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lecture 10: fix typo in definition of foldl'
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@@ -1357,7 +1357,7 @@ square ==> * ==> * ==> 16
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<div class="sourceCode" id="cb93"><pre class="sourceCode haskell"><code class="sourceCode haskell"><span id="cb93-1"><a href="#cb93-1" aria-hidden="true" tabindex="-1"></a><span class="ot">foldl' ::</span> (a <span class="ot">-></span> b <span class="ot">-></span> a) <span class="ot">-></span> a <span class="ot">-></span> [b] <span class="ot">-></span> a</span>
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<span id="cb93-2"><a href="#cb93-2" aria-hidden="true" tabindex="-1"></a>foldl' f z [] <span class="ot">=</span> z</span>
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<span id="cb93-3"><a href="#cb93-3" aria-hidden="true" tabindex="-1"></a>foldl' f z (x<span class="op">:</span>xs) <span class="ot">=</span> <span class="kw">let</span> z' <span class="ot">=</span> f z x</span>
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<span id="cb93-4"><a href="#cb93-4" aria-hidden="true" tabindex="-1"></a> <span class="kw">in</span> <span class="fu">seq</span> z' (<span class="fu">foldl</span> f z' xs)</span></code></pre></div>
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<span id="cb93-4"><a href="#cb93-4" aria-hidden="true" tabindex="-1"></a> <span class="kw">in</span> <span class="fu">seq</span> z' (foldl' f z' xs)</span></code></pre></div>
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<div class="sourceCode" id="cb94"><pre class="sourceCode haskell"><code class="sourceCode haskell"><span id="cb94-1"><a href="#cb94-1" aria-hidden="true" tabindex="-1"></a> foldl' (<span class="op">+</span>) <span class="dv">0</span> [<span class="dv">1</span>,<span class="dv">2</span>,<span class="dv">3</span>]</span>
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<span id="cb94-2"><a href="#cb94-2" aria-hidden="true" tabindex="-1"></a><span class="op">==></span> <span class="fu">seq</span> (<span class="dv">0</span><span class="op">+</span><span class="dv">1</span>) (foldl' (<span class="op">+</span>) (<span class="dv">0</span><span class="op">+</span><span class="dv">1</span>) [<span class="dv">2</span>,<span class="dv">3</span>]) <span class="co">-- seq forces first argument</span></span>
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<span id="cb94-3"><a href="#cb94-3" aria-hidden="true" tabindex="-1"></a> <span class="op">|</span> <span class="op">|</span></span>
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