文档介绍：16进制转换算成10进制程序
STC 单片机中的eeprom写入和读取6位的10进制数需要进行数据类型转换吗？具体是怎样现的呢？
2013-06-04 10:24 
满意回答
提问者采纳 2013-06-04 10:27
op_();_nop_();_nop_();
     _nop_();_nop_();_nop_();_nop_();
     _nop_();_nop_();_nop_();_nop_();
     _n
op_();_nop_();_nop_();_nop_();
 }
 unsigned char command(void) {
     TH0=-100;b_break=0;
     while (RI==0) {if (b_break==1) return(0);}
     RI=0;
     return(SBUF);
 }
 void earsure(unsigned char cpu) {
     switch (cpu) {
     case 1:        //89C51
     case 2:rst=p26=1;p27=p36=p37=0;nop();vpp=1;nop();
         ale=0;wait(110);ale=1;nop();
         break;
     case 3:
     case 4:break;
     case 5:
     case 6:break;
     }
 }
 void program(unsigned char cpu) {
     unsigned int bdata adds=0;
     unsigned char d;
     switch (cpu) {
     case 1:    //89C51
     case 2:
         p36=p37=1;rst=1;
         while (1) {
             TH0=-100;b_break=0;
             while (RI==0) {if (b_break==1) return;}
             RI=0;
             d=SBUF;
            
//address
             P0=adds%256;
             P2=adds/256;
             p27=1;
             //data
             P1=d;
             nop();    //48clcl
             //vpp
             vpp=1;
             nop();    //48clcl
             //ale
             ale=0;
             wait(1);//100uS
             ale=1;
             nop();    //10uS
             vpp=0;
             nop();    //48clcl
             p27=0;
             nop();    //48clcl
             P1=0xff;
             TH0=-100;b_break=0;
             while (d!=P1) {if (b_break==1) return;}    //data
polling
             SBUF=d;
             adds++;
         }
         break;
     case 3:
     case 4:
     case 5:
     case 6:break;
     }
 }
 void lock(unsigned char cpu) {
     unsigned char i;
     switch (cpu) {
     case 1:    //89c51
     case 2:
         //lock 1
         rst=p26=p36=p27=p37=1;nop();
         vpp=1;
         nop();