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相關如下
void LcdDisplay(int temp)得temp能用,但是void warn(int temp)說我的temp沒定義?
程序如下
void LcdDisplay(int temp)
{
unsigned char datas[] = {0, 0, 0, 0, 0};
unsigned char TempH,TempL;
if(temp&0x8000)
{
LcdWriteCom(0x85);
LcdWriteData('-');
temp=~temp;
temp +=1;
}
else
{
LcdWriteCom(0x85);
LcdWriteData('+');
}
TempH=temp>>4;
TempL=temp&0x0F;
TempL=TempL*6.25+0.5;
datas[0] = TempH/100%10;
datas[1] = TempH/10%10;
datas[2] = TempH%10;
datas[3] = TempL/10%10;
datas[4] = TempL%10;
LcdWriteCom(0x86);
LcdWriteData('0'+datas[0]);
LcdWriteCom(0x87);
LcdWriteData('0'+datas[1]);
LcdWriteCom(0x88);
LcdWriteData('0'+datas[2]);
LcdWriteCom(0x89);
LcdWriteData('.');
LcdWriteCom(0x8a);
LcdWriteData('0'+datas[3]);
}
void warn(int temp)
{
if(temp>=27)
{
beep=~beep;
delay(50);
}
}
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溫度報警.zip
2020-3-13 14:09 上傳
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32.8 KB, 下載次數: 4
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