C程序设计语言 （第二版） 练习 4-9

练习 4-9 以上介绍的getch与ungetch函数不能正确地处理压回的EOF。考虑压回EOF时应该如何处理？请实现你的设计方案。

注意：代码在win32控制台运行，在不同的IDE环境下，有部分可能需要变更。

IDE工具：Visual Studio 2010

 

代码块：

#include <stdio.h>
#include <stdlib.h>
#include <ctype.h>
#include <math.h>
#include <string.h>#define MAXOP 100
#define NUMBER '0'
#define MAXVAL 100
#define BUFSIZE 100
#define VAR '1'int sp = 0;
double val[MAXVAL];int buf[BUFSIZE];                    //此处原先为char型，改为int型
int bufp = 0;double variable[26];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;}
}void printTop(void){if(sp > 0){printf("Top: %g\n", val[sp-1]);}else{printf("Error! Stack Empty!\n");}
}void topCopy(void){if(sp > 0 || sp < MAXVAL){val[sp++] = val[sp-1];}else if(sp <= 0){printf("Error! Stack Empty!\n");}else{printf("Error! Stack Full!\n");}
}void swapTop(void){double temp;if(sp >= 2){temp = val[sp-1];val[sp-1] = val[sp-2];val[sp-2] = temp;}else{printf("Can't Swap Top Number!\n");}
}void emptyStack(void){for(int i = sp - 1; i >= 0; i--){val[i] = 0;}sp = 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;}
}int getop(char s[]){int i, c;while((s[0] = c = getch()) == ' ' || c == '\t');s[1] = '\0';if(c == 's'){int next1 = getch();int next2 = getch();if(next1 == 'i' && next2 == 'n'){return c;}}if(c == 'e'){int next1 = getch();int next2 = getch();if(next1 == 'x' && next2 == 'p'){return c;}}if(c == 'p'){int next1 = getch();int next2 = getch();if(next1 == 'o' && next2 == 'w'){return c;}}if(c >= 'a' && c <= 'z'){int next = getch();if(next == ' '){ungetch(next);return VAR;}}if(c == '-'){int next = getch();if(!isdigit(next) && next != '.'){ungetch(next);return c;}s[1] = c = next;i = 1;}else{i = 0;if(!isdigit(c) && c != '.'){return c;}}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;
}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 '%':op2 = pop();push((int)pop() % (int)op2);break;case 's':op2 = pop();push(sin(op2));break;case 'e':op2 = pop();push(exp(op2));break;case 'p':op2 = pop();push(pow(pop(), op2));break;case VAR:variable[s[0] - 'a'] = pop();push(variable[s[0] - 'a']);printf("%c = %lf\n", s[0], variable[s[0] - 'a']);break;case '\n':printf("\t%.8g\n", pop());break;default:printf("Error! Unknown Command %s\n", s);break;}}system("pause");return 0;
}