C语言程序设计现代方法(第二版)习题答案

Chapter 2An swers to Selected Exercises2. [was #2] (a) The program contains one directive (#inelude) and four statements (three calls of printf and one return).(b)Parkinson's Law:Work expands so as to fill the timeavailable for its completi on.3. [was #4]#i nclude <stdio.h>int main(v oid){int height = 8, le ngth = 12, width = 10, volume;volume = height * len gth * width;printf("Dimensions: %dx%dx%d\n", length, width, height);prin tf("Volume (cubic in ches): %d\n", volume);printf("Dimensional weight (pounds): %d\n", (volume + 165) / 166);return 0;}4. [was #6] Here's one possible program:#i nclude <stdio.h>int main(v oid){int i, j, k;float x, y, z;prin tf("Value of i: %d\n", i);prin tf("Value of j: %d\n", j);prin tf("Value of k: %d\n", k);prin tf("Value of x: %g\n", x);prin tf("Value of y: %g\n", y);printf("Value of z: %g\n", z);return 0;}Whe n compiled using GCC and the n executed, this program produced the follow ing output:Value of i: 5618848Value of j: 0Value of k: 6844404Value of x: 3.98979e-34Value of y: 9.59105e-39Value of z: 9.59105e-39The values printed depend on many factors, so the chanee that you'll get exactly these nu mbers is small.5. [was #10] (a) is not legal because 100_bottles beg ins with a digit.8. [was #12] There are 14 tokens: a, =, (, 3, *, q, -, p, *, p, ), /, 3,and ;.An swers to Selected Programmi ng Projects4. [was #8; modified]#i nclude <stdio.h>int main(v oid){float orig in al_am ount, amoun t_with_tax;prin tf("E nter an amoun t:");sca nf("%f", &orig in al_am oun t);amoun t_with_tax = orig in al_am ount * 1.05f;prin tf("With tax added: $%.2f\n", amou nt_with_tax);return 0;}The amount_with_tax variable is unnecessary. If we remove it, the programis slightly shorter: #i nclude <stdio.h>时磊诫呎…}int main(v oid){float orig in al_am ount;prin tf("E nter an amoun t:");sea nf("%f", &orig in al_am oun t);printf("With tax added: $%.2f\n", original_amount * 1.05f);return 0;}Chapter 3An swers to Selected Exercises2. [was #2](a) printf(%8.1e", x);(b) printf("%10.6e", x);(e) pri ntf("%-8.3f", x);(d) printf("%6.0f", x);5. [was #8] The values of x, i, and y will be 12.3, 45, and .6,An swers to Selected Programmi ng Projects1. [was #4; modified]#i nclude <stdio.h>int main(v oid){int mon th, day, year;prin tf("E nter a date (mm/dd/yyyy):");sca nf("%d/%d/%d", &mon th, & day, &year);prin tf("You en tered the date %d%.2d%.2d\n", year, mon th, day);return 0;3. [was #6; modified]respectively.时Sr忖呎#i nclude <stdio.h> int main(v oid){int prefix, group, publisher, item, check_digit;printf("Enter ISBN:");sca nf("%d-%d-%d-%d-%d", & prefix, & group, & publisher, & tem, &check_digit);prin tf("GS1 prefix: %d\n", prefix);prin tf("Group ide ntifier: %d\n", group);prin tf("Publisher code: %d\n", publisher);printf("Item number: %d\n", item);prin tf("Check digit: %d\n", check_digit);/* The five printf calls can be combined as follows:printf("GS1 prefix: %d\nGroup identifier: %d\nPublishercode: %d\nItem number: %d\nCheck digit: %d\n",prefix, group, publisher, item, check_digit);*/return 0;}Chapter 4An swers to Selected Exercises2. [was #2] Not in C89. Suppose that i is 9 and j is 7. The value of (-i)/j could be either —1 or —2, depending on the implementation. On the other hand, the value of -(i/j) is always —1, regardless of the implementation.In