Why step into next instrution with TD3 different value?

 Why step into next instrution with TD3 different value?

First:

#include <AT89X51.H>

unsigned int  TD3=0;//
void t0(void) interrupt 1 using 3
{

TL0=(65536-50000)%256;
TH0=(65536-50000)/256; //6MHz
TD3++;
}
void main()
{//1
TMOD=0x11; //
TL0=(65536-50000)%256;
TH0=(65536-50000)/256; //
ET0=1;
EA=1;
TR0=1;
while(1)
{
while(TD3<=18000); //
TD3=0;
}
}//1

Secondly:

#include <AT89X51.H>

unsigned int  TD3=0;//

void t0(void) interrupt 1 using 3

{
TH0=(65536-50000)/256; //

TL0=(65536-50000)%256;

TD3++;

}
void main()

{//1

TMOD=0x11; //

TH0=(65536-50000)/256; //

TL0=(65536-50000)%256;

ET0=1;

EA=1;

TR0=1;

while(1)

{
while(TD3<=18000); //

TD3=0;

}
}//1

TD3 counter limited all 18000.By the first way, TD3 counted to 18000 then step int TD3=0; .But , by the second way , TD3 counted only to 17920 then step into TD3=0; .Why did this happen?
Thanks.

 RE: Why step into next instrution with TD3 different value?

You are using an 8-bit processor. The processor is incapable of performing an atomic test of a 16-bit number, just as your 16-bit assign of the timer registers has to be splitted into two 8-bit assigns. The test is performed either low byte before high byte or the reverse - and if the variable gets incremented in-between, you get into troubles.

Why not let your interrupt perform the compare/reset, and set a flag?

You also have to think about this problem when you reload the timer in your interrupt handler - what happens when you assign just the low or the high part of the timer value?