C语言详解第五版答案

1、第1章1软件3位，字节，存储单元，主存储器，辅助存储器，LAN，WAN1a、b、c的值相加，把和存储到x；y除以z，将结果存储到x；c减b，然后加a，将结果存储到d；z加1，并将结果存储到z；celsius加，并将结果存储到kelvin。3源程序，编译器，编辑器（字处理器）1问题需求，分析，设计，实现，测试和验证，维护1算法细化(1) 获得以公里表示的距离。(2) 把距离转化成英里。2.1 用英里表示的距离是用公里表示的距离的0.621倍。(3) 显示用英里表示的距离。第2章节1a.void, double, returnb.printfc.MAX_ENTRIES, Gd.time, xyz1

2、23, this_is_a_long_onee.Sue's,part#2,"char", #include3预处理器，#define和#include-33double, int, char1Enter two integers> 5 7m = 10 n = 21 3My name is Jane Doe.I live in Ann Arbor, MIand I have 11 years of programming experience.1/* This is a comment? */* This one seems like a comment doe

3、snt it */1a. 22/7为37/22为0 22 % 7为1 7 % 22为7b. 16 / 15 为1 15 / 16为0 16 % 15为115 % 16为15c. 23 / 3 为7 3 / 23为0 23 % 3为2 3 % 23为3d. 16 / -3 为? -3 / 16为? 16 % -3为? -3 % 16为? （?意为结果是变化的）3a. 3 g. 未定义 m. b. ? h. 未定义 n. c. 1 i. ? o. 1d. j. 3 p. 未定义e. ? k. q. 3f. l. 9 r. （?意为结果是变化的）5a.white为c.orange为0e.lime为2

4、b.green为d.blue为f.purple为1printf("Salary is %10.2fn",salary);3xisiis100iis100xis1在调用scanf获得数据之前先调用printf显示提示。当调用scanf后有数据回应时调用printf。提示用于交互式程序而不用于批处理程序。批处理程序应该回显输入，交互程序也可以回显输入。第3章1问题输入double hours /* number of hours worked */double rate /* hourly rate of pay */ 问题输出double gross /* gross sal

5、ary */ 算法(1)输入工作时间hours和每小时所付的工资率rate。(2)计算总工资gross。2.1把hours * rate的值赋给gross。(3)显示总工资gross。3问题输入double reg_hours /* number of regular hours worked */double ot_hours /* number of overtime hours worked */double rate /* hourly rate of pay */ 问题输出double gross /* gross salary */ 算法(1)输入常规工作时数（reg_hours）

6、、加班时数（ot_hours）和每小时所付工资比率rate。(2)计算总工资gross。2.1 将reg_hours * rate的值赋给gross。2.2 前一步gross的值再加上ot_hours * 1.5 * rate。(3)显示总工资gross。1a. sqrt(u + v) * pow(w,2) b. log(pow(x,y) c.sqrt(pow(x y, 3) d. fabs(x * y w / z)1设计阶段1HI MOM被用较大的印刷体字母垂直地输出。1a. b. c.* * * * 31.42 * * * *d. e. 3函数参数被用来在一个程序的各个模块之间、主程序与其

7、模块之间传递信息。参数使函数更容易被其他函数或程序复用。带参数的函数是建造更大程序的基石。第4章3x = 3.0 y = 4.0 z = 2.0 flag = 0 ! ( flag | ( y + z >= x z ) 0 1 1 011（TRUE）0（FALSE）1（TRUE）1（TRUE）5ans为2。1a.not lessb.greater than1. if (x > y ) x = x + 10.0; printf("x Biggern"); else printf("x Smallern"); printf("y is

8、%.2fn",y); 3. if (engine_type = 'J') printf("Jet engine"); speed_category = 1; else printf("Propellers"); speed_category = 2; 1额外的程序常量:CAP_GALLONS 100 /* maximum number of gallons (in thousands) for basic fee */ 额外的程序变量:int excess /* number of gallons over CAP_GALLON

9、S */ comp_use_charge的改写算法：(1)used 为 current previous(2)if used > CAP_GALLONS excess 为used GAP_GALLONS use_charge 为 CAP_GALLONS * PRE_1000_CHG + excess * PER_1000_CHG * 2else use_charge 为 used * PER_1000_CHG1.语句部分工资税结果if (salary < 0.0 )?23500.00 < 0.00 为假else if (salary < 15000.00 )23500.

