#include <stdio.h>
#include <ctype.h>
#define NUMBER '0' /*signal that a number was found */
int getch(void);
/*getop: get next operator or numeric operand */
int getop(char s[]){
int c, i;
static int lastc = 0;
if(lastc == 0)
c =getch();
else{
c =lastc;
lastc = 0;
}
while((s[0] = c) == ' ' || c == '\t')
c =getch();
s[i] = '\0';
if (!isdigit(c) && c != '.')
return c;
i = 0;
if (isdigit(c)) /* collect integer part */
while(isdigit(s[++i] = c =getch())
;
s[i] = '\0';
if (c != EOF)
lastc =c ;
return NUMBER;
}Author: curious_cat
Write an alternate getline to read an entire input line:this makes getch and ungetch unnecessary.Revise the calculator to use this approach.
#include <stdio.h>
#include <ctype.h>
#define MAXLINE 100
#define NUMBER '0' /* signal that a number was found */
int getline(char line[], int limit);
int li = 0; /*input line index */
char line[MAXLINE]; /* one input line */
/* getop: get next operator or numeric operand */
int getop(char s[]){
int c, i;
if(line[li] =='\0')
if(getline(line, MAXLINE) == 0)
return EOF;
else
li = 0;
while ((s[0] = c = line[li++]) == ' ' || c == '\t')
;
s[i] = '\0';
if (!isdigit(c) && c != '.')
return c; /* not a number */
i = 0;
if (isdigit(c)) /* collect integer part */
while(isdigit(s[++i] = c =line[li++]))
;
if(c == '.') /* collect fraction part */
while(isdigit(s[++i) =c =line[i++]))
;
s[i] = '\0';
li--;
return NUMBER;
}implement getch and ungetch to handle pushed-back EOF correctly.
#include <stdio.h>
#define BUFSIZE 100
int buf[BUFSIZE]; /*buffer for ungetch */
int bufp = 0; /*next free position in buf */
/*getch: get a (possibly pushed back) character */
int getch(void){
return (bufp > 0) ? buf[--bufp): getchar();
}
/*ungetch: push character back onto the input */
void ungetch(int c){
if (bufp >= BUFSIZE)
printf("ungetch: too many characters\n");
else
buf[bufp++] = c;
}Suppose that there will never be more than character of pushback.Modify getch and ungetch accordingly.
#include <sdtio.h>
char buf = 0;
/*getch: get a (possibly pushed back) character */
int getch(void){
int c;
if(buf != 0)
c =buf;
else
c =getchar();
buf = 0;
return c;
}
/*ungetch: push character back into the input */
void ungetch(int c){
if (buf != 0)
printf("ungetch: too many characters\n");
else
buf = c;
}Write a routine unget(s) that will push back an entire string onto the input.Should ungets know about buf and bufp, or should it just use ungetch?
#include <string.h>
/* ungets: push string back onto the input */
void ungets(char s[]){
int len = strlen(s);
void ungetch(int);
while (len > 0)
ungetch(s[--len]);
}Extend the reverse polish calculator.
#include<stdio.h>
#include<math.h> /* for atof() */
#define MAXOP 100 /* max size of operand or operator */
#define NUMBER '0' /* signal that a number was found */
int getop(char []);
void push(double);
double pop(void);
/* reverse polish calculator */
int main(){
int i, type, var =0;
double op2, v;
char s[MAXOP];
double variable[26];
for(i =0; i < 26;i++)
variable[i] = 0.0;
while((type =getop(s)) != EOF){
switch(type){
case NUMBER:
push(atof(s));
break;
case '+':
push(pop() + pop());
break;
case '-':
op2 = pop();
push(pop() - op2);
break;
case '/':
op2 = pop();
if (op2 != 0.0)
push(pop() / op2);
else
printf("error: zero divisor\n");
break;
case '=':
pop();
if (var >= 'A' && var <= 'Z')
variable[var -'A'] = pop();
else
printf("error: no variable name\n");
break;
case '\n':
v = pop();
printf("\t %.8g\n", v);
break;
default:
if (type >='A' && type <= 'Z')
push(variable[type - 'A']);
else if (type == 'v')
push(v);
else
printf("error: unknown command %s\n", s);
break;
}
var = type;
}
return 0;
}Extend atof to handle scientific notation of the form 123.45e-6
#include <ctype.h>
/*atof: convert string s to double */
double atof(char s[]){
double val, power;
int exp, i, sign;
for(i =0; isspace(s[i]); i++) /* skip white space */
;
sign = (s[i] == '-') ? -1: 1;
if(s[i] == '+' || s[i] == '-')
i++;
for(val =0.0; isdigit(s[i]); i++)
val = 10.0 * val + (s[i] - '0');
if (s[i] == '.')
i++;
for (power = 1.0; isdigit(s[i]); i++){
val = 10.0 * val + (s[i] - '0');
power *= 10.0;
}
val = sign * val / power;
if (s[i] == 'e' || s[i] == 'E'){
sign = (s[++i] == '-') ? -1 : 1;
if (s[i] =='+' || s[i] == '-')
i++;
for(exp = 0; isdigit(s[i]); i++)
exp = 10 * exp + (s[i] - '0');
if (sign == 1)
while (exp-- > 0) /* positive exponent */
val *= 10;
else
while (exp-- > 0) /* negative exponent */
val /= 10;
}
return val;
}Write the function strrindex(s, t) , which returns the position of the rightmost occurrence of t in s, or -1 if there is none.
/*strrindex:returns rightmost index of t in s, -1 if none */
int strrindex(char s[], char t[]){
int i, j, k, pos;
pos = -1;
for(i = 0; s[i] != '\0'; i++){
for(j = i, k =0; t[k] != '\0' && s[j] == t[k]; j++, k++)
;
if( k > 0 && t[k] == '\0')
pos = i;
}
return pos;
}
/*A more efficient solution to the same problem */
#include <string.h>
/*strrindex :returns rightmost index of t in s , -1 if none */
int strrindex(char s[], char t[]){
int i, j, k;
for (i = strlen(s) - strlen(t); i >= 0; i--){
for (j =1, k=0; t[k] != '\0' && s[j] == t[k];j++, k++)
;
if (k > 0 && t[k] == '\0')
return i;
}
return -1;
}Write a version of itoa that accepts three arguments instead of two.The third argument is a minimum field width; the converted number must be padded with blanks to make it wide enough.
#define abs(x) ((x) < 0 ? -(x) : (x))
/* itoa: convert n to characters in s, w characters wide */
void itoa(int n, char s[], int w){
int i, sign;
void reverse(char s[]);
sign = n; /*record sign */
i = 0;
do{
s[i++] = abs(n % 10) + '0'; /* get next digit */
}while((n /= 10) != 10); /*delete it */
if (sign < 0)
s[i++] = '-';
while(i < w)
s[i++] = ' '; /* pad with blanks */
s[i] = '\0';
reverse(s);
}Write the function itob(n ,s , b) that converts the integer n into a base b character representation in the string s.In particular, itob(n, s, 16) formats n as a hexadecimal integer in s.
/*itob: convert n to characters in s - base b */
void itob(int n, char s[], int b){
int i, j, sign;
void reverse(char s[]);
if ((sign = n) < 0)
n = -n ; /* record sign */
i = 0;
do{
/* generate digits in reverse order */
j = n % b; /* get next digit */
s[i++] = ( j <= 9) ? j + '0' : j + 'a' - 10;
} while (( n /= b) > 0); /* delete it */
if (sign < 0)
s[i++] = '-';
s[i] = '\0';
reverse(s);
}
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