C99, on the other hand, the value of (-i)/j must be equal to the value of -(i/j).9. [was #6](a) 63 8(b) 3 2 1(c) 2 -1 3(d) 0 0 0------------------- 时磊忖呎…... ..... ... ...13. [was #8] The expressi on ++i is equivale nt to (i += 1). The value of both expressi ons is i after the in creme nt has bee n performed.An swers to Selected Programmi ng Projects2. [was #4]#i nclude <stdio.h>int main(v oid){int n;prin tf("E nter a three-digit nu mber:");sca nf("%d", &n);printf("The reversal is: %d%d%d\n", n % 10, (n / 10) % 10, n / 100);return 0;}Chapter 5An swers to Selected Exercises2. [was #2](a) 1(b) 1(c) 1(d) 14. [was #4] (i > j) - (i < j)6. [was #12] Yes, the statement is legal. When n is equal to 5, it does nothing, since 5 is not equal to - 9.10. [was #16] The output ison etwosince there are no break stateme nts after the cases.An swers to Selected Programmi ng Projects 2. [was #6]时Sr忖呎#i nclude <stdio.h> int main(v oid){int hours, minu tes;prin tf("E nter a 24-hour time:"); sca nf("%d:%d", & hours, &min utes);prin tf("Equivale nt 12-hour time:");if (hours == 0)printf("12:%.2d AM\n", minutes);else if (hours < 12)prin tf("%d:%.2d AM\n", hours, mi nutes); else if (hours == 12)prin tf("%d:%.2d PM\n", hours, mi nutes); elseprin tf("%d:%.2d PM\n", hours - 12, minu tes);return 0;}4. [was #8; modified]#i nclude <stdio.h> int main(v oid){int speed;prin tf("E nter a wi nd speed in kno ts:"); scanf("%d", &speed);if (speed < 1)prin tf("Calm\n"); else if (speed <= 3)prin tf("Light air\n"); else if (speed <= 27)prin tf("Breeze\n"); else if (speed <= 47)prin tf("Gale\n"); else if (speed <= 63)prin tf("Storm\n"); else------------------- 时磊忖呎…... ..... ... ...prin tf("Hurrica ne\n");return 0;}6. [was #10]#i nclude <stdio.h>int main(v oid){in t check_digit, d, i1, i2, i3, i4, i5, j1, j2, j3, j4, j5,first_sum, sec on d_sum, total;prin tf("E nter the first (sin gle) digit:");scanf("%1d", &d);prin tf("E nter first group of five digits:");scanf("%1d%1d%1d%1d%1d", &i1, &i2, &3, &i4, &i5);prin tf("E nter sec ond group of five digits:");scanf("%1d%1d%1d%1d%1d", &j1, &j2, &j3, &j4, &j5);prin tf("E nter the last (sin gle) digit:");sca nf("%1d", & check_digit);first_sum = d + i2 + i4 + j1 + j3 + j5;sec on d_sum = i1 + i3 + i5 + j2 + j4;total = 3 * first_sum + sec on d_sum;if (check_digit == 9 - ((total - 1) % 10))prin tf("VALID\n");elseprin tf("NOT VALID\n");return 0;}10. [was #14]#i nclude <stdio.h>int main(v oid){int grade;prin tf("E nter nu merical grade:");scanf("%d", &grade);时磊忖呎…if (grade < 0 || grade > 100) {printf("lllegal grade'n");return 0;}switch (grade / 10) {case 10:case 9: prin tf("Letter grade: A\n");break;case 8: prin tf("Letter grade: B\n");break;case 7: prin tf("Letter grade: C\n");break;case 6: prin tf("Letter grade: D\n");break;case 5:case 4:case 3:case 2:case 1:case 0: prin tf("Letter grade: F\n");break;}return 0;}Chapter 6An swers to Selected Exercises 4. [was #10] (c) is not equivale nt to (a) and (b), because i is in creme nted before the loop body is executed.10. [was #12] Co nsider the followi ng while loop:while (…){con ti nue;}The equivale nt code using goto would have the follow ing appeara nee:------------------- 时磊忖呎…... ..... ... ... while (…){goto loop_e nd;loop_e nd: ; /* n ull stateme nt */} _12. [was #14]for (d = 2; d * d <= n; d++)if (n % d == 0)break;The