MCU simulates I2C bus to read and write EEPROM (24CXX) program 2

The circuit remains unchanged. The following is a simulation circuit, except that the digital tube of port P2 is changed from observing which step the program is executed to checking the data in the receiving buffer.

The procedure is as follows:

#include
#define unit unsigned int
#define uchar unsigned char

uchar num=4;
uchar idata sendbuf[4]={0x96,0x84,0xd5,0x63};
uchar idata recivebuf[4];

sbit scl=P0^0;          
sbit sda=P0^1;

sbit led0=P2^0;
sbit led1=P2^1;
sbit led2=P2^2;
sbit led3=P2^3;
sbit led4=P2^4;
sbit led5=P2^5;
sbit led6=P2^6;
sbit led7=P2^7;

delay(void) //delay
{ 
  int i;
  for(i=0;i<1;i++);
}

start(void) //start
{
  sda=1;
  scl=1;
  delay();
  sda=0;
  delay();
  scl=0;
}

stop(void) //stop
{
  sda=0;
  scl=1;
  delay();
  sda=1;
  delay();
  scl=0;
}

answer(void) //answer
{
  sda=1;
  scl=1;
  delay();
  sda=0;
  scl=0;
}

noanswer(void) //no answer
{
  sda=1;
  scl=1;
  delay();
  sda=1;
  scl=0; 
}

checkanswer(void) //check answer
{
  sda=1;
  scl=1;
  F0=0;
  if(sda==1) F0=1;
  scl=0; 
}

sendabyte(uchar idata *saddress) //send a byte
{
  uchar n=8,temp=*saddress;
  while(n--)
  {
    if((temp&0x80)==0x80) sda=1;
 else sda=0;
   delay( );
   scl=1;
   delay();
   scl=0;
 temp=temp<<1;
  }
  checkanswer();
  if(F0==1) return;
}

reciveabyte(uchar idata *raddress) //recive a byte
{
  uchar n=8,temp;
  while(n--)
  {        
   scl=1;
 temp=temp<<1;
 if(sda==1)
   temp=temp|0x01 ;
 else
   temp=temp&0xfe;
 scl=0;
  }
  *raddress=temp;
}

sendnbyte(uchar n) //send n byte
{
  uchar idata *ps; 
  ps=&sendbuf[0];

  while(n--)
  {
   sendabyte(ps);
 ps++;
  }
  stop();
}

recivenbyte(uchar n) //recive n byte
{
  uchar idata *pr;
  pr=&recivebuf[0];
  while(n--)
  {
    reciveabyte(pr);
 answer();
 pr++;
  }
  noanswer();
  stop();
}

main(void) //MAIN
{
start();

sendabyte(0xa0);

sendabyte(0x00);

sendnbyte(num);

/*-----------------------*/

start();

sendabyte(0xa1);

recivenbyte(num);

P2 = recivebuf[7];
}

Program Description:

The num at the beginning of the program defines the number of data to be transmitted, as well as the send buffer and the receive buffer.

The operation of the main function is clear at a glance and will not be introduced in detail.

Problem: In this program, I defined the data in the receive buffer as 4, and indeed 4 data were received. However, the received data starts from recivebuf[4]. In this example, recivebuf[4] stores 0X96, recivebuf[5] stores 0X84, recivebuf[6] stores 0XD5, and recivebuf[7] stores 0X63.