怎么用C语言设计一个简单计算器

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怎么用C语言设计一个简单计算器,第1张

怎么用C语言设计一个简单计算器 #include<iostream>

#include<math.h>

using namespace std

main ()

{

char k

double s,g

k='-'

cout<<"求绝对值请按A求平方根请按B加减乘除清分别按+-*/" <<endl

cin>>g

while (1)

{

cin>>k

if (k=='=')

break

if (k=='A')

{

g=fabs (g)

break

}

if (k=='B')

{

g=sqrt (g)

break

}

cin>>s

if (k=='+')

g+=s

if (k=='-')

g-=s

if (k=='*')

g*=s

if (k=='/')

{

if (s==0)

cout<<"wrong"

else

g/=s

}

}

cout<<g

}

#include<stdio.h>void add(int a,int b,int c) { c=a+b printf("%d\t",c) printf("\n")} void minus(int a,int b,int c) { c=a-b printf("%d\t",c) printf("\n")} void multiplication(int a,int b,int c) { c=a*b printf("%d\t",c) printf("\n")} void div(int a,int b,int c) { c=(float)a/(float)b printf("%f\t",c) printf("\n")} main() { int a,b,c char p puts("input A:\n") scanf("%d",&a) puts("input B:\n") scanf("%d",&b) puts("input operation:\n") getchar() p=getchar() if(p=='+') add(a,b,c)else if(p=='-') minus(a,b,c)elseif(p=='*') multiplication(a,b,c)else if(p=='/') div(a,b,c)else puts("没有注册这个运算子号\n")} 以上是设计的一个简易计算器。可以进行相应的加减乘除。 简介:

C语言是一种计算机程式设计语言,它既具有高阶语言的特点,又具有组合语言的特点。它由美国贝尔研究所的D.M.Ritchie于1972年推出,1978年后,C语言已先后被移植到大、中、小及微型机上,它可以作为工作系统设计语言,编写系统应用程式,也可以作为应用程式设计语言,编写不依赖计算机硬体的应用程式。它的应用范围广泛,具备很强的资料处理能力,不仅仅是在软体开发上,而且各类科研都需要用到C语言,适于编写系统软体,三维,二维图形和动画,具体应用比如微控制器以及嵌入式系统开发。

求用C语言设计一个简单计算器

应该不难,就是按钮的讯息传递处理函式等,

用C语言怎么设计一个简单计算器?

#include<stdio.h>

void add(int a,int b,int c)

{

c=a+b

printf("%d\t",c)

printf("\n")

}

void minus(int a,int b,int c)

{

c=a-b

printf("%d\t",c)

printf("\n")

}

void multiplication(int a,int b,int c)

{

c=a*b

printf("%d\t",c)

printf("\n")

}

void div(int a,int b,int c)

{

c=(float)a/(float)b

printf("%f\t",c)

printf("\n")

}

main()

{

int a,b,c

char p

puts("input A:\n")

scanf("%d",&a)

puts("input B:\n")

scanf("%d",&b)

puts("input operation:\n")

getchar()

p=getchar()

if(p=='+') add(a,b,c)else

if(p=='-') minus(a,b,c)else

if(p=='*') multiplication(a,b,c)else

if(p=='/') div(a,b,c)else

puts("没有注册这个运算子号\n")

}

怎么用C语言程式设计一个简单计算器?

#include<<a href=":baidu./s?wd=stdio.h&tn=44039180_cpr&fenlei=mv6quAkxTZn0IZRqIHckPjm4nH00T1Y4ryfLuADkP1bYmvD3nhmz0ZwV5Hcvrjm3rH6sPfKWUMw85HfYnjn4nH6sgvPsT6KdThsqpZwYTjCEQLGCpyw9Uz4Bmy-bIi4WUvYETgN-TLwGUv3EPH6srjc4rH61" target="_blank" class="baidu-highlight">stdio.h</a>>

void main() { float x,y,zchar c

scanf("%f%c%f",&x,&c,&y)

switch ( c ) {

case '+': z=x+ybreak

case '-': z=x-ybreak

case '*': z=x*ybreak

case '/': z=( y==0 )?(0):(x/y)break

default: z=0break

}

printf("%f%c%f=%f\n",x,c,y,z)

}

C语言是一门通用计算机程式语言,应用广泛。C语言的设计目标是提供一种能以简易的方式编译、处理低阶储存器、产生少量的机器码以及不需要任何执行环境支援便能执行的程式语言。

尽管C语言提供了许多低阶处理的功能,但仍然保持着良好跨平台的特性,以一个标准规格写出的C语言程式可在许多电脑平台上进行编译,甚至包含一些嵌入式处理器(微控制器或称MCU)以及超级电脑等作业平台。