10、00 < 15000.00 为假else if (salary < 30000.00 )23500.00 < 30000.00 为真tax = (salary 15000.00)* 0.18 3.if ( pH > 7 ) if (pH < 12 ) printf("Alkaline"); else printf("Very alkaline"); else if (pH = 7 ) printf("Neutral"); else if (pH > 2) printf("Acidic&qu

11、ot;); else printf("Very acidic");1.redblue yellow第5章1a. 计数循环(1) 初始化sum为0。(2) 设置lcv为0。(3)while lcv<35(4)获得下一个考试成绩。(5)把考试成绩与sum相加。(6)把lcv的值加1。3控制文件结束的循环(1)初始化count为0。(2)获得第一个温度并存储输入状态。(3) while 输入状态并不表示已到达文件末尾(4) if温度>100, count加1；(5) 获得下一个温度并存储输入状态。10 101 92 83 74 65 51a. Enter an int

12、eger> 5 5 25 125 625 b. Enter an integer> 6 6 36 216 1296c. Enter an integer> 7 7 49 343 2401 一般地，该循环显示n,n2,n3和n4。3count = 0; sum = 0; while (count < 5) printf("Next number> "); scanf("%d", &next_num); sum += next_num; count += 1; printf("%d numbers were a

13、dded; ", count); printf("their sum is %d.n",sum);1当n = 8时:语 句addsum作 用sum = 0; 0初始化sum为0odd = 1; 1初始化odd为1odd < n; 1<8为真sum += odd; 1sum = 0 + 1odd +=23odd = 1 + 2odd < n; 3<8 为真sum += odd; 4sum = 1 + 3odd +=25odd = 3 + 2odd < n; 5<8 为真sum +=odd; 9sum = 4 + 5odd +=27

14、odd = 5 + 2odd < n; 7<8为真sum += odd; 16sum = 9 + 7odd += 29odd = 7 + 2odd < n; 9<8 为假退出循环printf("Sum ofOutput: Sum of positive odd numbers less than8 is 16.3两个问题的答案都为0。5+i -j; n = i * j; m = i + j; j-; p = i + j;7.a.1 10 2 8 3 6 4 4 5 2b. j = 10; for ( i = 0; i < 5; +i) printf(&q

15、uot;%d %dn", i+1,j); j -= 2; 1. 任何小于8 000桶的初始化供给。3.Number of barrels currently in tank> 0 barrels are available. Enter number of gallons removed> After removal of 7581.00 gallons (180.50 barrels), 8170.30 barrels are available. Enter number of gallons removed> After removal of 7984.20

16、gallons (190.10 barrels),Only 7980.20 barrels are left. * WARNING *Available Supply is less than 10 percent of tank's 80000.00-barrel capacity.1步骤a初始化循环控制变量。 步骤c循环重复的条件。 步骤e循环控制变量的更新。1a. * b. *1while循环较好，因为do-while在每次迭代时对相同的条件检测两次。1for (status = fscanf(hdd_file, "%d", &next_hdd); s

17、tatus = 1; status = fscanf(hdd_file, "%d",&next_hdd) if (next_hdd > heat_deg_days) heat_deg_days = next_hdd; coldest_mon = ct; +ct; 1for ( count = 0; count <= n; +count) printf("DEBUG* * * count = %dn", count); sum += count; printf("DEBUG* * * sum = %dn", sum)

18、; 第6章1.void sum_n_avg(double n1, /* input numbers */ double n2, double n3, double *sump, /* output sum of the three numbers */double *avgp) /* output average of the numbers */3引 用何处合法数据类型值&manymain int *指向带灰底的单元valpsubdouble *指向带灰底的单元codemainchar'g'&codemainchar *指向不带底纹的单元countpsubin

19、t *指向带网格的单元*countpsubint14*valpsubdoubleletpsubchar *指向不带底纹的单元&xmaindouble *指向带灰底的单元1函数调用num1num2num3order(&num3, &num2);81210order(&num2, &num1);128order(&num3, &num2);108对函数order的这种调用顺序将num1、num2、num3变成降序（从最大到最小）。节1void onef(int dat, int *outlp, int *out2p) int tmp; two

20、f(dat, &tmp, out2p); . void twof(int indat, int *result1p, int *result2p)1因为有很多值需要返回，这只能通过使用输出参数实现。第7章1表示误差出现在当一个double类型变量的尾数包含的位（二进制位）数不足以精确表示某个分数时；抵消误差发生在当两个在数量上相差悬殊的数进行某种运算时，其结果是较小的数丢失。3 x y m n10.5 7.2 5 2a. x / (double)m b. x / m c. (double)(n * m) d. (double)(n / m) + y e. (double)(n / m)

21、 1a. 3 b. 'E' c. -11typedef enum monday, tuesday, wednesday, thursday, friday, saturday, sunday dat_t;a.0b.3c.0（FALSE）d.fridaye.wednesdayf.1（TRUE）节12.0.3.0就是一个这样的区间。第8章节1x3是一个有效的变量名，x3是对数组x第4个元素的引用。3double sq_root11; int cube11;1之前 节x0x1x2x3x4x5x6x7之后 x0x1x2x3x4x5x6x71#include<math.h>

