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copied!<blockquote> <p>user=> (def l '(a 1 2)) user=> ((first l))</p> </blockquote> <p>Turn this into:</p> <blockquote> <p>user=> (def l `(~a 1 2))</p> </blockquote> <p>The ~ here resolves the symbol a to its corresponding var, and the backtick makes unquoting work.</p> <p>In general, you must understand the difference between vars (which are bound to something) and symbols (which are never bound to anything).</p> <p>I'll try to explain it (in the hope that my exaplanation does not confuse you further):</p> <pre><code>user=> (def v "content") #'user/content </code></pre> <p>-> defines a var in the current namespace under the symbol 'v (fully qualified 'user/v, assuming this is the current namespace), and binds it (the var, not the symbol) to the object "content".</p> <pre><code>user=> v "content" </code></pre> <p>-> resolves v to the var, and gets the bound value</p> <pre><code>user=> #'v #'user/v </code></pre> <p>-> resolves to the var itself</p> <pre><code>user=> 'v v </code></pre> <p>-> does not resolve anything, just a plain symbol (unfortunately, the REPL does not indicate this, printing 'v as v)</p> <pre><code>user=> `v user/v </code></pre> <p>-> as you already quoted, resolves to the symbol in the current context (namespace), but the result is still a symbol (fully qualified), not the var user/v</p> <pre><code>user=> '(v) (v) </code></pre> <p>-> plain quoting, does not resolve anything</p> <pre><code>user=> `(v) (user/v) </code></pre> <p>-> syntax-quote, same as quoting, but resolves symbols to namespace-qualified symbols</p> <pre><code>user=> `(~v) ("content") </code></pre> <p>-> resolve the symbol to its var (which is implicitely dereferenced), yielding its bound object</p>

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