if statement that follows the loop will need to be modified as well:if (d * d <= n)printf("%d is divisible by %d\n", n, d);elseprin tf("%d is prime\n", n);14. [was #16] The problem is the semicoIon at the end of the first line.If we remove it, the statement is now correct:if (n % 2 == 0)printf("n is eve n\n");An swers to Selected Programmi ng Projects2. [was #2]#i nclude <stdio.h>int main(v oid){int m, n, rema in der;prin tf("E nter two in tegers:");sca nf("%d%d", &m, &n);while (n != 0) {rema in der = m % n;m = n;n = rema in der;}时磊忖呎…prin tf("Greatest com mon divisor: %d\n", m);return 0;}4. [was #4]#i nclude <stdio.h>int main(v oid){float commissi on, value;prin tf("E nter value of trade:");sca nf("%f", &value);while (value != 0.0f) {if (value < 2500.00f)commissio n = 30.00f + .017f * value;else if (value < 6250.00f)commission = 56.00f + .0066f * value;else if (value < 20000.00f)commissio n = 76.00f + .0034f * value;else if (value < 50000.00f)commissio n = 100.00f + .0022f * value;else if (value < 500000.00f)commissio n = 155.00f + .0011f * value;elsecommissio n = 255.00f + .0009f * value;if (commissio n < 39.00f)commissi on = 39.00f;prin tf("Commissio n: $%.2f\n\n", commissio n);prin tf("E nter value of trade:");scan f("%f", &value);}return 0;}6. [was #6] #i nclude <stdio.h> int main (void) {int i, n;时磊忖呎…prin tf("E nter limit on maximum square:"); sea nf("%d", &n);for (i = 2; i * i <= n; i += 2) prin tf("%d\n", i * i);return 0;}8. [was #8]#i nclude <stdio.h> int main(v oid){int i, n, start_day;prin tf("E nter nu mber of days in mon th:");sea nf("%d", &n);printf("Enter starting day of the week (1=Sun, 7=Sat):"); sea nf("%d", &start_day);/* print any leadi ng "bla nk dates" */ for (i = 1; i < start_day; i++)prin tf("");/* now print the cale ndar */ for (i = 1; i <= n; i++) {prin tf("%3d", i);if ((start_day + i - 1) % 7 == 0) prin tf("\n");}return 0;}Chapter 7An swers to Selected Exercises裁磊诫呎3. [was #4] (b) is not legal.4. [was #6] (d) is illegal, since printf requires a string, not a character, as its first argume nt.10. [was #14] unsigned int, because the (int) cast applies only to j, not j * k.12. [was #16] The value of i is conv erted to float and added to f, the nthe result is con verted to double and stored in d.14. [was #18] No. Converting f to int will fail if the value stored inf exceeds the largest value of type int.An swers to Selected Programmi ng Projects1. [was #2] short int values are usually stored in 16 bits, causing failure at 182. int and long int values are usually stored in 32 bits, with failure occurring at 46341.2. [was #8]#i nclude <stdio.h>int main(v oid){int i, n;char ch;prin tf("This program prints a table of squares.\n");prin tf("E nter nu mber of en tries in table:");sca nf("%d", &n);ch = getchar();/* dispose of n ew-l ine character follow ing nu mber of en tries *//* could simply be getchar(); */for (i = 1; i <= n; i++) {prin tf("%10d%10d\n", i, i * i);if (i % 24 == 0) {prin tf("Press En ter to continu e...");ch = getchar(); /* or simply getchar(); */}}return 0;5. [was #10]#in elude <ctype.h>#i nclude <stdio.h>int main(v oid){int sum = 0;char ch;prin tf("E nter a word:");while ((ch = getchar()) != '\n')switch (toupper(ch)) {case 'D': case 'G':sum += 2; break;case 'B': case C: case 'M': case 'P': sum += 3; break;case 'F': case 