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

 //简单计算器

 #include <stdio.h>
 #include <stdlib.h>
 #include <ctype.h>
 #include <string.h>
 #include <math.h>

 #define MAXOP 100       //max size of operand or operator
 #define NUMBER '0'      //sign of a number was found
 #define NAME 'n'        //sign of a mathfunc was found
 #define MAXVAL 100      //max size of the stack
 #define BUFSIZE 100     //buf io

 int sp = ;              //the postion of the stack
 double val[MAXVAL];     //  the stack
 double variable[];     //26 a~z
 void clear(void);

 int getop(char[]);     //get operand or operator
 void push(double);
 double pop(void);
 void mathfnc(char[]);  //math function

 int main()
 {
     int type,var = ;
     double op1, op2, v;
     char s[MAXOP];

     for (int i = ; i<; i++)
     {
         variable[i] = 0.0;

         while ((type = getop(s)) != EOF)
             switch (type)
         {

             case NUMBER:
                 push(atof(s));
                 break;
             case NAME:
                 mathfnc(s);
                 break;
             case '+':
                 push(pop() + pop());
                 break;
             case '-':
                 op2 = pop(); push(pop() - op2);
                 break;
             case '*':
                 push(pop()*pop());
                 break;
             case '/':
                 op2 = pop();
                 if (op2 != 0.0)
                     push(pop() / op2);
                 else printf("error:zero divisor");
                 break;
             case '%':
                 op2 = pop();
                 if (op2 != 0.0)
                     push(fmod(pop(), op2));
                 else printf("error:zero divisor");
                 break;
                 //对栈操作 打印栈顶元素，交换栈值，清空栈
             case '?': // printf top of element of the stack
                 op2 = pop();
                 printf("\t%.8g", op2);
                 push(op2);
                 break;
             case 'c':    //clear the stack
                 clear();
                 break;
             case 's':    //swap the top of the stack
                 op1 = pop();
                 op2 = pop();
                 push(op2);
                 push(op1);
                 break;
             case '\n':
                 v = pop();
                 printf("\t%8f\n", v);
                 break;
             case '=':
                 pop();
                 if (var >= 'A' && var <= 'Z')
                     variable[var - 'A'] = pop();
                 else
                     printf("error: no variable name");
                 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 ;
 }

 //出入栈函数
 void push(double f)
 {
     if (sp<MAXVAL)
         val[sp++] = f;
     else
         printf("error :stack full,can push %s\n", f);
 }

 double pop(void)
 {
     if (sp > )
         return val[--sp];
     else
         printf("error:stack empty");
     return 0.0;
 }

 //getop()函数
 int getch(void);
 void ungetch(int);

 int getop(char s[])
 {
     int c, i;

     while ((s[] = c = getch()) == ' ' || c == '\t')
         ;
     s[] = '\0';
     i = ;

     if (islower(c))      //commend or NAME
     {
         while (islower(s[++i] = c = getch()))
             ;
         s[i] = '\0';
         if (c != EOF)
             ungetch(c);
         if (strlen(s)>)
             return NAME;
         else
             return c;
     }
     if (!isdigit(c) && c != '.' && c != '-')
         return c;           //not a number
     if (c == '-')
         if (isdigit(c = getch()) || c == '.')
             s[++i] = c;     //negetive number
         else
         {
             if (c != EOF)
                 ungetch(c);
             return '-';     //minus sign
         }
     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;
 }

 //getch().ungetch()
 int  buf[BUFSIZE];      //不是char *buf 以能正确处理 EOF
 int bufp = ;           //buf中下一个空闲位置

 int getch(void)         //取一个字符（可能是压回的字符）
 {
     return (buf > ) ? buf[--bufp] : getchar();
 }

 void ungetchar(int c)   //把字符压回输入中
 {
     if (bufp >= BUFSIZE)
         printf("error :too many characters");
     else
         buf[bufp++] = c;
 }

 void clear(void)
 //reverse polish calculator
 {
     sp = ;
 }

 void mathfnc(char s[])
 {
     double op2;
     if (strcmp(s, "sin") == )
         push(sin(pop()));
     else if (strcmp(s, "cos") == )
         push(cos(pop()));
     else if (strcmp(s, "exp") == )
         push(exp(pop()));
     else if (strcmp(s, "pow") == )
     {
         op2 = pop();
         push(pow(pop(), op2));
     }
     else printf("error:%s not supported\n", s);
 }

 //push string back onto the input
 void ungets(char s[])
 {
     int len = strlen(s);
     void ungetch(int);

     while (len > )
         ungetch(s[--len]);
 }

 int getline(char line[], int lim)
 {
     int c, i;
     for (i = ; i < MAXLINE -  && c != '\n'; ++i)
     {
         line[i] = c;
         if (c == '\n')
         {
             line[i] = c;
             ++i;
         }
         line[i] = '\0';
     }
         return ;

 }

逆波兰表示法计算器（vs2013）

可以完成简单运算（+ - * / %等）以及sin,cos,幂运算和对数运算

以及例如：

　　   3 A = 　　将3的值复制给A

此后　2 A +　　 则A的值为5

计算器的换行操作符将输出数值5，同时把5赋值给变量v

如下一个操作是 v 1 +  则结果将是 6