second operand: 256
Now enter the first operand: 438
438 % 256 is 182
Enter next number for first operand (<= 0 to quit): 1234567
1234567 % 256 is 135
Enter next number for first operand (<= 0 to quit): 0
Done
#include
int main(void)
{
int first, last, result;
printf("This program computes moduli.\n");
printf("Enter an integer to serve as the second operand: ");
scanf("%d", &first);
printf("Now enter the first operand: ");
scanf("%d", &last);
while(last > 0)
{
result = last % first;
printf("%d %% %d is %d\n", last, first, result);
printf("Enter next number for first operand (<= 0 to quit): ");
scanf("%d", &last);
}
printf("Done\n");
return 0;
}
#include
void Temperatures(double tem);
int main(void)
{
double Fahrenheit;
printf("输入一个华氏温度(非数字时,程序结束):");
while( scanf("%lf",&Fahrenheit) == 1)
{
Temperatures(Fahrenheit);
printf("输入一个华氏温度(非数字时,程序结束):");
}
printf("End\n");
return 0;
}
void Temperatures(double tem)
{
double Centigrade, Kelvin;
const double h_convert_c1 = 5.0 / 9.0, h_convert_c2 = 32.0, c_convert_k = 273.16;
Centigrade = h_convert_c1 * (tem - h_convert_c2);
Kelvin = Centigrade + c_convert_k;
printf("华氏温度:%.2lf 摄氏温度:%.2lf 开氏温度:%.2lf\n", tem, Centigrade, Kelvin);
}