discount = DISCOUNT_RATE * cost_total; else discount = ;
final_total = cost_total + shipping - discount; printf(\
printf(\lb_artichokes, price_artichokes, cost_artichokes);
printf(\lb_beets, price_beets, cost_beets); printf(\carrots at $%.2f per pound: $%.2f\\n\price_carrots, cost_carrots); printf(\
vegetables: $%.2f\\n\ printf(\printf(\
printf(\return 0; }
Chapter 8 Programming Exercises
PE 8--‐1
/* Programming Exercise 8-1 */ #include <>
int main(void) { int ch; int ct = 0; while ((ch = getchar()) != EOF) ct++;
printf(\ return 0; }
PE 8--‐3
/* Programming Exercise 8-3 */
/* Using eliminates need to assume consecutive coding */ #include <> #include <> int main(void) { int ch; unsigned long uct = 0; unsigned long lct = 0; unsigned long oct = 0;
while ((ch = getchar()) != EOF) if (isupper(ch)) uct++; else if (islower(ch)) lct++; else oct++;
printf(\
printf(\other characters read\\n\ return 0; }
/* or you could use if
(ch >= 'A' && ch <= 'Z') uct++;
else if (ch >= 'a' && ch <= 'z') lct++; else oct++;
*/
PE 8--‐5
/* Programming Exercise 8-5 */
/* -- an improved number-guesser */
/* but relies upon truthful, correct responses */ #include <> #include <> int
main(void) { int high = 100; int low = 1; int guess = (high + low) / 2; char response;
printf(\
printf(\h if it is high, and with an l if it is low.\\n\number %d?\\n\
while ((response = getchar()) != 'y') /* get response */ {
if (response == '\\n') continue;
if (response != 'h' && response != 'l') {
printf(\printf(\ }
if (response == 'h') high = guess - 1; else if (response == 'l') low = guess + 1; guess = (high + low) / 2;
printf(\ }
printf(\return 0; }
PE 8--‐7
/* Programming Exercise 8-7 */ #include <> #include <> #include <>
#define BASEPAY1 */ switch (response) {
case 'a': pay = BASEPAY1; break; case 'b': pay = BASEPAY2; break; case 'c': pay = BASEPAY3; break; case 'd': pay = BASEPAY4; break;
default : printf(\menu();
continue; f; taxes: $%.2f; net: $%.2f\\n\gross, taxes, net); menu(); } printf(\ return 0; }
void menu(void) {
printf(\
\
printf(\ \
printf(\BASEPAY2);
printf(\BASEPAY4); printf(\
printf(\ \}
int getfirst(void) { int ch;
ch = getchar(); while (isspace(ch)) ch = getchar(); while (getchar() != '\\n') continue; return ch; }
Chapter 9 Programming Exercises
PE 9--‐1
/* Programming Exercise 9-1 */ #include <>
double min(double, double); int main(void) {
double x, y; printf(\numbers (q to quit): \(scanf(\
{ printf(\
min(x,y)); printf(\\ }
printf(\ return 0; }
double min(double a, double b) {
return a < b ? a : b; }
/* alternative implementation double min(double a, double b) { if (a < b) return a; else return b; } */
PE 9--‐3
/* Programming Exercise 9-3 */ #include <>
void chLineRow(char ch, int c, int r); int main(void)
{ char ch; int col, row;
printf(\while ( (ch = getchar()) != '#') { if (ch == '\\n') continue;
printf(\if (scanf(\chLineRow(ch, col, row);
printf(\ }
printf(\ return 0; }
n\ }
printf(\ return 0; }
void larger_of(double *p1, double *p2) { if (*p1 > *p2) *p2 = *p1; else *p1 = *p2; }
Enter q\&x, &n) == 2)
{ xpow = power(x,n); /* function call */ printf(\xpow); printf(\quit.\\n\
} printf(\bye!\\n\
} double power(double a, int b) /* function definition */
{ double pow = 1; int i; if (b == 0) { if (a == 0)
printf(\pow = ; } else if (a == 0) pow = ; else if (b > 0) for(i = 1; i <= b; i++) pow *= a; else /* b < 0 */ pow = / power(a, - b);
return pow; /* return the value of pow */ }
PE 9--‐10
/* Programming Exercise 9-10 */ #include <> void to_base_n(int x, int base); int
main(void) { int number; int b; int count; printf(\quit):\\n\1)
{ printf(\\&b))== 1
&& (b < 2 || b > 10)) {
printf(\ } if
(count != 1) break;
printf(\to_base_n(number, b); putchar('\\n');
printf(\ }
printf(\return 0; }
void to_base_n(int x, int base) /* recursive function */ { int r; r = x % base; if (x >= base) to_base_n(x / base, base); putchar('0' + r); return; }
Chapter 10
Programming Exercises
PE 10--‐1
/* Programming Exercise 10-1 */ #include <>
#define MONTHS 12 f inches.\\n\\n\AVERAGES:\\n\\n\
printf(\printf(\
for (month = 0; month < MONTHS; month++)
{ /* for each month, sum rainfall over years */ for (year = 0, subtot =0; year < YRS; year++) subtot += *(*(rain + year) + month); printf(\ }
printf(\return 0; }
PE 10--‐3
/* Programming Exercise 10-3 */ #include <>
#define LEN 10
int max_arr(const int ar[], int n); void show_arr(const int ar[], int n);
C Primer Plus第6版编程练习答案
