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 = 0;              //the postion of the stack
double val[MAXVAL];     //  the stack
double variable[26];     //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 = 0;
    double op1, op2, v;
    char s[MAXOP];

    for (int i = 0; i<26; 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("	%.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 '
':
                v = pop();
                printf("	%8f
", 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
", s);
                break;

        }
        var = type;
    }
    return 0;
}

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

double pop(void)
{
    if (sp > 0)
        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[0] = c = getch()) == ' ' || c == '	')
        ;
    s[1] = '';
    i = 0;

    if (islower(c))      //commend or NAME
    {
        while (islower(s[++i] = c = getch()))
            ;
        s[i] = '';
        if (c != EOF)
            ungetch(c);
        if (strlen(s)>1)
            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] = '';
    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 = 0;           //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 > 0) ? 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 = 0;
}

void mathfnc(char s[])
{
    double op2;
    if (strcmp(s, "sin") == 0)
        push(sin(pop()));
    else if (strcmp(s, "cos") == 0)
        push(cos(pop()));
    else if (strcmp(s, "exp") == 0)
        push(exp(pop()));
    else if (strcmp(s, "pow") == 0)
    {
        op2 = pop();
        push(pow(pop(), op2));
    }
    else printf("error:%s not supported
", s);
}


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

    while (len > 0)
        ungetch(s[--len]);
}

int getline(char line[], int lim)
{
    int c, i;
    for (i = 0; i < MAXLINE - 1 && c != '
'; ++i)
    {
        line[i] = c;
        if (c == '
')
        {
            line[i] = c;
            ++i;
        }
        line[i] = '';
    }
        return 0;

}

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