二十世纪八十年代,为了避免各开发厂商用的C语言语法产生差异,由美国国家标准局为C语言制定了一套完整的美国国家标准语法,称为ANSI C,作为C语言最初的标准。[1] 目前2011年12月8日,国际标准化组织(ISO)和国际电工委员会(IEC)释出的C11标准是C语言的第三个官方标准,也是C语言的最新标准,该标准更好的支援了汉字函式名和汉字识别符号,一定程度上实现了汉字程式设计。

用C语言编一个简单计算器

已经为你实现了加减乘除了。 main.c 简易计算器 Created by dushengduan on 15/6/29. Copyright (c) 2015年 dushengduan. All rights reserved.#include <stdio.h>double plus()double minus()double multi()double divide()int main(int argc, const char * argv[]) { insert code here...double a, b char c while(1 == 1) {printf("简易计算器\n") printf("输入第一个数:\n") scanf("%lf", &a) printf("输入运算子:\n") scanf("%s", &c) printf("输入第二个数:\n") scanf("%lf", &b) switch (c) {case '+':printf("%lf %c %lf = %lf", a, c, b, plus(a, b)) break case '-':printf("%lf %c %lf = %lf", a, c, b, minus(a, b)) break case '*':printf("%lf %c %lf = %lf", a, c, b, minus(a, b)) break case '/':printf("%lf %c %lf = %lf", a, c, b, minus(a, b)) break default:printf("运算子号输入错误!\n") break }}return 0}加法double plus(double a, double b){return a + b}double minus(double a, double b){return a - b}double multi(double a, double b){return a * b}double divide(double a, double b){return a / b}

设计一个简单计算器

#include <stdio.h>

int main(void)

{

int data1, data2

char op

printf("please input data1 op data2:")

scanf("%d %c %d", &data1 , &op , &data2)

switch (op)

{

case '+':

printf("%d + %d = %.0f\n", data1, data2, (double)data1 + (double)data2)

break

case '-':

printf("%d - %d = %.0f\n", data1, data2, (double)data1 - (double)data2)

break

case '*':

printf("%d * %d = %.0f\n", data1, data2, (double)data1 * (double)data2)

break

case '/':

if (data2 == 0)

{

printf("Error! chu shu wei 0.")

}

else

{

printf("%d / %d = %.0f\n", data1, data2, (double)data1 / (double)data2)

}

break

case '%':

if (data2 == 0)

{

printf("Error! chu shu wei 0.")

}

else

{

printf("%d %% %d=%d3\n", data1, data2, data1 % data2)

}

break

default:

printf("Unknown operator\n")

}

return 0

}如果要求是浮点数的话换一下资料型别就可以了

用c语言程式设计一个简单计算器,求其原始码

/*

2013年12月23日 12:43:46

目的:计算器的实现

*/

# include <stdio.h>

# include <ctype.h>

# include <math.h>

char get_choice(void)获取使用者输入的选项,并建立目

char get_first(void)获取使用者输入的选项,并剔除错误输入

float get_int(void)获取使用者输入的计算值

float add(void)定义加法函式

float subtraction(void)定义减法函式

float multiplication(void)定义乘法函式

float division(void)定义除法函式

float extract(void)定义开方函式

float square(void)定义平方函式

float cube(void)定义立方函式

int count = 0

int main(void)

{

char choice

printf("***欢迎使用由小钱制作的计算器***\n")

choice = get_choice()

while(choice != 'q')

{

switch(choice)

{

case 'a':

add()break

case 'b':

subtraction()break

case 'c':

multiplication()break

case 'd':

division()break

case 'e':

extract()break

case 'f':

square()break

case 'g':

cube()break

default :

printf("您输入有误,请重新输入:")break

}

fflush(stdin)

choice = get_choice()

}

printf("bye")

return 0

}

获取使用者输入的选项,并建立目录

char get_choice(void)

{

char ch

int a = 0

建立目录

printf("\n--------------------------------\n")

printf("a. 加法\t\t\tb. 减法\nc. 乘法\t\t\td. 除法\n")

printf("e. 开方\t\t\tf. 平方\ng. 立方\t\t\tq. 退出\n")

printf("--------------------------------\n")

printf("请输入你的选项:")

ch = get_first()

while(ch == ' ' || ch == '\n' || ch == '\t')

ch = get_first()

判断使用者输入的选项是否有误

while((ch<'a' || ch>'g') &&ch !='q')

{

putchar(ch)

printf(" 你输入的选项有误,请重新输入:")

ch = get_first()

}

return ch

}

获取使用者输入的选项,并剔除错误输入

char get_first(void)

{

char ch

ch = getchar()

剔除由使用者输入选项时产生的换行符

while(ch == '\n')

{

ch = getchar()

}

return ch

}

获取使用者输入的计算值

float get_int(void)

{

float input

char ch

int a

if(count == 0)

printf("亲!请输入数值:")

if(count == 1)

printf("亲!请输入第一个数值:")

if(count == 2)

printf("亲!请输入第二个数值:")

a = scanf("%f", &input)

