基于C語言的簡單計算器的加法程序

ID:151228 · 發表于 2016-11-30 15:04

做了個簡單的計算器練手
單片機計算器仿真原理圖：

單片機源程序預覽：

#include<reg52.h>
sbit in1 = P2^0; //按鍵P2.0輸出
sbit in2 = P2^1;
sbit in3 = P2^2;
sbit in4 = P2^3;
sbit out1 = P2^4; //動態行變換
sbit out2 = P2^5; //動態行變換
sbit out3 = P2^6; //動態行變換
sbit out4 = P2^7; //動態行變換
unsigned char code LedChar[]={
0xC0, 0xF9, 0xA4, 0xB0, 0x99, 0x92, 0x82, 0xF8,
0x80, 0x90, 0x88, 0x83, 0xC6, 0xA1, 0x86, 0x8E
};
unsigned char KeySta[4][4] = {
{1,1,1,1},{1,1,1,1},{1,1,1,1},{1,1,1,1}
};
unsigned char LedBuff[6] = {
0xFF, 0xFF, 0xFF, 0xFF, 0xFF, 0xFF
};
unsigned char code KeyCodeMap[4][4] = { //矩陣按鍵編號到標準鍵盤鍵碼的映射表
{ 0x31, 0x32, 0x33, 0x26 }, //數字鍵1、數字鍵2、數字鍵3、向上鍵
{ 0x34, 0x35, 0x36, 0x25 }, //數字鍵4、數字鍵5、數字鍵6、向左鍵
{ 0x37, 0x38, 0x39, 0x28 }, //數字鍵7、數字鍵8、數字鍵9、向下鍵
{ 0x30, 0x1B, 0x0D, 0x27 } //數字鍵0、ESC鍵、回車鍵、向右鍵
};
void KeyDriver();
void main()
{
EA = 1; //使能總中斷
ET0 = 1; //使能T0中斷
TMOD = 0x01; //設置T0為模式1
TH0 = 0xDC;
TL0 = 0x00;
TR0 = 1; //啟動T0
P1=0xFF;
LedBuff[0] = LedChar[0]; //上電顯示0
while (1)
{
KeyDriver(); //調用按鍵驅動函數
}
}
void ShowNumber(unsigned long num)
{
signed char i;
unsigned char buf[6];
for(i=0; i<6; i++)
{
buf[i] = num % 10;
num = num / 10;
}
for(i=5; i>=1; i--)
{
if(buf[i] == 0)
{
LedBuff[i] = 0xFF;
}
else
break;
}
for(; i>=0; i--)
{
LedBuff[i] = LedChar[buf[i]];
}
}
void KeyAction(unsigned char keycode)
{
static unsigned long result = 0;
static unsigned long addend = 0;
if((keycode >= 0x30) &&(keycode <= 0x39))
{
addend = (addend *10) + (keycode - 0x30);
ShowNumber(addend);
}
else if(keycode == 0x26)
{
result += addend;
addend = 0;
ShowNumber(result);
}
else if(keycode == 0x0D)
{
result += addend;
addend = 0;
ShowNumber(result);
}
else if(keycode == 0x1B)
{
addend = 0;
result = 0;
ShowNumber(addend);
}
}
void KeyDriver()
{
unsigned char i, j;
static unsigned char backup [4][4] = {
{1,1,1,1},{1,1,1,1},{1,1,1,1},{1,1,1,1}
};
for(i=0; i<4; i++)
{
for(j=0; j<4; j++)
{
if(backup[i][j] != KeySta[i][j])
{
if(backup[i][j] == 0)
{
KeyAction(KeyCodeMap[i][j]);
}
backup[i][j] = KeySta[i][j];
}
}
}
}
/* 按鍵掃描函數，需在定時中斷中調用，推薦調用間隔1ms */
void KeyScan()
{
unsigned char i;
static unsigned char keyout = 0; //矩陣按鍵掃描輸出索引
static unsigned char keybuf[4][4] = { //矩陣按鍵掃描緩沖區
{0xFF, 0xFF, 0xFF, 0xFF}, {0xFF, 0xFF, 0xFF, 0xFF},
{0xFF, 0xFF, 0xFF, 0xFF}, {0xFF, 0xFF, 0xFF, 0xFF}
};
//將一行的4個按鍵值移入緩沖區
keybuf[keyout][0] = (keybuf[keyout][0] << 1) | in1;
keybuf[keyout][1] = (keybuf[keyout][1] << 1) | in2;
keybuf[keyout][2] = (keybuf[keyout][2] << 1) | in3;
keybuf[keyout][3] = (keybuf[keyout][3] << 1) | in4;
//消抖后更新按鍵狀態
for (i=0; i<4; i++) //每行4個按鍵，所以循環4次
{
if ((keybuf[keyout][i] & 0x0F) == 0x00)
{ //連續4次掃描值為0，即4*4ms內都是按下狀態時，可認為按鍵已穩定的按下
KeySta[keyout][i] = 0;
}
else if ((keybuf[keyout][i] & 0x0F) == 0x0F)
{ //連續4次掃描值為1，即4*4ms內都是彈起狀態時，可認為按鍵已穩定的彈起
KeySta[keyout][i] = 1;
}
}
//執行下一次的掃描輸出
keyout++; //輸出索引遞增
keyout = keyout & 0x03; //索引值加到4即歸零
switch (keyout) //根據索引，釋放當前輸出引腳，拉低下次的輸出引腳
{
case 0: out1 = 0; out2 = 1; out3 = 1;out4 = 1; ; break; //當我按下第一行的按鍵，只有keyout==0時符合條件
case 1: out1 = 1; out2 = 0; out3 = 1;out4 = 1; ; break;
case 2: out1 = 1; out2 = 1; out3 = 0;out4 = 1; ; break;
case 3: out1 = 1; out2 = 1; out3 = 1;out4 = 0; ; break;
default: break;
}
}
/* 數碼管動態掃描刷新函數，需在定時中斷中調用 */
void LedScan()
{
static unsigned char i = 0; //動態掃描的索引
P0 = 0xFF; //顯示消隱
switch (i)
{
case 0:P1=0xFE; i++; P0=LedBuff[0];break;
case 1:P1=0xFD; i++; P0=LedBuff[1];break;
case 2:P1=0xFB; i++; P0=LedBuff[2];break;
case 3:P1=0xF7; i++; P0=LedBuff[3];break;
case 4:P1=0xEF; i++; P0=LedBuff[4];break;
case 5:P1=0xDF; i=0; P0=LedBuff[5];break;
default:break;
}
}
/* T0中斷服務函數，用于數碼管顯示掃描與按鍵掃描 */
void InterruptTimer0() interrupt 1
{
TH0 = 0xDC;
TL0 = 0x00;
LedScan();
KeyScan();
}

復制代碼

所有資料打包下載：

簡單的計算器加法程序.zip (79.82 KB, 下載次數: 21)

ID:1 · 發表于 2016-12-3 23:45

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基于C語言的簡單計算器的加法程序

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