用C语言编写一个简单的计算器

Question

要求以实现十进制浮点数的连续四则运算，并能够对括号的嵌套进行正确解析，能够解析“减号”和“负号”

Answer 1

花了一个多小时纯手工帮你写好了，你只需要加上你的学号姓名即可交差，如果满意，还请赏个采纳吧，谢谢

// Q7937325.cpp : 定义控制台应用程序的入口点。
//
//#include "stdafx.h"

#include <stdio.h>
#include <stdlib.h>
#include <conio.h>
#include <string.h>

/*****************************************/
/* 将数字字符转化成浮点型实数进行计算 */
/*****************************************/
double readnum(char f[], int*i)
{
    double x = 0.0;
    int k = 0;
    while (f[*i] >= '0'&&f[*i] <= '9')
    {
        x = x * 10 + (f[*i] - '0');
        (*i)++;
    }
    if (f[*i] == '.')
    {
        (*i)++;
        while (f[*i] >= '0'&&f[*i] <= '9')
        {
            x = x * 10 + (f[*i] - '0');
            (*i)++;
            k++;
        }
    }
    while (k-->0)
    {
        x = x / 10.0;
    }
    return (x);
}
/*******************************/
/* 计算后缀表达式的值 */
/*******************************/
double evalpost(char f[])
{
    double obst[10];
    int top = 0;
    int i = 0;
    double x1, x2;
    while (f[i] != '=')
    {
        if (f[i] >= '0'&&f[i] <= '9')
        {
            obst[top] = readnum(f, &i); top++;
        }
        else if (f[i] == ' ')
            i++;
        else if (f[i] == '+')
        {
            x1 = obst[--top];
            x2 = obst[--top];
            obst[top] = x1 + x2;
            i++;
            top++;
        }
        else if (f[i] == '-')
        {
            x1 = obst[--top];
            x2 = obst[--top];
            obst[top] = x2 - x1;
            i++;
            top++;
        }
        else if (f[i] == '*')
        {
            x1 = obst[--top];
            x2 = obst[--top];
            obst[top] = x1*x2;
            i++;
            top++;
        }
        else if (f[i] == '*')
        {
            x1 = obst[--top];
            x2 = obst[--top];
            obst[top] = x1*x2;
            i++;
            top++;
        }
        else if (f[i] == '/')
        {
            x1 = obst[--top];
            x2 = obst[--top];
            obst[top] = x2 / x1;
            i++;
            top++;
        }
    }
    return obst[0];
}
/***********************************/
/* 判断字符是否为操作字符 */
/***********************************/
int is_operation(char op)
{
    switch (op)
    {
    case'^':
    case'K':
    case'+':
    case'-':
    case'*':
    case'/': return 1;
    default: return 0;
    }
}
/*****************************/
/* 判断字符的优先级 */
/*****************************/
int priority(char op)
{
    switch (op)
    {
    case'=': return -1;
    case'(': return 0;
    case'+':
    case'-': return 1;
    case'*':
    case'/': return 2;
    default: return -1;
    }
}
/******************************/
/* 中缀表达式转化成后缀表达式*/
/******************************/
void postfix(char e[], char f[])
{
    int i = 0, j = 0, k = 0;
    char opst[100];
    int top = 0;
    opst[0] = '='; top++;
    while (e[i] != '=')
    {
        if ((e[i] >= '0'&&e[i] <= '9') || e[i] == '.')
            f[j++] = e[i];
        else if (e[i] == '(')
        {
            opst[top] = e[i]; top++;
        }
        else if (e[i] == ')')
        {
            k = top - 1;
            while (opst[k] != '(') { f[j++] = opst[--top]; k = top - 1; }
            top--;
        }
        else if (is_operation(e[i]))
        {
            f[j++] = ' ';
            while (priority(opst[top - 1]) >= priority(e[i]))
                f[j++] = opst[--top];
            opst[top] = e[i];