判断使用者的输入是否为一个数值

while(a != 1)

{

剔除使用者输入错误的字元

while((ch = getchar()) != '\n')

{

putchar(ch)

printf(" 不是一个数值,请输入例如3、111.2、或者-1")

a = scanf("%f", &input)

}

}

return input

}

定义加法函式

float add(void)

{

float i, j, sum

count = 0

count = count+1

i = get_int()

count = count+1

j = get_int()

sum = i + j

printf("%.2f + %.2f = %.2f\n", i, j, sum)

return sum

}

定义减法函式

float subtraction(void)

{

float i, j, sum

count = 0

count = count+1

i = get_int()

count = count+1

j = get_int()

sum = i - j

printf("%.2f - %.2f = %.2f\n", i, j, sum)

return sum

}

定义乘法函式

float multiplication(void)

{

float i, j, sum

count = 0

count = count+1

i = get_int()

count = count+1

j = get_int()

sum = i * j

printf("%.2f * %.2f = %.2f\n", i, j, sum)

return sum

}

定义除法函式

float division(void)

{

float i, j, sum

count = 0

count = count+1

i = get_int()

count = count+1

j = get_int()

判断除数是否为0

while(j == 0)

{

printf("除数不能为0\n请重新输入!!!\n")

j = get_int()

}

sum = i / j

printf("%.2f / %.2f = %.2f\n", i, j, sum)

return sum

}

定义开方函式

float extract(void)

{

float i, sum

count = 0

i = get_int()

判断开方数是否小于0,如果小于0,则让使用者重新输入

while(i <0)

{

printf("请输入大于0的数值\n")

i = get_int()

}

sum = sqrt(i)

printf("%.2f的开方等于%.2f\n", i, sum)

return sum

}

定义平方函式

float square(void)

{

float i, sum

count = 0

i = get_int()

sum = i * i

printf("%.2f的平方等于%.2f\n", i, sum)

return sum

}

定义立方函式

float cube(void)

{

float i, sum

count = 0

i = get_int()

sum = i * i * i

printf("%f的立方等于%.3f\n", i, sum)

return sum

}

#include<<a href="https://www.baidu.com/s?wd=stdio.h&tn=44039180_cpr&fenlei=mv6quAkxTZn0IZRqIHckPjm4nH00T1Y4ryfLuADkP1bYmvD3nhmz0ZwV5Hcvrjm3rH6sPfKWUMw85HfYnjn4nH6sgvPsT6KdThsqpZwYTjCEQLGCpyw9Uz4Bmy-bIi4WUvYETgN-TLwGUv3EPH6srjc4rH61" target="_blank" class="baidu-highlight">stdio.h</a>>

void main() { float x,y,zchar c

scanf("%f%c%f",&x,&c,&y)

switch ( c ) {

case '+': z=x+ybreak

case '-': z=x-ybreak

case '*': z=x*ybreak

case '/': z=( y==0 )?(0):(x/y)break

default: z=0break

}

printf("%f%c%f=%f\n",x,c,y,z)

}

C语言是一门通用计算机编程语言,应用广泛。C语言的设计目标是提供一种能以简易的方式编译、处理低级存储器、产生少量的机器码以及不需要任何运行环境支持便能运行的编程语言。

尽管C语言提供了许多低级处理的功能,但仍然保持着良好跨平台的特性,以一个标准规格写出的C语言程序可在许多电脑平台上进行编译,甚至包含一些嵌入式处理器(单片机或称MCU)以及超级电脑等作业平台。

二十世纪八十年代,为了避免各开发厂商用的C语言语法产生差异,由美国国家标准局为C语言制定了一套完整的美国国家标准语法,称为ANSI C,作为C语言最初的标准。[1]  目前2011年12月8日,国际标准化组织(ISO)和国际电工委员会(IEC)发布的C11标准是C语言的第三个官方标准,也是C语言的最新标准,该标准更好的支持了汉字函数名和汉字标识符,一定程度上实现了汉字编程。

简单的用c语言写一个计算器程式,加减乘除能用就好 #include"stdio.h"

void main()

{

float a,b,c

char e

printf("input a,e,b\n")/*输入两个数和符号,例如3+8*/

scanf("%f%c%f",&a,&e,&b)

switch(e)

{

case '+':c=a+bbreak

case '-':c=a-bbreak

case '*':c=a*bbreak

case '/':

if(b==0.0) printf("error\n")

else c=a/bbreak

}

printf("%f%c%f=%f",a,e,b,c)

}

如何用vc++编写一个简单的(只有加减乘除)计算器程式?