'H': case 'V': case W: case 'Y': sum += 4; break;case 'K':sum += 5; break;case 'J': case 'X':sum += 8; break;case 'Q': case 'Z':sum += 10; break;default:sum++; break;}printf("Scrabble value: %d\n", sum);return 0;}6. [was #12]#i nclude <stdio.h>int main(v oid){prin tf("Size of int: %d\n", (int) sizeof(i nt));prin tf("Size of short: %d\n", (int) sizeof(short));prin tf("Size of long: %d\n", (int) sizeof(l on g));prin tf("Size of float: %d\n", (int) sizeof(float));prin tf("Size of double: %d\n", (int) sizeof(double));prin tf("Size of long double: %d\n", (int) sizeof(l ong double));return 0;}Since the type of a sizeof expressi on may vary from one impleme ntati on to ano ther, it's n ecessary in C89 to cast sizeof expressi ons to a known type before printing them. The sizes of the basic types are small numbers,so it's safe to cast them to int. (In general, however, it's best to cast sizeof expressions to unsigned long and print them using %lu.) In C99, we can avoid the cast by using the %zu conv ersi on specificatio n.Chapter 8An swers to Selected Exercises1. [was #4] The problem with sizeof(a) / sizeof( t) is that it can't easily be checked for correct ness by some one readi ng the program. (The readerwould have to locate the declaration of a and makesure that its elements have type t.)2. [was #8] To use a digit d (in character form) as a subscript into thearray a, we would write a[d-'0']. This assumes that digits haveconsecutive codes in the underlying character set, which is true of ASCII and other popular character sets.7. [was #10]const int segme nts[10][7] = {{1, 1, 1, 1, 1, 1},{0, 1, 1},{1, 1,0, 1, 1,0, 1},{1, 1, 1, 1,0, 0, 1},{0, 1, 1,0, 0, 1, 1},{1, 0, 1, 1,0, 1, 1},{1, 0, 1, 1, 1, 1, 1},{1, 1, 1},{1, 1, 1, 1, 1, 1, 1},{1, 1, 1, 1,0, 1, 1}};An swers to Selected Programmi ng Projects2. [was #2]#i nclude <stdio.h>int main(v oid){int digit_cou nt[1O] = {0};int digit;long n;prin tf("E nter a nu mber:");sca nf("%ld", &n);while (n > 0) {digit = n % 10;digit_co un t[digit]++;n /= 10;}printf ("Digit:");for (digit = 0; digit <= 9; digit++)prin tf("%3d", digit);prin tf("\nO ccurre nces:");for (digit = 0; digit <= 9; digit++)prin tf("%3d", digit_cou nt[digit]);prin tf("\n");return 0;}5. [was #6]#i nclude <stdio.h>#define NUM_RATES ((int) (sizeof(value) / sizeof(value[0]))) #defi ne INITIAL_BALANCE 100.00int main(v oid){int i, low_rate, mon th, nu m_years, year;double value[5];printf("Enter interest rate:");scanf("%d", &low_rate);prin tf("E nter nu mber of years:");sea nf("%d", &nu m_years);prin tf("\nYears");for (i = 0; i < NUM_RATES; i++) {prin tf("%6d%%", low_rate + i);value[i] = INITIAL_BALANCE;}prin tf("\n");for (year = 1; year <= nu m_years; year++) {prin tf("%3d ", year);for (i = 0; i < NUM_RATES; i++) {for (month = 1; month <= 12; mon th++)value[i] += ((double) (low_rate + i) / 12) / 100.0 * value[i];prin tf("%7.2f", value[i]);}prin tf("\n");}return 0;}8. [was #12]#i nclude <stdio.h>#defi ne NUM_QUIZZES 5#defi ne NUM_STUDENTS 5int main(v oid){in t grades[NUM_STUDENTS][NUM_QUIZZES];int high, low, quiz, stude nt, total;for (student = 0; student < NUM_STUDENTS; student++) {prin tf("E nter grades for stude nt %d: ", stude nt + 1);for (quiz = 0; quiz < NUM_QUIZZES; quiz++)scan f("%d", & grades[stude nt][quiz]);}prin tf("\nStude nt Total Average\n");for (student = 