Exercises 4-6
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#include <stdio.h>#include <stdlib.h> /* for atof() */#include <math.h>#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);void clear(void);/* reverse Polish calculator */main(){ int type,var,i; double op1; 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 '*': 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 '%': op2 = pop(); if (op2 != 0.0) push(fmod(pop(),op2)); else printf("error: zero divisor\n"); break; case '?': op2=pop(); printf("\t%.8g\n", op2); push(op2); break; case 'c': clear(); break; case 'd': op2=pop(); push(op2); push(op2); break; case 's': op1=pop(); op2=pop(); push(op1); push(op2); break; case '\n': v=pop(); printf("\t%.8g\n", v); break; case '=': pop();/*弹出变量符号*/ if(var>='A'&&var<='Z') variable[var-'A']=pop();/*将值赋给=前面的变量*/ else printf("error:no variable name"); break; default: if(type>='A'&&type<='Z') push(variable[type-'A']); /*save variable*/ else if(type=='v') push(v); else printf("error: unknown command %s\n", s); break; } var=type; } return 0;}#define MAXVAL 100 /* maximum depth of val stack */int sp = 0; /* next free stack position */double val[MAXVAL]; /* value stack *//* push: push f onto value stack */void push(double f){ if (sp < MAXVAL) val[sp++] = f; else printf("error: stack full, can't push %g\n", f);}/* pop: pop and return top value from stack */double pop(void){ if (sp > 0) return val[--sp]; else { printf("error: stack empty\n"); return 0.0; }}void clear(){ sp=0;}#include <ctype.h>int getch(void);void ungetch(int);/* getop: get next character or numeric operand */int getop(char s[]){ int i, c; while ((s[0] = c = getch()) == ' ' || c == '\t')/*ignore white space*/ ; s[1] = '\0'; if (!isdigit(c) && c != '.'&&c!='-') return c; /* not a number */ i = 0; if(c=='-') { if(isdigit(c=getch())||c=='.') s[++i]=c; else { if(c!=EOF) ungetch(c); return '-'; } } if (isdigit(c)) /* collect integer part */ while (isdigit(s[++i] = c = getch())) ; if (c == '.') /* collect fraction part */ while (isdigit(s[++i] = c = getch())) ; s[i] = '\0'; //这一步可以将可以忽略掉上面操作数中的非数字,因为它并不属于操作数的一部分,比如123+,+是操作符,故忽略掉。 if (c != EOF) ungetch(c); //将非数字字符保存起来,下次判断,如果是操作符则返回,并做相关运算,否则出错了。 return NUMBER;}/*getch和ungetch函数的配对使用是为了解决去除操作数中含有操作符的问题,因为当收集操作数时,结尾的符号是一个非数字(因为遇到非数字循环才结束),可以用ungetch先存起来,等到下一次操作再用getch读出,*/#define BUFSIZE 100char buf[BUFSIZE]; /* buffer for ungetch */int bufp = 0; /* next free position in buf */int getch(void) /* get a (possibly pushedback)character */{ return (bufp > 0) ? buf[--bufp]: getchar();}void ungetch(int c) /* push character back on input */{ if (bufp >= BUFSIZE) printf("ungetch: too many characters\n"); else buf[bufp++] = c;}