先设定介面如下

加法按钮程式码

void CMy03Dlg::OnBnClickedButton1()

{

TODO:在此新增控制元件通知处理程式程式码

UpdateData(TRUE)

m_Nub3=m_Nub1+m_Nub2

UpdateData(FALSE)

}

减法按钮程式码

void CMy03Dlg::OnBnClickedButton2()

{

TODO:在此新增控制元件通知处理程式程式码

UpdateData(TRUE)

m_Nub3=m_Nub1-m_Nub2

UpdateData(FALSE)

}

乘法按钮程式码

void CMy03Dlg::OnBnClickedButton3()

{

TODO:在此新增控制元件通知处理程式程式码

UpdateData(TRUE)

m_Nub3=m_Nub1*m_Nub2

UpdateData(FALSE)

}

除法按钮程式码

void CMy03Dlg::OnBnClickedButton4()

{

TODO:在此新增控制元件通知处理程式程式码

UpdateData(TRUE)

if(m_Nub2!=0)

m_Nub3=m_Nub1 / m_Nub2

else

AfxMessageBox("被除数不能为0")

UpdateData(FALSE)

}

清除按钮程式码

void CMy03Dlg::OnBnClickedButton5()

{

TODO:在此新增控制元件通知处理程式程式码

UpdateData(TRUE)

m_Nub3=0

m_Nub1=0

m_Nub2=0

UpdateData(FALSE)

}

结束按钮程式码

void CMy03Dlg::OnBnClickedButton6()

{

TODO:在此新增控制元件通知处理程式程式码

CDialog::OnOK()

}

如果只允许在输入框中输入资料应该怎样处理?

制作托盘程式

目的:在工作列中建立一个图示,使该程式永远驻留在记忆体中。例如邮件检查程式可以作为驻留程式,一旦有邮件来了,就可以接收邮件。

Shell_NotifyIcon函式传送讯息来增加、删除、修改工作列的图示

BOOL TrayMessage(HWND hWnd, DWORD dwMessage, HICON hIcon, PSTR pszTip)

{

BOOL res

NOTIFYICONDATA tnd

tnd.cbSize = sizeof(NOTIFYICONDATA)

tnd.hWnd = hWnd

tnd.uID = IDI_ICON1

tnd.uFlags = NIF_MESSAGE|NIF_ICON|NIF_TIP

tnd.uCallbackMessage = WM_MY_TRAY_NOTIFICATION

tnd.hIcon = hIcon

lstrcpyn(tnd.szTip, pszTip, sizeof(tnd.szTip))

res = Shell_NotifyIcon(dwMessage, &tnd)dwMessage为NIM_ADD从工作列中新增图示、NIM_DELETE从工作列中删除图示、NIM_MODIFY改变工作列中图示

if (hIcon)

DestroyIcon(hIcon)

return res

}

定义一个回拨讯息:WM_MY_TRAY_NOTIFICATION

在DLG的CPP档案中,

#define WM_MY_TRAY_NOTIFICATION WM_USER+100

为对话方块新增讯息对映ON_MESSAGE(WM_MY_TRAY_NOTIFICATION,OnTrayNotification)

在DLG的标头档案中应该有

public:

long m_Nub1

float m_Nub3

CBitmapButton Button

afx_msg void OnBnClickedButton1()

long m_Nub2

afx_msg void OnBnClickedButton2()

afx_msg void OnBnClickedButton4()

afx_msg void OnBnClickedButton5()

afx_msg void OnBnClickedButton3()

afx_msg void OnBnClickedButton6()

afx_msg void OnBnClickedButton7()

afx_msg LRESULT OnTrayNotification(WPARAM wparam, LPARAM lparam)

在DLG的CPP档案中应该有

BEGIN_MESSAGE_MAP(CMailCheckDlg, CDialog)

……

ON_MESSAGE(WM_MY_TRAY_NOTIFICATION,OnTrayNotification)

……

END_MESSAGE_MAP()

并定义一个回拨讯息函式

LRESULT CMailCheckDlg::OnTrayNotification(WPARAM wparam, LPARAM lparam)

{

switch (lparam )

{

case WM_RBUTTONUP:

case WM_LBUTTONDBLCLK:修改不同的按钮处理事件,以观察图示退出效果。

ShowWindow(SW_SHOW)

TrayMessage(m_hWnd, NIM_DELETE, NULL, "")从工作列中删除图示

}

return 0

}

在对话方块视窗上新增“驻留”按钮,双击按钮新增程式码

void CMailCheckDlg::OnBnClickedButton1()

{

TODO:在此新增控制元件通知处理程式程式码

下面程式向工作列新增图示

TrayMessage(m_hWnd, NIM_ADD, NULL, "计算器程式")

TrayMessage(m_hWnd, NIM_MODIFY, m_hIcon, "计算器程式")

ShowWindow(SW_HIDE)