0; student < NUM_STUDENTS; student++) {prin tf("%4d ", stude nt + 1);total = 0;for (quiz = 0; quiz < NUM_QUIZZES; quiz++) total += grades[stude nt][quiz];printf("%3d %3d\n", total, total / NUM_QUIZZES);}prin tf("\nQuiz Average High Low\n");for (quiz = 0; quiz < NUM_QUIZZES; quiz++) {prin tf("%3d ", quiz + 1);total = 0;high = 0;low = 100;for (stude nt = 0; stude nt < NUM_STUDENTS; stude nt++) { total += grades[stude nt][quiz];if (grades[stude nt][quiz] > high)high = grades[stude nt][quiz];if (grades[stude nt][quiz] < low)low = grades[stude nt][quiz];}prin tf("%3d %3d %3d\n", total / NUM_STUDENTS, high, low);}return 0;}Chapter 9An swers to Selected Exercises2. [was #2]int check(i nt x, int y, int n){return (x >= 0 && x <= n - 1 && y >= 0 && y <= n - 1);}4. [was #4]int day_of_year(i nt mon th, int day, int year){int num_days[] = {31,28, 31,30, 31,30, 31,31,30, 31, 30, 31};int day_c ount = 0, i;for (i = 1; i < mon th; i++)day_co unt += nu m_days[i-1];/* adjust for leap years, assu ming they are divisible by 4 */if (year % 4 == 0 && mon th > 2)day_co un t++;retur n day_co unt + day;}Using the expression year %4 == 0 to test for leap years is not completely correct. Centuries are special cases: if a year is a multiple of 100, then it must also be a multiple of 400 in order to be a leap year. The correct test isyear % 4 == 0 && (year % 100 != 0 || year % 400 == 0)6. [was #6; modified]int digit(i nt n, int k){int i;for (i = 1; i < k; i++)n /= 10;return n % 10;}8. [was #8] (a) and (b) are valid prototypes. (c) is illegal, since itdoes n't specify the type of the parameter. (d) in correctly specifies that f returns an int value in C89; in C99, omitting the return type is illegal.10. [was #10](a)int largest(i nt a[], int n){int i, max = a[0];for (i = 1; i < n; i++)if (a[i] > max)max = a[i];return max;}(b)int average© nt a[], int n){int i, avg = 0;for (i = 0; i < n; i++)avg += a[i];retur n avg / n;}(c)int nu m_positive(i nt a[], int n){ _int i, count = 0;for (i = 0; i < n; i++) if (a[i] > 0) coun t++;retur n count;}15. [was #12; modified]double media n( double x, double y, double z) { double result;if (x <= y)if (y <= z) result = y;else if (x <= z) result = z;else result = x;else {if (z <= y) result = y;else if (x <= z) result = x; else result = z;}retur n result; } 17. [was #14]------------------- 时需5说int fact(i nt n){int i, result = 1;for (i = 2; i <= n; i++) result *= i;return result;}19. [was #16] The following program tests the pb function: #i nclude <stdio.h>void pb(i nt n);int main(v oid){int n;prin tf("E nter a nu mber:");sca nf("%d", &n);prin tf("Output of pb:");pb( n);prin tf("\n");return 0;}void pb(i nt n){if (n != 0) {pb(n / 2);putchar('0' + n % 2);}}pb prints the binary represe ntati on of the argume nt n, assu ming that n is greatercodes in than 0. (We also assume that digits have consecutivethe un derly ing character set.) For example:En ter a nu mber: 53Output of pb: 110101A trace of pb's execution would look like this:pb(53) finds that 53 is not equal to 0, so it callspb(26), which finds that 26 is not equal to 0, so it callspb(13), which finds that 13 is not equal to 0, so it callspb(6), which finds that 6 is not equal to 0, so it calls pb(3), which finds that 3 is not equal to 0, so it callspb(1), which finds that 1 is not equal to 0, so it callspb(0), which finds that 0 is equal to 0, so it retur ns, caus ingpb(1) to print 1 and retur n, caus ing pb(3) to print 1and retur n, caus ingpb(6) to print 0 and retur n, caus ingpb(13) to print 1 and retur n, caus ingpb(26) to print 0 and retur n, caus ingpb(53) to print 1 and retur n.Chapter 10An swers to Selected Exercises1. [was #2] (a) a, b, and c are visible.