22、#define MAX_SIZE 11. double cubeMAX_SIZE; int sq_rootMAX_SIZE; int i; for (i = 0; i < MAX_SIZE; +i) sq_rooti = sqrt(double)i); cubei = i * i * i; 1seg_len = sqrt(pow(xi+1 xi, 2) + pow(yi+1 yi, 2);3sum = 0; for (i = 0; i < LIST_SIZE; i += 2) sum += listi;1如果一个数组的几个元素都被某个函数处理，传递整个数组数据比传递单个的元素更好。

23、3/* * Gets data to place in dbl_arr until value of sentinel* is encountered in the input.* Returns number of values stored through dbl_sizep.* Stops input prematurely if there are more than dbl_max data* values before the sentinel or if invalid data is encountered.* Pre: sentinel and dbl_max are def

24、ined and dbl_max* is the declared size of dbl_arr* Post: returns 1 for no error, and 0 for any error conditions.*/intfill_to_sentinel(int dbl_max, /* input declared size of dbl_arr */double sentinel,/* input end of data value in input list */double dbl_arr,/* output array of data */int *dbl_sizep)/*

25、 output number of data values stored in dbl_arr */ double data;int i, status; int result = 1; /* Sentinel input loop */ i = 0; status = scanf("%lf",&data); while (status = 1 && data != sentinel && i < dbl_max) dbl_arri = data; i+; status = scanf("%lf",&

26、data); /* Issues error message on premature exit */ if (status != 1) printf("n* Error in data format *n"); printf("* Using first %d data values *n", i); result = 0; else if (data != sentinel) printf("n* Error: too much data before sentinel *n"); printf("* Using fir

27、st %d data values *n", i); result = 0; /* Sends back size of used portion of array, and error status */ *dbl_sizep = i; return (result); 5.| | $ | $ - | $ ch is -.节1a.n-1作为匹配的位置返回。 b. 第一个匹配的位置被返回。3为实现降序排序，用index_of_max代替所出现的每个index_of_min，用largest代替所有的smallest来改写选择排序算法。节1a. int i, j, k; for (i

28、= 0; i < MAXCRS; +i) printf("Processing course number %d: n", i); for (j = 0; j < 5; +j) printf(" Campus %dn", j); for (k = 0; k < 4; +k) printf(" Enter number of "); switch (k) case 0 : printf("Freshmen > "); break; case 1 ： printf("Sophomore

29、s > "); break; case 2 : printf("Juniors > "); break; case 3 : printf("Seniors > "); break; scanf("%d", &enrollijk); b. int i, j, jcnt; /* Number of juniors. */ jcnt = 0; for (i = 0; i < MAXCRS; +i) for (j = 0; j < 5; +j) jcnt += enrollij2;c. /* *

30、 Compute the number of students in a course who have a * specific rank. * returns -1 if rank or course is out of range. */ int find_students (int enroll54, int rank, int course) int i, cnt = 0; if (rank >= 0 && rank <= 3) && (course >= 0 && course < MAXCRS) for (i

31、 = 0; i < 5; +i) cnt += enrollcourseirank; else cnt = -1; return (cnt); 用 printf("Number of sophomores in course 2 is %dn", find_students(enroll, 1, 2);d. int i, j, total, upper; total = 0; upper = 0; for (j = 0; j < 5; +j) for (i = 0; i < MAXCRS; +i) upper += enrollij2 + enrollij

32、3; printf("Number of upperclass students on campus "); printf("%d is %d.n", j, upper); total += upper; printf("Total upperclass students on all campuses is %d.n",total);1.a. sales1fall, sales1winter, sales1spring, sales1summerb. sales0spring, sales1spring, sales2spring,

33、 sales3spring, sales4springc. sales0fall, sales1fall, sales2fall, sales3fall, sales4fall sales0winter, sales1winter, sales2winter, sales3winter, sales4winter sales0spring, sales1spring, sales2spring, sales3spring, sales4spring sales0summer, sales1summer, sales2summer, sales3summer, sales4summer第9章1b

34、3 char blanks = ""或 char blanks30 = ""1Ad John Quincy Join1John Adams1a. if (strcmp(name1,name2) = 0) printf("Names match.n"); else printf("Names do not match.n");b. if (strcmp(w1,w2) < 0) strcpy(word, w1); else strcpy(word, w2);c. int i, len; len = strlen(s1); i = strlen(s2); if (i < len) len = i; /* Since len will never be larger than STR_LEN, no need to check for overflow of strings. */ for (i = 0; i < len && s1i = s2i; +i) mtchi = s1i