用MFC编写一个简单的加减乘除计算器

我有程式,加31782771群

c语言计算器程式设计包含加减乘除简单的函式运算

实用计算器之程式设计

[摘 要]多用计算器的构思及设计程式码

[关键词]多用计算器;设计

数值计算可以说是日常最频繁的工作了,WIN98提供了“计算器”软体供使用者使用,该软体可以处理一般的一步四则运算,例如:3+2、5/3等等,但在日常中使用者经常遇到多步四则运算问题,例如:3+4*5-4/2,45*34/2+18*7等等,那么该个计算器就无法胜任了,作者制作了一个实用的计算器,该计算器新增不少功能:(程式介面如图)

1.可以实现连续的四则运算

2.可以实现输入式子的显示

3.可以方便计算个人所得税

4.滑鼠、键盘均可输入资料

5.操作介面友好

6.击键可发声

构建该个计算器所需研究及解决的核心问题有如下几个:1、连乘求值?2、字元显示 3、键盘输入?4、击键发声?5、个人所得税法规,为了使大家对程式有更一步认识,现将程式码提供给读者参考:

*定义阵列及窗体变数

Dim number2(0 To 50) As Double

Dim number(0 To 50) As Double

Dim z As Integer

Dim k As Integer, r As Integer

Dim j As Integer

Dim str As String

*呼叫名为“playsound”的API函式

Private Declare Function PlaySound Lib "winmm.dll" Alias "PlaySoundA" (ByVal lpszName As String, ByVal hModule As Long, ByVal dwFlags As Long) As Long

Private Const SND_FILENAME = &H20000?

Private Const SND_ASYNC = &H1?

Private Const SND_SYNC = &H0

*判断通用过程

Sub pianduan(p As String)

r = 0

Dim i As Integer, l As Integer, h As Integer

h = 0

i = 1

If InStr(Trim$(p), "*") <>0 Then

k = k + 1

End If

If InStr(Trim$(p), "/") <>0 Then

r = r + 1

End If

End Sub

*连乘通用过程(略)

*各按钮事件过程

Private sub Command1_Click(Index As Integer)

PlaySound App.Path &"\start.wav", 0, SND_SYNC

Text1.Text = Text1.Text + Command1(Index).Caption

Text2.Text = Text2.Text + Command1(Index).Caption

Text1.SetFocus

End Sub

rivate sub Command10_Click()

PlaySound App.Path &"\start.wav", 0, SND_SYNC

str = Text3.Text

End Sub

Private sub Command11_Click()

PlaySound App.Path &"\start.wav", 0, SND_SYNC

Text3.Text = str

End Sub

rivate sub Command2_Click()

PlaySound App.Path &"\start.wav", 0, SND_SYNC

Dim totle As Double

Dim n As Integer

Call pianduan(Text1.Text)

If k >= 1 Or r >= 1 Then

Call liancheng(totle)

number2(z) = totle

If Mid$(Trim$(Text1.Text), 1, 1) = "-" Then

number2(z) = -totle

End If

k = 0: r = 0

Else

number2(z) = Val(Text1.Text)

End If

Text1.Text = ""

Text2.Text = Text2 + "+"

z = z + 1

Text1.SetFocus

End Sub

rivate sub Command3_Click()

PlaySound App.Path &"\start.wav", 0, SND_SYNC

Dim totle As Double

Dim n As Integer

Call pianduan(Text1.Text)

If k >= 1 Or r >= 1 Then

Call liancheng(totle)

number2(z) = totle

If Mid$(Trim$(Text1.Text), 1, 1) = "-" Then

number2(z) = -totle

End If

k = 0: r = 0

Else

number2(z) = Val(Text1.Text)

End If

Text1.Text = ""

Text2.Text = Text2 + "-"

Text1.Text = Text1.Text &"-"

z = z + 1

Text1.SetFocus

End Sub

Private sub Command4_Click()

PlaySound App.Path &"\start.wav", 0, SND_SYNC

Text2.Text = Text2.Text + "*"

Text1.Text = Text1.Text + "*"

Text1.SetFocus

End Sub

rivate sub Command5_Click()

PlaySound App.Path &"\start.wav", 0, SND_SYNC

Text2.Text = Text2 + "/"

Text1.Text = Text1 + "/"

Text1.SetFocus

End Sub

Private sub Command6_Click()

PlaySound App.Path &"\sound.wav", 0, SND_SYNC

Dim totle As Double

Dim n As Integer

Call pianduan(Text1.Text)

If k >= 1 Or r >= 1 Then

Call liancheng(totle)

number2(z) = totle

If Mid$(Trim$(Text1.Text), 1, 1) = "-" Then

number2(z) = -totle

End If

k = 0: r = 0

Else

number2(z) = Val(Text1.Text)

End If

Text1.Text = ""

z = z + 1

Dim dengyu As Double

Dim v As Integer

For v = 0 To 50

dengyu = dengyu + number2(v)

Next v

If dengyu <0 Then

Text3.ForeColor = &HFF&

Else

Text3.ForeColor = &HFF0000

End If

Text3.Text = dengyu

Text1.SetFocus

If Len(Text3.Text) >= 14 Then

calcresult.Show

End If

End Sub

rivate sub Command7_Click()

PlaySound App.Path &"\start.wav", 0, SND_SYNC

z = 0: k = 0: r = 0: j = 0

Dim i As Integer

For i = 0 To 50

number(i) = 0

number2(i) = 0

Next i

Text1.Text = ""

Text2.Text = ""

Text3.Text = ""

Text1.SetFocus

End Sub

rivate sub Command8_Click()

PlaySound App.Path &"\start.wav", 0, SND_SYNC

If Val(Text3.Text) = 0 Then

MsgBox "除数不能为0!"