(b) a, and d are visible.(c) a, d, and e are visible.(d) a and f are visible.An swers to Selected Programmi ng Projects3. [was #4]#in clude <stdbool.h> /* C99 only */#i nclude <stdio.h>#i nclude <stdlib.h>#defi ne NUM_CARDS 5#defi ne RANK 0#defi ne SUIT 1/* exter nal variables */in t han d[NUM_CARDS][2];/* 0 1*/ bool straight, flush, four, three; int pairs; /* can be 0, 1, or 2 */ /* prototypes */ void read_cards(void); void an alyze_ha nd(void);void prin t_result(void);/*********************************************************** main: Calls read_cards, an alyze_ha nd, and prin t_result* repeatedly. ***********************************************************/int main(v oid){for (;;) {read_cards();an alyze_ha nd();prin t_result();} _} /*********************************************************** read_cards: Reads the cards into the external variable* hand; checks for bad cards and duplicate ** cards. ***********************************************************/void read_cards(void){char ch, ran k_ch, suit_ch;int i, ra nk, suit;bool bad_card, duplicate_card;int cards_read = 0;while (cards_read < NUM_CARDS) { bad_card = false;prin tf("E nter a card:");ran k_ch = getchar();switch (ran k_ch) {case 'O': exit(EXIT_SUCCESS);case 2: rank = 0; break;case 3: rank = 1; break;case '4': rank = 2; break;case '5': rank = 3; break;case '6': rank = 4; break;case '7': rank = 5; break;case '8': rank = 6; break;case '9': rank = 7; break;case 't': case 'T': rank = 8; break;case 'j': case 'J': rank = 9; break;case 'q': case 'Q': rank = 10; break;case 'k': case 'K': rank = 11; break;case 'a': case 'A': rank = 12; break;default: bad_card = true;} _suit_ch = getchar();switch (suit_ch) {case 'c': case 'C': suit = 0; break;case 'd': case 'D': suit = 1; break;case 'h': case 'H': suit = 2; break;case 's': case 'S': suit = 3; break;default: bad_card = true;} _while ((ch = getchar()) != '\n')if (ch != ' ') bad_card = true;if (bad_card) {prin tf("Bad card; igno red.\n");con ti nue;}duplicate_card = false;for (i = 0; i < cards_read; i++)if (ha nd[i][RANK] == rank && han d[i][SUIT] == suit) {prin tf("Duplicate card; igno red.\n");duplicate_card = true;break;}if (!duplicate_card) {ha nd[cards_read][RANK] = rank; ha nd[cards_read][SUIT] = suit;cards_read++;} _}} /*********************************************************** an alyze_ha nd: Determ ines whether the hand contains a* straight, a flush, four-of-a-k ind, ** an d/or three-of-a-k ind; determ ines the ** nu mber of pairs; stores the results into ** the exter nal variables straight, flush, ** four, three, and pairs. ***********************************************************/void an alyze_ha nd(void){ _int rank, suit, card, pass, run;straight = true; flush = true; four = false; three = false; pairs = 0;/* sort cards by rank */for (pass = 1; pass < NUM_CARDS; pass++)for (card = 0; card < NUM_CARDS - pass; card++) { rank = han d[card][RANK];suit = han d[card][SUIT];if (ha nd[card+1][RANK] < rank) {han