User`s manual

Dynamic C User’s Manual digi.com 115

7.3.2 Example: Delay Loop

An important detail about MS_TIMER is that it overflows (“rolls over”) approximately every 49 days, 17

hours. This behavior causes the following delay loop code to fail:

/* THIS CODE WILL FAIL!! */

endtime = MS_TIMER + delay;

while (MS_TIMER < endtime) {

//do something

}

If “MS_TIMER + delay” overflows, this returns immediately. The correct way to code the delay loop so

that an overflow of MS_TIMER does not break it, is this:

endtime = MS_TIMER + delay;

while ((long)MS_TIMER - endtime < 0) {

//do something

}

The interval defined by the subtraction is always correct. This is true because the value of the interval is

based on the values of MS_TIMER and “endtime” relative to one another, so the actual value of these vari-

able does not matter.

One way to conceptualize why the second code snippet is always correct is to consider a number circle like

the one in Figure 7.1. In this example, delay=5. Notice that the value chosen for MS_TIMER will “roll

over” but that it is only when MS_TIMER equals or is greater than “endtime” that the while loop will eval-

uate to false.

Figure 7.1 “delay=5”

Another important point to consider is that the interval is cast to a signed number, which means that any

number with the high bit set is negative. This is necessary in order for the interval to be less than zero

when MS_TIMER is a large number.

MS_TIMER

endtime