Exit Sub

End If

Text3.Text = 1 / Val(Text3.Text)

End Sub

Private sub Command9_Click()

PlaySound App.Path &"\start.wav", 0, SND_SYNC

Text3.ForeColor = &HFF0000

Text3.Text = Val(Text3.Text) * Val(Text3.Text)

End Sub

rivate sub muninter_Click()

Dim i

i = Shell("C:\Program Files\InterExplorer\iexplore.exe", vbMaximizedFocus)

End Sub

rivate sub munmp3_Click()

Dim i

i = Shell("C:\Program Files\Windows Media Player\mplayer2", vbNormalNoFocus)

End Sub

Private sub mun *** _Click()

Dialog.Show

End Sub

rivate sub muntax_Click()

tax.Show

End Sub

rivate sub munver_Click()

ver.Show

End Sub

rivate sub notepad_Click()

Dim i

i = Shell("c:\windows\notepad", vbNormalFocus)

End Sub

Private sub Text1_KeyPress(KeyAscii As Integer)

PlaySound App.Path &"\start.wav", 0, SND_SYNC

Dim num As Integer

num = Val(KeyAscii)

If num >47 And num <58 Then

Text1.Text = Text1.Text + CStr(num - 48)

Text2.Text = Text2.Text + CStr(num - 48)

End If

If num = 46 Then

Text1.Text = Text1.Text + "."

Text2.Text = Text2.Text + "."

End If

If KeyAscii = 43 Then

Dim totle As Double

Dim n As Integer

Call pianduan(Text1.Text)

If k >= 1 Or r >= 1 Then

Call liancheng(totle)

number2(z) = totle

If Mid$(Trim$(Text1.Text), 1, 1) = "-" Then

number2(z) = -totle

End If

k = 0: r = 0

Else

number2(z) = Val(Text1.Text)

End If

Text1.Text = ""

Text2.Text = Text2 + "+"

z = z + 1

End If

If KeyAscii = 45 Then

Call pianduan(Text1.Text)

If k >= 1 Or r >= 1 Then

Call liancheng(totle)

number2(z) = totle

If Mid$(Trim$(Text1.Text), 1, 1) = "-" Then

number2(z) = -totle

End If

k = 0: r = 0

Else

number2(z) = Val(Text1.Text)

End If

Text1.Text = ""

Text2.Text = Text2 + "-"

Text1.Text = Text1.Text &"-"

z = z + 1

End If

If KeyAscii = 42 Then

Text2.Text = Text2.Text + "*"

Text1.Text = Text1.Text + "*"

End If

If KeyAscii = 47 Then

Text2.Text = Text2.Text + "/"

Text1.Text = Text1.Text + "/"

End If

If KeyAscii = vbKeyReturn Then

PlaySound App.Path &"\sound.wav", 0, SND_SYNC

Call pianduan(Text1.Text)

If k >= 1 Or r >= 1 Then

Call liancheng(totle)

number2(z) = totle

If Mid$(Trim$(Text1.Text), 1, 1) = "-" Then

number2(z) = -totle

End If

k = 0: r = 0

Else

number2(z) = Val(Text1.Text)

End If

Text1.Text = ""

z = z + 1

Dim dengyu As Double

Dim v As Integer

For v = 0 To 50

dengyu = dengyu + number2(v)

Next v

If dengyu <0 Then

Text3.ForeColor = &HFF&

Else

Text3.ForeColor = &HFF0000

End If

Text3.Text = dengyu

End If

If KeyAscii = vbKeyEscape Then

z = 0: k = 0: r = 0: j = 0

Dim i As Integer

For i = 0 To 50

number(i) = 0

number2(i) = 0

Next i

Text1.Text = ""

Text2.Text = ""

Text3.Text = ""

Text1.SetFocus

End If

If Len(Text3.Text) >= 14 Then

calcresult.Show

End If

End Sub

rivate sub Text3_Change()

tax2.Text1 = Text3.Text

End Sub

用c语言编写能运算加减乘除的计算器程式,用到栈

#include "stdio.h"

#include "string.h"

#include "ctype.h"

