有很多人用三菱的PLC做定時停機，主要是為了收尾款。

一般都是通過比較當前時間，和PLC里設定的起始時間，然后進行判斷

這樣用估計已經有20年歷史了

今天我突然想到一個BUG，也不知道應該不應該說出來

比如PLC 在2020年1月是到期了，不能用了。

我也解不了密碼。

但我另外拿一個觸摸屏，接上原來的PLC，修改一下PLC的時間，改向前半年。

那不是又能正常使用了么？

寫的一手騙流量的好帖，你這這些只能偏偏新手；給你一段觸摸屏的加密代碼你看看，看看有沒BUG：
各種觸摸屏的腳本語言格式可能不是很一樣，我這個是Kinco的，把頭文件和部分關鍵字換一下可以上維綸和昆侖；
           #include "macrotypedef.h"                
#include "math.h"                
                
/*                
    Read,Write Local address function:            
      int ReadLocal( const char *type, int addr, int nRegs, void *buf, int flag  );            
    int WriteLocal( const char *type, int addr, int nRegs, void *buf , int flag );            
                
    Parameter:     type     is the string of "LW","LB" etc;            
                
                
                
                
    return value : 1  ,Operation success            
                
                
     eg: read the value of local lw200 and write it to the lw202,with the codetype BIN,            
        The code is :        
                
        short buf[2] = {0};            
        ReadLocal("LW", 200, 2, (void*)buf, 0);        
        WriteLocal("LW", 202, 2, (void*)buf, 0);        
*/                
int MacroEntry()                
{                
    short Date[3]={0};            
    short Lic[8]={0};            
    short Cd[16]={0};            
    short Subb[6]={0};            
    short i,j;            
    short YTen,YGe,MTen,MGe,DTen,DGe,Y,M,D;            
    //short Year,Month,Dat;            
    ReadLocal("LW",10003,3,(void*)Date,0);  //獲取現在日期            
     int temp=Date[2]*10000+Date[1]*100+Date[0];            
    CuDG=temp;    //當前日期組合            
                
                
    ReadLocal("RW",200,8,(void*)Lic,0);    //獲得8個字的授權號            
    for( i=0,j=0;i<16;i++)            
        {        
            if(i%2==0)    
                Cd=Lic[j] & 0xff;
            else    
                {
                
                
                }
        if(!(Cd>='A' && Cd<= 'Z' ) )        
            {    
                Fail=1;
                return 0;
            }    
                
        }        
WriteLocal("LW", 20, 16, (void*)Cd, 0);  //將拆解后的字符放到LW20-35中                
    if(Date[1]%2==0) //假如是偶數月            
        {        
            YTen=Cd[2]-'H';        
            YGe=Cd[4]-'E';        
            MTen=Cd[6]-'I';      
            MGe=Cd[8]-'J';        
            DTen=Cd[10]-'K';      
            DGe=Cd[12]-'F';      
        }        
    else            
        {            
            YTen=Cd[1]-'F';      
            YGe=Cd[5]-'E';        
            MTen=Cd[9]-'M';      
            MGe=Cd[11]-'H';      
            DTen=Cd[13]-'U';    
            DGe=Cd[15]-'M';      
        }        
Subb[0]=YTen;                
Subb[1]=YGe;                
Subb[2]=MTen;                
Subb[3]=MGe;                
Subb[4]=DTen;                
Subb[5]=DGe;                
                
WriteLocal("LW", 40, 6, (void*)Subb, 0);                
                
Y=YTen*10+YGe;                
M=MTen*10+MGe;                
D=DTen*10+DGe;                
Rq=20000000+Y*10000+M*100+D;                
Fail=0;                
                
if(Y>30 || Y<16 || M>12 || M <1 || D>31 || D<1 || Rq<=CuDG)                
    {            
        Fail=1;        
                
    }            
    else            
    {            
        Fail=0;        
        if((2000+Y-Date[2])>=2)        
            KGH=1;    
                
    }                                                  
    return 0;            
}