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    copied!<p>Okay, let's step through your program line by line. I'm also assigning unique numbers (think of them as object addresses, if you're used to C-like languages) for each newly-created object so you can see what's what. :-)</p> <pre><code>(define x (list 1 2 3)) ; =&gt; #1 = (1 . #2), #2 = (2 . #3), #3 = (3 . ()) (define y (list 4 5)) ; =&gt; #4 = (4 . #5), #5 = (5 . ()) (define z (cons (car x) (cdr y))) ; =&gt; #6 = (1 . #5) (define w (append y z)) ; =&gt; #7 = (4 . #8), #8 = (5 . #6) (define v (cons (cdr x) (cdr y))) ; =&gt; #9 = (#2 . #5) (set-car! x 6) ; =&gt; #1 = (6 . #2) (set-car! y 7) ; =&gt; #4 = (7 . #5) (set-cdr! (cdr x) (list 8)) ; =&gt; #2 = (2 . #10), #10 = (8 . ()) </code></pre> <p>Now, let's look at your values (for each reference, use the last assigned value):</p> <pre><code>x ; #1 =&gt; (6 . #2) =&gt; (6 . (2 . #10)) =&gt; (6 2 8) y ; #4 =&gt; (7 . #5) =&gt; (7 5) z ; #6 =&gt; (1 . #5) =&gt; (1 5) w ; #7 =&gt; (4 . #8) =&gt; (4 . (5 . #6)) =&gt; (4 . (5 . (1 . #5))) =&gt; (4 5 1 5) v ; #9 =&gt; (#2 . #5) =&gt; ((2 . #10) 5) =&gt; ((2 8) 5) </code></pre> <hr> <p>Edit: I'm adding a diagram to explain my answer, since you can't have diagrams in a comment. I don't have time to make a diagram showing the values above, but this hopefully explains a few things.</p> <p><img src="https://i.stack.imgur.com/YoHf6.png" alt="Expression tree"></p> <p>Each pair has two "slots", <code>car</code> and <code>cdr</code>, represented as the left and right boxes in the diagram above. Each of these slots, as you see, has three possible things:</p> <ol> <li>An atom (a number in your examples, or symbols in the diagram, such as <code>let</code>, <code>s5</code>, and <code>sqrt</code>)</li> <li>A reference (represented as an arrow in the diagram)</li> <li>Null (represented as a black box in the diagram)</li> </ol> <p>You can put any of these in any of the slots. So, in my explanation above, each of the <code>#</code> items is an arrow, each of the non-<code>#</code> numbers is an atom, and each of the <code>()</code> is a black box. So, in the line</p> <pre><code>(define v (cons (cdr x) (cdr y))) </code></pre> <p>you are creating a pair, where the left-hand slot has the same contents as the right-hand slot of <code>x</code> (i.e., an arrow going to pair 2), and the right-hand slot has the same contents as the right-hand slot of <code>y</code> (an arrow going to pair 5). In other words, both boxes in <code>v</code> contain arrows, each going off a different pair.</p> <p>Hope this makes more sense. :-)</p>
 

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