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    copied!<p>There's a difference between normal order evaluation an lazy evaluation (as in Haskell).</p> <pre><code>square x = x * x </code></pre> <p>Evaluating the following expression...</p> <pre><code>square (square (square 2)) </code></pre> <p>... with eager evaluation:</p> <pre><code>&gt; square (square (2 * 2)) &gt; square (square 4) &gt; square (4 * 4) &gt; square 16 &gt; 16 * 16 &gt; 256 </code></pre> <p>... with normal order evaluation:</p> <pre><code>&gt; (square (square 2)) * (square (square 2)) &gt; ((square 2) * (square 2)) * (square (square 2)) &gt; ((2 * 2) * (square 2)) * (square (square 2)) &gt; (4 * (square 2)) * (square (square 2)) &gt; (4 * (2 * 2)) * (square (square 2)) &gt; (4 * 4) * (square (square 2)) &gt; 16 * (square (square 2)) &gt; ... &gt; 256 </code></pre> <p>... with lazy evaluation:</p> <pre><code>&gt; (square (square 2)) * (square (square 2)) &gt; ((square 2) * (square 2)) * ((square 2) * (square 2)) &gt; ((2 * 2) * (2 * 2)) * ((2 * 2) * (2 * 2)) &gt; (4 * 4) * (4 * 4) &gt; 16 * 16 &gt; 256 </code></pre> <p>That's because lazy evaluation looks at the syntax tree and does tree-transformations...</p> <pre><code>square (square (square 2)) || \/ * / \ \ / square (square 2) || \/ * / \ \ / * / \ \ / square 2 || \/ * / \ \ / * / \ \ / * / \ \ / 2 </code></pre> <p>... whereas normal order evaluation only does textual expansions.</p> <p>That's why we, when using lazy evaluation, get more powerful (evaluation terminates more often then other strategies) while the performance is equivalent to eager evaluation (at least in O-notation).</p>
 

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