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    <p>In Common Lisp <code>LAMBDA</code> is two different things: a macro and a symbol which can be used in a LAMBDA expression.</p> <p>The LAMBDA expression:</p> <pre><code>(function (lambda (x) (foo x))) </code></pre> <p>shorter written as</p> <pre><code>#'(lambda (x) (foo x)) </code></pre> <p>An applied lambda expression is also valid:</p> <pre><code>((lambda (x) (+ x x)) 4) </code></pre> <p>Above both forms are part of the core syntax of Common Lisp.</p> <p>Late in the definition of Common Lisp a macro called <code>LAMBDA</code> has been added. Confusingly enough, but with good intentions. ;-) It is documented as <a href="http://www.lispworks.com/documentation/HyperSpec/Body/m_lambda.htm" rel="noreferrer">Macro LAMBDA</a>.</p> <pre><code>(lambda (x) (+ x x)) </code></pre> <p>expands into</p> <pre><code>(function (lambda (x) (+ x x)) </code></pre> <p>It makes Common Lisp code look slightly more like Scheme code and then it is not necessary to write</p> <pre><code>(mapcar #'(lambda (x) (+ x x)) some-list) </code></pre> <p>With the <code>LAMBDA</code> macro we can write</p> <pre><code>(mapcar (lambda (x) (+ x x)) some-list) </code></pre> <p>Your example fails because</p> <pre><code>((my-lambda (x) (* x x)) 2) </code></pre> <p>is not valid Common Lisp syntax.</p> <p>Common Lisp expects either</p> <ul> <li>a data object</li> <li>a variable</li> <li>a function call in the form <code>(function args...)</code></li> <li>a function call in the form <code>((lambda (arglist ...) body) args...)</code></li> <li>a macro form like <code>(macro-name forms...)</code></li> <li>a special form using one of the built-in special operators like <code>FUNCTION</code>, <code>LET</code>, ... defined in the list of <a href="http://www.lispworks.com/documentation/HyperSpec/Body/03_ababa.htm#clspecialops" rel="noreferrer">special operators</a> in Common Lisp</li> </ul> <p>As you can see a syntax of</p> <pre><code>((macro-name forms...) forms...) </code></pre> <p>is not a part of Common Lisp.</p> <p>It is possible to read the character <code>λ</code> as <code>LAMBDA</code>:</p> <pre><code>(defun λ-reader (stream char) (declare (ignore char stream)) 'LAMBDA) (set-macro-character #\λ #'λ-reader) </code></pre> <p>Example:</p> <pre><code>CL-USER 1 &gt; ((λ (x) (* x x)) 3) 9 CL-USER 2 &gt; '(λ (x) (* x x)) (LAMBDA (X) (* X X)) </code></pre>
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    1. USRainer Joswig
    1. USRainer Joswig
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    1. COIt's not the `function` in the expansion of `lambda` that is a problem, it's the hard-wired behavior of function applications with `lambda`. So in your explanation I'd change: "An applied lambda expression is also valid" into "An applied `lambda` expression is also valid (only for `lambda` and not for macros that expand to it)".
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      1. USEli Barzilay
    2. COIt makes sense now, thanks! So there's no way to get (λ ...) to emulate a lambda expression?
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    3. CO@qrl: the ((λ (x) ...) 4) syntax is possible with a read-macro in CL implementations which support that character. The Lisp reader would have to expand λ into LAMBDA.
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