001
09.06.2004, 21:03 Uhr
mike
Pinguinhüpfer
(Operator)
|
Hi!
C++ Lösung habe ich loider keine, vielleicht hilft der ne C Lösung:
| C++: |
/*Polish notation 1.0
written by Dennis M. Ritchie
Copy & Paste by mike :-) */
#include <stdio.h>
#include <stdlib.h>
#include <ctype.h>
#define MAXOP 100
#define NUMBER '0'
int getop(char []);
void push(double);
void ungetch(int c);
int getch(void);
double pop(void);
int main() {
int type;
double op2;
char s[MAXOP];
while ((type = getop(s)) != EOF) {
switch (type) {
case NUMBER: push(atof(s)); break;
case '+': push(pop() + pop()); 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 '\n': printf("\t%.8g\n", pop()); break;
default: printf("error: unknown command %s\n", s); break;
}
}
return 0;
}
#define MAXVAL 100
int sp = 0;
double val[MAXVAL];
void push(double f) {
if (sp < MAXVAL)
val[sp++] = f;
else printf("error: stack full, can't push %g\n", f);
}
double pop(void) {
if (sp > 0)
return val[--sp];
else {
printf("error: stack empty\n");
return 0.0;
}
}
int getop(char s[]) {
int i, c;
while ((s[0] = c = getch()) == ' ' || c == '\t') ;
s[1] = '\0';
if (!isdigit(c) && c != '.')
return c;
i = 0; if (isdigit(c))
while (isdigit(s[++i] = c = getch())) ;
if (c == '.')
while (isdigit(s[++i] = c = getch())) ;
s[i] = '\0';
if (c != EOF)
ungetch(c);
return NUMBER;
}
#define BUFSIZE 100
char buf[BUFSIZE];
int bufp = 0;
int getch(void) {
return (bufp > 0) ? buf[--bufp] : getchar();
}
void ungetch(int c) {
if (bufp >= BUFSIZE) printf("ungetch: too many characters\n");
else buf[bufp++] = c;
}
|
mfg
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