Question

在51单片机中如何利用宋老师的简易计算器程序改成可以实现连加连减

Answer 1

uchar code LedChar[]={
0x3f,0x06,0x5b,0x4f,
0x66,0x6d,0x7d,0x07,
0x7f,0x6f}; //0~9数码管缓冲区
uchar LedBuff[] = {
0x00,0x00,0x00,0x00,
0x00,0x00};
uchar code KeyCodeMap[4][4] = {
{ 0x31, 0x32, 0x33, 0x26 }, //数字键1、数字键2、数字键3、向上键
{ 0x34, 0x35, 0x36, 0x25 }, //数字键4、数字键5、数字键6、向左键
{ 0x37, 0x38, 0x39, 0x28 }, //数字键7、数字键8、数字键9、向下键
{ 0x30, 0x1B, 0x0D, 0x27 } //数字键0、ESC键、 回车键、 向右键
};
uchar KeySta[4][4] = {
{1,1,1,1},{1,1,1,1},
{1,1,1,1},{1,1,1,1}
};
void keydriver();
void main()
{
EA = 1;
TMOD = 0X01;
TH0 = 0XFC;
TL0 = 0X67;
TR0 = 1;
ET0 = 1;
LedBuff[0] = LedChar[0];

while(1)
{
keydriver();
}
}
void shownumber(signed long num)
{
unsigned char sign;
uchar buf[6];
signed char i;
if (num < 0) //首先提取并暂存符号位
{
sign = 1;
num =-num;
}
else
{
sign = 0;
}

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] = 0x00;
}
else
{
break;
}
}
if (sign != 0) //负数时，需在最前面添加负号
{
if (i < 5) //当有效位数小于6位时添加负号，否则显示结果将是错的
{
LedBuff[i+1] = 0x40;
}
}

for( ;i>=0;i--)
{
LedBuff[i] = LedChar[buf[i]];
}
}
void keyaction(uchar keycode)
{
static char oprt = 0;
static signed long result = 0;
static signed long addend = 0;

if((keycode >= 0x30) && (keycode <= 0x39))
{
addend = (addend*10) + (keycode - 0x30);
shownumber(addend);
}

else if(keycode == 0x26) //+
{
oprt = 0;
result += addend;
addend = 0;
shownumber(result);
}

else if(keycode == 0x28) //-
{
oprt = 1;
if(result == 0 )
{
result = addend -result;
}
else
{
result -= addend;
}
addend = 0;
shownumber(result);
}

else if(keycode == 0x25) //x
{
oprt = 2;
if(result == 0)
{
result = addend - result;
}
else
{
result *= addend;
//result = result * 1;
}
addend = 0;
shownumber(result);
}

else if(keycode == 0x27) //%
{
oprt = 3;
if(result == 0)
{
result = addend - result;
}
else
{
result /= addend;
}
addend = 0;
shownumber(result);
}

else if(keycode == 0x0d)
{
if (oprt == 0)
{
result += addend; //进行加法运算
}
else if(oprt == 1)
{
result -= addend; //进行减法运算
}

else if(oprt == 2)
{
result *= addend;
}

else if(oprt == 3)
{
result /= addend;
}
addend = 0;
shownumber(result); //运算结果显示到数码管
}

else if(keycode == 0x1b)
{
result = 0;
addend = 0;
shownumber(addend);
}
}

Answer 2

得到木头[数字]；GreedIsGood