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Correct the descriptions of WHILE...END and RROT. Thanks to Lalo Martins

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for pointing out these errors.

llvm-svn: 21683
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Reid Spencer 2005-05-04 15:43:40 +00:00
parent dc2ab16e6f
commit d06dc0f95d
1 changed files with 21 additions and 17 deletions
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@ -787,7 +787,7 @@ using the following construction:</p>
</tr>
<tr><td>RROT</td>
<td>RROT</td>
<td>w1 w2 w3 -- w2 w3 w1</td>
<td>w1 w2 w3 -- w3 w1 w2</td>
<td>Reverse rotation. Like ROT, but it rotates the other way around.
Essentially, the third element on the stack is moved to the top
of the stack.</td>
@ -946,24 +946,28 @@ using the following construction:</p>
executed. In either case, after the (words....) have executed, execution continues
immediately following the ENDIF. </td>
</tr>
<tr><td>WHILE (words...) END</td>
<td>WHILE (words...) END</td>
<tr><td>WHILE word END</td>
<td>WHILE word END</td>
<td>b -- b </td>
<td>The boolean value on the top of the stack is examined. If it is non-zero then the
"words..." between WHILE and END are executed. Execution then begins again at the WHILE where another
boolean is popped off the stack. To prevent this operation from eating up the entire
stack, you should push on to the stack (just before the END) a boolean value that indicates
whether to terminate. Note that since booleans and integers can be coerced you can
use the following "for loop" idiom:<br/>
<code>(push count) WHILE (words...) -- END</code><br/>
<td>The boolean value on the top of the stack is examined (not popped). If
it is non-zero then the "word" between WHILE and END is executed.
Execution then begins again at the WHILE where the boolean on the top of
the stack is examined again. The stack is not modified by the WHILE...END
loop, only examined. It is imperative that the "word" in the body of the
loop ensure that the top of the stack contains the next boolean to examine
when it completes. Note that since booleans and integers can be coerced
you can use the following "for loop" idiom:<br/>
<code>(push count) WHILE word -- END</code><br/>
For example:<br/>
<code>10 WHILE DUP &gt;d -- END</code><br/>
This will print the numbers from 10 down to 1. 10 is pushed on the stack. Since that is
non-zero, the while loop is entered. The top of the stack (10) is duplicated and then
printed out with &gt;d. The top of the stack is decremented, yielding 9 and control is
transfered back to the WHILE keyword. The process starts all over again and repeats until
the top of stack is decremented to 0 at which the WHILE test fails and control is
transfered to the word after the END.</td>
<code>10 WHILE &gt;d -- END</code><br/>
This will print the numbers from 10 down to 1. 10 is pushed on the
stack. Since that is non-zero, the while loop is entered. The top of
the stack (10) is printed out with &gt;d. The top of the stack is
decremented, yielding 9 and control is transfered back to the WHILE
keyword. The process starts all over again and repeats until
the top of stack is decremented to 0 at which point the WHILE test
fails and control is transfered to the word after the END.
</td>
</tr>
<tr><th colspan="4"><b>INPUT &amp; OUTPUT OPERATORS</b></th></tr>
<tr>