單片機源程序如下:
void key_Service()
{
switch(key_scan(170))
{
case 1:
switch(wnd)
{
case 0:T1Cnt=epxs_short; parm1 = Read_EEPROM_u16(0x0400); if(parm1<1000) parm1 = parm1+5; else {parm1++;} if(parm1<30) parm1=30;if (parm1>1050) parm1=1050; Save_EEPROM_u16(0x0400, parm1); break;//psi:0~150,0~100加5,到100退位顯示100~150加1
case 1:T1Cnt=epxs_short; parm2 = Read_EEPROM_u16(0x0500); parm2++; if(parm2<20) parm2=20;if(parm2>103) parm2=103; Save_EEPROM_u16(0x0500, parm2); break;
case 2:T1Cnt=epxs_short; parm3 = Read_EEPROM_u16(0x0600); parm3 = parm3+5; if(parm3<20) parm3=20;if(parm3>995) parm3=995; Save_EEPROM_u16(0x0600, parm3); break;
case 3:T1Cnt=epxs_short; parm4 = Read_EEPROM_u16(0x0700); parm4++; if(parm4<20) parm4=20;if(parm4>103) parm4=103; Save_EEPROM_u16(0x0700, parm4); break;
}
FMQCnt = FMQ_short; //按鍵聲音觸發，滴一聲就停。
break;
//case 2:第一次按下進來后執行T1Cnt=0; ，第二次按下進來后執行wnd++;if(wnd>3) wnd=0;如此循環，如何實現？
case 2:
T1Cnt=0;
wnd++;if(wnd>3) wnd=0;
FMQCnt = FMQ_short; //按鍵聲音觸發，滴一聲就停。
break;
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