重读The C programming Lanuage 笔记三：简单计算器程序

  1 //简单计算器
  2 
  3 #include <stdio.h>
  4 #include <stdlib.h>
  5 #include <ctype.h>
  6 #include <string.h>
  7 #include <math.h>
  8 
  9 #define MAXOP 100       //max size of operand or operator
 10 #define NUMBER ‘0‘      //sign of a number was found
 11 #define NAME ‘n‘        //sign of a mathfunc was found
 12 #define MAXVAL 100      //max size of the stack
 13 #define BUFSIZE 100     //buf io
 14 
 15 int sp = 0;              //the postion of the stack
 16 double val[MAXVAL];     //  the stack
 17 double variable[26];     //26 a~z
 18 void clear(void);
 19 
 20 int getop(char[]);     //get operand or operator
 21 void push(double);
 22 double pop(void);
 23 void mathfnc(char[]);  //math function
 24 
 25 int main()
 26 {
 27     int type,var = 0;
 28     double op1, op2, v;
 29     char s[MAXOP];
 30 
 31     for (int i = 0; i<26; i++)
 32     {
 33         variable[i] = 0.0;
 34 
 35         while ((type = getop(s)) != EOF)
 36             switch (type)
 37         {
 38 
 39             case NUMBER:
 40                 push(atof(s));
 41                 break;
 42             case NAME:
 43                 mathfnc(s);
 44                 break;
 45             case ‘+‘:
 46                 push(pop() + pop());
 47                 break;
 48             case ‘-‘:
 49                 op2 = pop(); push(pop() - op2);
 50                 break;
 51             case ‘*‘:
 52                 push(pop()*pop());
 53                 break;
 54             case ‘/‘:
 55                 op2 = pop();
 56                 if (op2 != 0.0)
 57                     push(pop() / op2);
 58                 else printf("error:zero divisor");
 59                 break;
 60             case ‘%‘:
 61                 op2 = pop();
 62                 if (op2 != 0.0)
 63                     push(fmod(pop(), op2));
 64                 else printf("error:zero divisor");
 65                 break;
 66                 //对栈操作 打印栈顶元素，交换栈值，清空栈
 67             case ‘?‘: // printf top of element of the stack
 68                 op2 = pop();
 69                 printf("\t%.8g", op2);
 70                 push(op2);
 71                 break;
 72             case ‘c‘:    //clear the stack
 73                 clear();
 74                 break;
 75             case ‘s‘:    //swap the top of the stack
 76                 op1 = pop();
 77                 op2 = pop();
 78                 push(op2);
 79                 push(op1);
 80                 break;
 81             case ‘\n‘:
 82                 v = pop();
 83                 printf("\t%8f\n", v);
 84                 break;
 85             case ‘=‘:
 86                 pop();
 87                 if (var >= ‘A‘ && var <= ‘Z‘)
 88                     variable[var - ‘A‘] = pop();
 89                 else
 90                     printf("error: no variable name");
 91                 break;
 92             default:
 93                 if (type >= ‘A‘ || type <= ‘Z‘)
 94                     push(variable[type - ‘A‘]);
 95                 else if (type == ‘v‘)
 96                     push(v);
 97                 else
 98                     printf("error:unknown command %s\n", s);
 99                 break;
100 
101         }
102         var = type;
103     }
104     return 0;
105 }
106 
107 //出入栈函数
108 void push(double f)
109 {
110     if (sp<MAXVAL)
111         val[sp++] = f;
112     else
113         printf("error :stack full,can push %s\n", f);
114 }
115 
116 double pop(void)
117 {
118     if (sp > 0)
119         return val[--sp];
120     else
121         printf("error:stack empty");
122     return 0.0;
123 }
124 
125 
126 //getop()函数
127 int getch(void);
128 void ungetch(int);
129 
130 int getop(char s[])
131 {
132     int c, i;
133 
134     while ((s[0] = c = getch()) == ‘ ‘ || c == ‘\t‘)
135         ;
136     s[1] = ‘\0‘;
137     i = 0;
138 
139     if (islower(c))      //commend or NAME
140     {
141         while (islower(s[++i] = c = getch()))
142             ;
143         s[i] = ‘\0‘;
144         if (c != EOF)
145             ungetch(c);
146         if (strlen(s)>1)
147             return NAME;
148         else
149             return c;
150     }
151     if (!isdigit(c) && c != ‘.‘ && c != ‘-‘)
152         return c;           //not a number
153     if (c == ‘-‘)
154         if (isdigit(c = getch()) || c == ‘.‘)
155             s[++i] = c;     //negetive number
156         else
157         {
158             if (c != EOF)
159                 ungetch(c);
160             return ‘-‘;     //minus sign
161         }
162     if (isdigit(c))
163         while (isdigit(s[++i] = c = getch()))
164             ;
165     if (c == ‘.‘)
166         while (isdigit(s[++i] = c = getch()))
167             ;
168     s[i] = ‘\0‘;
169     if (c != EOF)
170         ungetch(c);
171     return NUMBER;
172 }
173 
174 //getch().ungetch()
175 int  buf[BUFSIZE];      //不是char *buf 以能正确处理 EOF
176 int bufp = 0;           //buf中下一个空闲位置
177 
178 int getch(void)         //取一个字符（可能是压回的字符）
179 {
180     return (buf > 0) ? buf[--bufp] : getchar();
181 }
182 
183 void ungetchar(int c)   //把字符压回输入中
184 {
185     if (bufp >= BUFSIZE)
186         printf("error :too many characters");
187     else
188         buf[bufp++] = c;
189 }
190 
191 
192 void clear(void)
193 //reverse polish calculator
194 {
195     sp = 0;
196 }
197 
198 void mathfnc(char s[])
199 {
200     double op2;
201     if (strcmp(s, "sin") == 0)
202         push(sin(pop()));
203     else if (strcmp(s, "cos") == 0)
204         push(cos(pop()));
205     else if (strcmp(s, "exp") == 0)
206         push(exp(pop()));
207     else if (strcmp(s, "pow") == 0)
208     {
209         op2 = pop();
210         push(pow(pop(), op2));
211     }
212     else printf("error:%s not supported\n", s);
213 }
214 
215 
216 //push string back onto the input
217 void ungets(char s[])
218 {
219     int len = strlen(s);
220     void ungetch(int);
221 
222     while (len > 0)
223         ungetch(s[--len]);
224 }

 1 int getline(char line[], int lim)
 2 {
 3     int c, i;
 4     for (i = 0; i < MAXLINE - 1 && c != ‘\n‘; ++i)
 5     {
 6         line[i] = c;
 7         if (c == ‘\n‘)
 8         {
 9             line[i] = c;
10             ++i;
11         }
12         line[i] = ‘\0‘;
13     }
14         return 0;
15 
16 }