#include "math.h"

expression evaluate

#define iMUL 0

#define iDIV 1

#define iADD 2

#define iSUB 3

#define iCap 4

#define LtKH 5

#define RtKH 6

#define MaxSize 100

void iPush(float)

float iPop()

float StaOperand[MaxSize]

int iTop=-1

char Srcexp[MaxSize]

char Capaexp[MaxSize]

char RevPolishexp[MaxSize]

float NumCapaTab[26]

char validexp[]="*/+-()"

char NumSets[]="0123456789"

char StackSymb[MaxSize]

int operands

void NumsToCapas(char [], int , char [], float [])

int CheckExpress(char)

int PriorChar(char,char)

int GetOperator(char [], char)

void counterPolishexp(char INexp[], int slen, char Outexp[])

float CalcRevPolishexp(char [], float [], char [], int)

void main()

{

int ilen

float iResult=0.0

printf("enter a valid number string:\n")

memset(StackSymb,0,MaxSize)

memset(NumCapaTab,0,26)A--NO.1, B--NO.2, etc.

gets(Srcexp)

ilen=strlen(Srcexp)

printf("source expression:%s\n",Srcexp)

NumsToCapas(Srcexp,ilen,Capaexp,NumCapaTab)

printf("Numbers listed as follows:\n")

int i

for (i=0i<operands++i)

printf("%.2f ",NumCapaTab[i])

printf("\nCapaexp listed in the following:\n")

printf("%s\n",Capaexp)

ilen=strlen(Capaexp)

counterPolishexp(Capaexp,ilen,RevPolishexp)

printf("RevPolishexp:\n%s\n",RevPolishexp)

ilen=strlen(RevPolishexp)

iResult=CalcRevPolishexp(validexp, NumCapaTab, RevPolishexp,ilen)

printf("\ncounterPolish expression:\n%s%.6f\n",Srcexp,iResult)

}

void iPush(float value)

{

if(iTop<MaxSize) StaOperand[++iTop]=value

}

float iPop()

{

if(iTop>-1)

return StaOperand[iTop--]

return -1.0

}

void NumsToCapas(char Srcexp[], int slen, char Capaexp[], float NumCapaTab[])

{

char ch

int i, j, k, flg=0

int sign

float val=0.0,power=10.0

i=0j=0k=0

while (i<slen)

{

ch=Srcexp[i]

if (i==0)

{

sign=(ch=='-')?-1:1

if(ch=='+'||ch=='-')

{

ch=Srcexp[++i]

flg=1

}

}

if (isdigit(ch))

{

val=ch-'0'

while (isdigit(ch=Srcexp[++i]))

{

val=val*10.0+ch-'0'

}

if (ch=='.')

{

while(isdigit(ch=Srcexp[++i]))

{

val=val+(ch-'0')/power

power*=10

}

} end if

if(flg)

{

val*=sign

flg=0

}

} end if

write Capaexp array

write NO.j to array

if(val)

{

Capaexp[k++]='A'+j

Capaexp[k++]=ch

NumCapaTab[j++]=valA--0, B--1,and C, etc.

}

else

{

Capaexp[k++]=ch

}

val=0.0

power=10.0

i++

}

Capaexp[k]='\0'

operands=j

}

float CalcRevPolishexp(char validexp[], float NumCapaTab[], char RevPolishexp[], int slen)

{

float sval=0.0, op1,op2

int i, rt

char ch

recursive stack

i=0

while((ch=RevPolishexp[i]) &&i<slen)

{

switch(rt=GetOperator(validexp, ch))

{

case iMUL: op2=iPop()op1=iPop()

sval=op1*op2

iPush(sval)

break

case iDIV: op2=iPop()op1=iPop()

if(!fabs(op2))

{

printf("overflow\n")

iPush(0)

break

}

sval=op1/op2

iPush(sval)

break

case iADD: op2=iPop()op1=iPop()

sval=op1+op2

iPush(sval)

break

case iSUB: op2=iPop()op1=iPop()

sval=op1-op2

iPush(sval)

break

case iCap: iPush(NumCapaTab[ch-'A'])

break

default:

}

++i

}

while(iTop>-1)

{

sval=iPop()

}

return sval

}

int GetOperator(char validexp[],char oper)

{

int oplen,i=0

oplen=strlen(validexp)

if (!oplen) return -1

if(isalpha(oper)) return 4

while(i<oplen &&validexp[i]!=oper) ++i

if(i==oplen || i>=4) return -1

return i

}

int CheckExpress(char ch)

{

int i=0

char

while((=validexp[i]) &&ch!=) ++i

if (!)

return 0

return 1

}

int PriorChar(char curch, char stach)

{

栈外优先顺序高于(>)栈顶优先顺序时,才入栈

否则(<=),一律出栈

if (curch==stach) return 0等于时应该出栈

else if (curch=='*' || curch=='/')

{

if(stach!='*' &&stach!='/')

return 1

}

else if (curch=='+' || curch=='-')

