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    <p>Rebol's default dialect (the <code>do</code> dialect) does not support the notion of a call to a function having a variable number of arguments. If you want to break a rule as fundamental as this, then you need your own dialect. Nothing stopping you from making:</p> <pre><code>tweet [Hello World How Are You Today?] </code></pre> <p>But the idea of using <code>word!</code> instead of <code>string!</code> in this case is a little dodgy, as many common tweets aren't valid for the Rebol parser:</p> <pre><code>tweet [LOL! :)] </code></pre> <p>Neglecting that issue, note that by default you won't get any expression evaluation. So this <code>tweet</code> dialect would have to choose a way to show where you want an evaluation. You might use get-word elements to do variable substitution, and parentheses for more general evaluations:</p> <pre><code>&gt;&gt; a: 10 &gt;&gt; b: 20 &gt;&gt; tweet [When you add :a and :b you get (a + b), LOL ":)"] "When you add 10 and 20 you get 30, LOL :)" </code></pre> <p>BTW, <code>take-n</code> in Rowland's example isn't returning a block. Not directly, I mean. It's perhaps better understood with parentheses, and by addressing the implicit "do" that's there on every interpretation:</p> <pre><code>do [do (take-n 4) 1 2 3 4] </code></pre> <p><code>take-n</code> works with only one parameter (the "n") and then returns a function which takes n parameters. Let's call that function f, so step one of this evaluation turns into something equivalent to:</p> <pre><code>do [f 1 2 3 4] </code></pre> <p>When the second do kicks in, that function gets run...and it's returning a block. In practice, I doubt you're going to want to be counting the parameters like this.</p>
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    1. PORebol simulating unlimited args any example from what is said here?
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    1. USHostileFork
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    1. PORebol simulating unlimited args any example from what is said here?
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    1. COThanks a lot hostilefork your answers are always thoughtfull, I will have a long week-end to read all of them thorougly :)
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