series issues
[1/2] from: mgkiourt:otenet:gr at: 1-Oct-2001 20:23
Could you please elucidate the followingcode
A. do["hello"]=="hello"
do[print "hello"]==hello
do "1"+"2"==3
B.>>set[number num ten]10
==10
>>print [number num ten]
10 10 10
>>set[one two three][1 2 3]
>>print three
3
3
[2/2] from: joel::neely::fedex::com at: 1-Oct-2001 14:20
I'm not totally sure what the question(s) is(are), but I'll
try...
mgkiourt wrote:
> Could you please elucidate the followingcode
> A. do["hello"]=="hello"
> do[print "hello"]==hello
> do "1"+"2"==3
>
DO applied to a block causes all expressions in the block to
be evaluated, with the value of the last expression serving
as the value of the entire DO [...] expression.
>> do ["hello"]
== "hello"
Simply evaluating a string produces the string itself. The
double equal sign printed by the interpreter ("==") indicates
that the following output is the value of an expression.
>> do [print "hello"]
hello
The result of printing a string is that the string show up in
the output (console window, in this case). The absence of a
double equal is a clue that we're looking at *output* (a side
effect of PRINT) rather than a value. Since PRINT has no
resulting value (it is UNSET) there's no double equal at all.
>> do "1" + "2"
** Script Error: Cannot use add on string! value
** Where: halt-view
** Near: do "1" + "2"
>> do "1"+"2"
** Script Error: Cannot use add on string! value
** Where: halt-view
** Near: do "1" + "2"
None of these are legal REBOL syntax.
>> do "1 + 2"
== 3
>>
The value of DO applied to a string is obtained by LOADing the
string (converting it to an internal REBOL structure) and then
applying DO to that structure.
>> load "1 + 2"
== [1 + 2
]
LOADing the string "1 + 2" creates a block containing the
expression
1 + 2
which evaluates to 3 when DOne.
> B.>>set[number num ten]10
> ==10
<<quoted lines omitted: 6>>
> >>print[one two three]
> 1 2 3
When you give SET a block as its first argument, and a single
(non-block) value as the second argument, it SETs all words in
the first block to the same value (the second argument). In
your first expression
set [number num ten] 10
all three words were set to the integer value ten.
When you give SET two blocks, the words in the first block
are set to the corresponding values from the second block.
HTH!
-jn-
--
This sentence contradicts itself -- no actually it doesn't.
-- Doug Hofstadter
joel<dot>neely<at>fedex<dot>com
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