Specifications

BASIC Stamp II

Page 262 • BASIC Stamp Programming Manual 1.8 • Parallax, Inc.

If you want For...Next to count by some amount other than 1, you can

specify a stepVal. For example, change the previous example to count

down by 3:

reps var nib ' Counter for the FOR/NEXT loop.

FOR reps = 10 to 1 STEP 3 ' Repeat with reps = 10, 7...1.

debug dec ? reps ' Each rep, show values of reps.

Note that even though you are counting down, stepVal is still positive.

For...Next takes its cue from the relationship between start and end, not

the sign of stepVal. In fact, although PBASIC2 won’t squawk if you use

a negative entry for stepVal, its positive-integer math treats these values

as large positive numbers. For example, –1 in two’s complement is

65535. So the following code executes only once:

reps var word ' Counter for the FOR/NEXT loop.

FOR reps = 1 to 10 STEP -1 ' Actually FOR reps = 1 to 10 step 65535

debug dec ? reps ' Executes only once.

This brings up a good point: the instructions inside a For...Next loop

always execute once, no matter what start, end and stepVal values are

assigned.

There is a potential bug that you should be careful to avoid. PBASIC

uses unsigned 16-bit integer math to increment/decrement the counter

variable and compare it to the stop value. The maximum value a 16-bit

variable can hold is 65535. If you add 1 to 65535, you get 0 as the 16-bit

maximum mileage it can display).

If you write a For...Next loop whose step value is larger than the

difference between the stop value and 65535, this rollover will cause the

loop to execute more times than you expect. Try the following example:

reps var word ' Counter for the loop.

FOR reps = 0 to 65500 STEP 3000 ' Each loop add 3000.

debug dec ? reps ' Show reps in debug window.

NEXT ' Again until reps>65500.

The value of reps increases by 3000 each trip through the loop. As it

approaches the stop value, an interesting thing happens: 57000, 60000,

63000, 464, 3464... It passes the stop value and keeps going. That’s

because the result of the calculation 63000 + 3000 exceeds the maximum