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Conditional branching must again be used
only within a colon definition and use the form:
IF (True Part)
Forth-words
THEN
IF (True Part)
Forth-words
ELSE (False Part)
Forth-words
THEN
These conditional statement rely on
testing the top number on the stack and decide whether to execute the True Part
or the False Part of the condition. If the top item on the stack is true
(non-zero) then the True Part will be executed. If the top item is false (zero)
then the true part will be skipped and execution of the False Part will take
place. If the ELSE part is missing then
execution skips to just after the THEN
statement. ENDIF is a synonym of THEN.
There are several mathematical
operations which will leave either a true (non-zero) flag or a false (zero) flag
on the stack to be tested for by IF.
0<
This will leave a true flag on the stack
if the number on the top of the stack is less than zero, otherwise it leaves a
false flag.
e.g.
-4 0< <CR>
will leave a true flag (non zero). To
see this type
. <CR>
to print the top number on the stack
which is the flag. This will print
-1
to show the true flag
914 0< . <CR>
will print a 0 (false flag).
0=
This will leave a true flag if the
number on the top of the stack is equal to zero, otherwise it leaves a false
flag.
<
This will leave a true flag if the
second number on the stack is less than the top number, otherwise it leaves a
false flag.
e.g.
40 25 < . <CR>
will print 0 (false flag).
If we look at the stack during this
operation we will see
|
Operation
|
TOS
|
|
40
|
40
|
|
25
|
40
25
|
|
<
|
0
|
|
.
|
empty
|
>
This will leave a true flag if the
second number on the stack is greater than the top number, else a false flag
will be left.
e.g.
40 25 > . <CR>
will print -1
(true flag).
=
This will leave a true flag if the top
two numbers are equal, otherwise it will leave a false flag.
Now for some exampes using the
conditional branching structure, type:
: TEST= <CR>
= IF ." BOTH ARE EQUAL " THEN <CR>
." FINISHED " ; <CR>
Now key in two numbers followed by TEST=
and carriage return.
e.g.
11 119 TEST= <CR>
this will print
FINISHED
119 119 TEST= <CR>
will print
BOTH ARE EQUAL FINISHED
Now key in
: TEST1= <CR>
= IF ." EQUAL " ELSE ." UNEQUAL " THEN <CR>
CR ."
FINISHED " ; <CR>
Now key in:
249 249 TEST1= <CR>
this will print
EQUAL
FINISHED
Try:
249 249 TEST1= <CR>
this will print
UNEQUAL
FINISHED
Notice how the part after THEN
was
executed in both cases.
Two more loops structure will now be
discussed
BEGIN
Forth-words
UNTIL
BEGIN
Forth-words
WHILE
Forth-words
REPEAT
Going to BEGIN
UNTIL the value at the
top of the stack is test upon searching
UNTIL. If the flag is false (zero) then
the loop starting from
BEGIN is repeated. If the value is true (non zero) then
an exit from the loop occours.
Try typing the following examples:
: COUNT-DOWN DECIMAL 100 <CR>
BEGIN 1 DUP DUP . CR 0= UNTIL <CR>
." DONE " ;
<CR>
Now key in
COUNT-DOWN <CR>
This will print:
99
98
97
2
1
0
DONE
The BEGIN
WHILE
REPEAT structure
uses the
WHILE condition to abort a loop in the middle of that loop.
WHILE will
test the flag left on the top of the stack and if that flag is true, will
continue with the execution of words up to
REPEAT which then branches always
(unconditionally) back to
BEGIN. If the flag is false then
WHILE will cause
execution to skip the words up to REPEAT
and thus exit from the loop.
We will now build a program to print out
the cubes of numbers from 1 upwards, until the cube is greater than 3000. The
colon definition could be as follows:
: CUBE DECIMAL
0 BEGIN 1+ <CR>
DUP DUP DUP
DUP * * DUP <CR>
3000 < WHILE
." THE CUBE OF " <CR>
SWAP . ." IS "
. CR REPEAT <CR>
DROP DROP DROP
." ALL DONE " CR ; <CR>
You may get an error message MSG#4
appearing on the screen: this means that the word you have just created already
exists. This is not a problem, since the new-word will be created and all
actions referencing the word
CUBE will be directed to the latest definition
using that word.
Try run this by keying in:
CUBE <CR>
and watch the results.
Now to follow what is happening by
writing down the values on the stack at each iteration, if you are having any
difficulty in doing this, the stack values are:
|
STACK
|
|
Operation
|
Display
|
|
|
Emtpy
|
|
DECIMAL
|
|
|
|
0
|
|
0
|
|
|
|
0
|
|
BEGIN
|
|
|
|
1
|
|
1+
|
|
|
|
N
N
|
|
DUP
|
|
Lets now refer to the
|
|
N
N N
|
|
DUP
|
|
number of the stack as N
|
|
N
N N N
|
|
DUP
|
|
|
|
N
N N N N
|
|
DUP
|
|
|
|
N
N N N2
|
|
*
|
|
|
|
N
N N3
|
|
*
|
|
|
|
N
N N3
N3
|
|
DUP
|
|
|
|
N
N N3
N3 3000
|
|
3000
|
|
|
|
N
N N3
N3 flag
|
|
<
|
|
|
|
IF
TRUE:
|
|
|
|
|
|
N
N N3
|
|
WHILE
|
|
|
|
N
N N3
|
|
." THE CUBE OF "
|
THE CUBE OF
|
|
|
N
N3 N
|
|
SWAP
|
|
|
|
N
N3
|
|
.
|
N
|
|
|
N
N3
|
|
." IS "
|
IS
|
|
|
N
|
|
.
|
N3
|
|
|
N
|
|
CR
|
carriage return
|
|
|
N
|
|
REPEAT
|
|
|
|
IF
FALSE:
|
|
|
|
|
|
N
N
|
|
DROP
|
|
|
|
N
|
|
DROP
|
|
|
|
empty
|
|
DROP
|
|
|
|
|
|
." ALL DONE "
|
ALL DONE
|
|
|
|
|
CR
|
|
|
|
|
|
;
|
|
|
In fact, it is a good idea to check the
stack contents during the execution of any new Forth word to make sure that it
is working correctly. (Note that DROP merely clears the top number from the
stack)
Finally, one extra construct has been
added to circunvent the problem of deeply nested IF
THEN
ELSE structure.
This is
CASE OF structure. It takes the general form:
n1
CASE
n1 OF forth-words
ENDOF
n2 OF forth-words
ENDOF
nk OF forth-words
ENDOF
forth-words
ENDCASE
For example type:
: TEST4 CASE
<CR>
1 OF ."
FIRST CASE " ENDOF <CR>
2 OF ."
SECOND CASE " ENDOF <CR>
3 OF ."
THIRD CASE " ENDOF <CR>
." OTHER CASE " <CR>
ENDCASE ; <CR>
Now type:
1 TEST4 CR 2 TEST4 CR 3 TEST4 CR <CR>
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