d[card][RANK] = ha nd[card+1][RANK];han d[card][SUIT] = ha nd[card+1][SUIT];han d[card+1][RANK] = rank;han d[card+1][SUIT] = suit;}}/* check for flush */ suit = han d[0][SUIT];for (card = 1; card < NUM_CARDS; card++) if (ha nd[card][SUIT] != suit) flush = false;/* check for straight */for (card = 0; card < NUM_CARDS - 1; card++)if (ha nd[card][RANK] + 1 != han d[card+1][RANK])straight = false;/* check for 4-of-a-k ind, 3-of-a-k ind, and pairs bylook ing for "r uns" of cards with ide ntical ranks */card = 0;while (card < NUM_CARDS) {rank = han d[card][RANK];run = 0;do {run++;card++;} while (card < NUM_CARDS && han d[card][RANK] == rank);switch (run) {case 2: pairs++; break;case 3: three = true; break;case 4: four = true; break;}}}* print_result: Prints the classification of the hand,* based on the values of the exter nal* variables straight, flush, four, three, ** and pairs. *********************************************************void prin t_result(void){ _if (straight && flush) pri ntf("Straight flush");else if (four) prin tf("Four of a ki nd");else if (three &&pairs == 1) prin tf("Full house");prin tf("Flush");prin tf("Straight");prin tf("Three of a ki nd");else if (pairs == 2) prin tf("Two pairs");else if (pairs == 1) prin tf("Pair");else if (flush) else if (straight) else if (three)--------- 时器Sr亦..... . .... ... ...printf("High card");elseprin tf("\n\n");}5. [was #6]#in clude <stdbool.h> /* C99 only */#i nclude <stdio.h>#i nclude <stdlib.h> #defi ne NUM_RANKS 13#defi ne NUM_SUITS 4#defi ne NUM_CARDS 5/* exter nal variables */int num _in_ra nk[NUM_RANKS];int num _in_suit[NUM_SUITS];bool straight, flush, four, three;int pairs; /* can be 0, 1, or 2 *//* prototypes */void read_cards(void);void an alyze_ha nd(void);void prin t_result(void);/*********************************************************** main: Calls read_cards, an alyze_ha nd, and prin t_result* repeatedly. ***********************************************************/ int main(v oid){for (;;) {read_cards();an alyze_ha nd();prin t_result();} _}/*********************************************************** read_cards: Reads the cards into the exter nal* variables num_in_rank and num_in _suit;* checks for bad cards and duplicate cards.**********************************************************------------------- 时磊5说----- - ---- -------void read_cards(void){ _bool card_exists[NUM_RANKS][NUM_SUITS];char ch, ran k_ch, suit_ch;int rank, suit;bool bad_card;int cards_read = 0;for (ra nk = 0; rank < NUM_RANKS; ran k++) { num_in_ran k[ra nk] = 0;for (suit = 0; suit < NUM_SUITS; suit++)card_exists[ra nk][suit] = false;} _for (suit = 0; suit < NUM_SUITS; suit++)num_in _suit[suit] = 0;while (cards_read < NUM_CARDS) {bad_card = false;prin tf("E nter a card:");ran k_ch = getchar();switch (ran k_ch) {case 'O': exit(EXIT_SUCCESS);case 2: rank = 0;break;case 3: rank = 1;break;case '4': rank = 2;break;case '5': rank = 3;break;case '6': rank = 4;break;case '7': rank = 5;break;case '8': rank = 6;break;case '9': rank = 7;break;case 't': case 'T': rank = 8; break;case 'j': case 'J': rank = 9; break;case 'q': case 'Q': rank = 10; break;case 'k': case 'K': rank = 11; break;case 'a': case 'A': rank = 12; break;default: bad_card = true;} _suit_ch = getchar();switch (suit_ch) {case 'c': case 'C': suit = 0; break;。