</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>
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 >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 >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 >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 >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 & OUTPUT OPERATORS</b></th></tr>
<tr>
src="http://www.w3.org/Icons/valid-html401" alt="Valid HTML 4.01!"></a>
<a href="mailto:rspencer@x10sys.com">Reid Spencer</a><br>
- <a href="http://llvm.cs.uiuc.edu">LLVM Compiler Infrastructure</a><br>
+ <a href="http://llvm.org">LLVM Compiler Infrastructure</a><br>
Last modified: $Date$
</address>