{

if (stach=='(' || stach==')')

return 1

}

else if (curch=='(')

{

if (stach==')')

return 1

}

return 0

}

void counterPolishexp(char INexp[], int slen, char Outexp[])

{

int i, j, k,pr

char t

i=0

j=k=0

while (INexp[i]!='=' &&i<slen)

{

if (INexp[i]=='(')

StackSymb[k++]=INexp[i]

iPush(*(INexp+i))

else if(INexp[i]==')')

{

if((t=iPop())!=-1)

while((t=StackSymb[k-1])!='(')

{

Outexp[j++]=t

k--

}

k--

}

else if (CheckExpress(INexp[i])) is oparator

{

printf("operator %c k=%d\n",INexp[i],k)

while (k)

{

iPush(*(INexp+i))

if(pr=PriorChar(INexp[i],StackSymb[k-1]))

break

else

{

if ((t=iPop())!=-1)

t=StackSymb[k-1]k--

Outexp[j++]=t

}

} end while

StackSymb[k++]=INexp[i]mon process

}

else if() 变数名

{

printf("operand %c k=%d\n",INexp[i],k)

Outexp[j++]=INexp[i]

}

i++

}

while (k)

{

t=StackSymb[k-1]k--

Outexp[j++]=t

}

Outexp[j]='\0'

}

注:程式源于“百度知道”

用verilog编写一个最简单的加减乘除的计算器的程式

verilog是有加法器乘法器的。也直接识别 + - * / 符号。

module kjasdja(a,option,b,result)

input option,a,b

output result

always @(a,b,option)

begin

result_r=0结果暂存器清零

case(option)

+:result_r=a+b

-:result_r=a-b

*:result_r=a*b

/:result_r=a/b

assign result =result_r

endmodule

大概演算法就这样。写的仓促,语法可能有误。另外除法reg型别只能储存整数部分,小数通过移位操作实现,比较麻烦。比如3/5=0.6

做的时候先3=30,然后30/5=6,然后对6在数码管的显示进行调整就好。把6显示在小数点后面1位就好

用vb编写一个计算器程式,实现加减乘除,

Dim v As Boolean

Dim s As Integer

Dim X As Double

Dim Y As Double

Private Sub Command1_Click(Index As Integer)

If Form1.Tag = "T" Then

If Index = 10 Then

Text1.Text = "0"

Else

Text1.Text = Command1(Index).Caption

End If

Form1.Tag = ""

Else

Text1.Text = Text1.Text &Command1(Index).Caption

End If

End Sub

Private Sub Command2_Click(Index As Integer)

Form1.Tag = "T"

If v Then

X = Val(Text1.Text)

v = Not v

Else

Y = Val(Text1.Text)

Select Case s

Case 0

Text1.Text = X + Y

Case 1

Text1.Text = X - Y

Case 2

Text1.Text = X * Y

Case 3

If Y <>0 Then

Text1.Text = X / Y

Else

MsgBox ("不能以0为除数")

Text1.Text = X

v = False

End If

Case 4

Y = 0

v = False

End Select

X = Val(Text1.Text)

End If

s = Index

End Sub

Private Sub Frame1_DragDrop(Source As Control, X As Single, Y As Single)

End Sub

控制元件自己新增吧,空间名要和程式码名一致

求一简单的加减乘除计算器c++程式

#include<stdio.h>

#include<math.h>

void main()

{

float a,b

char C

while(1)

{

scanf("%f%c%f",&a,&C,&b)

if((C!='+')&&(C!='-')&&(C!='*')&&(C!='/'))

break

switch(C)

{

case '+': printf("%f+%f=%f",a,b,a+b)

break

case '-': printf("%f-%f=%f",a,b,a-b)

break

case '*': printf("%f*%f=%f",a,b,a*b)

break

case '/': printf("%f/%f=%f",a,b,a/b)

break

}

}

}

想改成按1 2 3 4分别为加减乘除,只需要将程式中的+ - * / 改成1 2 3 4即可。按除了+ - * / 以外的键就会退出。

用c++语言编写一个简单的计算器程式,会加减乘除就行,本人初学不太会,特训求帮助

这个是最简单,简陋的计算器。很多情况没考虑进去,例如除数不能为0之类的,真要写完整的话程式码还要更多。

程式码如下: #include <iostream>using namespace stdint main(){ float a, b, result char operation cout <<"请输入算式,如1+2并回车:" <<endl cin >>a >>operation >>b switch(operation) { case '+': result = a + bbreak case '-': result = a - bbreak case '*': result = a * bbreak case '/': result = a / bbreak default: cout <<"输入非法,程式退出!" <<endlreturn -1 } cout <<endl <<"结果为:" <<endl <<a <<operation <<b <<"=" <<result <<endl return 0}

知道switch函式 吗 用这个就行

建俩个int型变数 一个字元型变数