            top++;
        }
        i++;
    }
    while (top) f[j++] = opst[--top]; f[j] = '\0';
}
void print_1(char val[])
{
    system("cls");
    printf("| ______________________________________ | \n");
    printf("| |                                    | | \n");
    printf("| |        欢迎使用多功能计算器        | | 本计算器能够进行10进制数 \n");
    printf("| |____________________________________| | +，/，*，/，()的连续运算\n");
    printf("| |____________________________________| | \n");
    printf("| |       设计人：这里放你的学号姓名   | | 输入所要经计算的表达式 \n");
    printf("| |____________________________________| | 如：a*b/(c-d)\n");
    printf("|                                        | \n");
    printf("|   %s", val);
    for (int i = 0; i < 35 - strlen(val); i++) 
        printf(" ");
    printf("  | \n");
    printf("| ___  ___  ___  ___  ___  ___  ___  ___ | \n");
    printf("|  ________ ________ ________ ________   | \n");
    printf("|  |      | |      | |      | |      |   | \n");
    printf("|  |  ⑨  | |  ⑧  | |  ⑦  | |  ×  |   | \n");
    printf("|  |______| |______| |______| |______|   | \n");
    printf("|  ________ ________ ________ ________   | \n");
    printf("|  |      | |      | |      | |      |   | \n");
    printf("|  |  ⑥  | |  ⑤  | |  ④  | |  －  |   | \n");
    printf("|  |______| |______| |______| |______|   | \n");
    printf("|  ________ ________ ________ ________   | \n");
    printf("|  |      | |      | |      | |      |   | \n");
    printf("|  |  ③  | |  ②  | |  ①  | |  ＋  |   | 继续计算/退出？1/0\n");
    printf("|  |______| |______| |______| |______|   | \n");
    printf("|  ________ ________ ________ ________   | \n");
    printf("|  |      | |      | |      | |      |   | \n");
    printf("|  |  〇  | |   ＝ | |  AC  | |  ÷  |   | \n");
    printf("|  |______| |______| |______| |______|   | \n");
    printf("|________________________________________| \n");
    getch();
    system("cls");
}
void printf_2()
{
    system("cls");
    printf("\n\n\n\n\n\n\n\n\t\t\t ##############################\n");
    printf("\t\t\t # #\n");
    printf("\t\t\t #----------谢谢使用----------#\n");
    printf("\t\t\t # #\n");
    printf("\t\t\t ##############################\n");
    printf("\t\t\t --制作\n ");
}
/****************/
/* 转化 */
/****************/
void zhuanhuan(char g[], char e[])
{
    int k, i, j = 0;
    for (i = 0; g[i] != '='; i++)
    {
        k = i + 1;
        if (g[i] == '('&&g[k] == '-')
        {
            e[j++] = g[i];
            e[j++] = '0';
        }
        else e[j++] = g[i];
    }
    e[j] = '=';
}
int main()
{
    int wei;
    char e[100], f[100], g[100];
    int sign; int flag;
    print_1("0");
    do
    {
        printf("输入所要经计算的表达式(如：a*b/(c-d)=)：\n");
        scanf("%s", g);
        zhuanhuan(g, e);
        postfix(e, f);
        printf("输出保留几位小数：\n");
        scanf("%d", &wei);
        char resstr[100];
        sprintf(resstr, "%s%.*lf", g, wei, evalpost(f));
        print_1(resstr);
        while (1)
        {
            flag = 3;
            printf("继续计算/退出？1/0?");
            sign = getch();
            printf("%c\n", sign);
            switch (sign)
            {
            case '1':flag = 1; getch(); break;
            case '0':flag = 0; getch(); break;
            default: printf("非法输入，请重新输入：\n");
            }
            if (flag == 1 || flag == 0)break;
        }
    } while (flag == 1);
    printf_2();
    return 0;
}

开发截图

以下是程序运行效果