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<h1 id="borrowing">Borrowing</h1>
<p>Time: 10 minutes</p>
<ul>
<li><p>Goal: convert <code>strcat</code> function below (<a href="https://play.rust-lang.org/?code=pub%20fn%20main%28%29%20%7B%0A%20%20%20%20let%20str1%20%3D%20format%21%28%22fellow%20%22%29%3B%0A%20%20%20%20let%20str2%20%3D%20format%21%28%22Rustaceans%22%29%3B%0A%20%20%20%20let%20str3%20%3D%20strcat%28str1%2C%20str2%29%3B%0A%20%20%20%20println%21%28%22%7B%7D%22%2C%20str3%29%3B%0A%7D%0A%0A%2F%2F%2F%20Concatenate%20%60suffix%60%20onto%20the%20end%20of%20%60prefix%60.%0Afn%20strcat%28mut%20prefix%3A%20String%2C%20suffix%3A%20String%29%20-%3E%20String%20%7B%0A%20%20%20%20for%20ch%20in%20suffix.chars%28%29%20%7B%0A%20%20%20%20%20%20%20%20prefix.push%28ch%29%3B%0A%20%20%20%20%7D%0A%20%20%20%20prefix%0A%7D%0A%0A%2F%2F%20Challenge%3A%20Convert%20%60strcat%60%20to%20use%20borrowing%2C%20not%20ownership.%0A%0A%2F%2F%20Question%3A%20Now%20that%20you%27ve%20converted%20%60strcat%60%2C%20what%20happens%20if%20you%0A%2F%2F%20call%20%60strcat%60%20using%20the%20same%20string%20for%20%60prefix%60%20and%20%60suffix%60%3F%0A%2F%2F%20Why%3F&version=nightly">playpen</a>) so that it uses borrowing, not ownership.</p>
<p>(As part of this, we no longer want to build up a whole new string; so get rid of <code>str3</code> and make the code just change <code>str1</code> in-place.)</p>
Hint: Getting the syntax right can be a bit tricky if you've never written in Rust before. Click on the hidden text below for some guidance.
<div id="hint2" class="hint" onclick="var h = document.getElementById('hint2'); h.style.color = (h.style.color == 'inherit') ? 'transparent' : 'inherit';">
<p>You want to change the signature of <code>join_words</code> as follows: <code>fn strcat(prefix: &mut String, suffix: &String) { ... }</code></p>
<p>Now <code>prefix</code> is a mutable reference to some <code>String</code> on the caller's side. We need a mutable reference so we can push new content onto the string.</p>
<p><code>suffix</code> is a shared reference; a shared reference suffices because we will only <em>read</em> from <code>suffix</code>.</p>
<p>Note the return value has also changed; since we are going to be mutating <code>prefix</code> in place, we no longer need to return anything.</p>
</div></li>
<li><p>Question (also in code): now that you've converted <code>strcat</code>, what happens if you call <code>strcat</code> using the same string instance for <code>prefix</code> and <code>suffix</code>, i.e. <code>strcat(s, s)</code>? Why?</p></li>
</ul>
<div class="sourceCode"><pre class="sourceCode rust"><code class="sourceCode rust"><span class="kw">pub</span> <span class="kw">fn</span> main() {
<span class="kw">let</span> str1 = <span class="pp">format!</span>(<span class="st">"fellow "</span>);
<span class="kw">let</span> str2 = <span class="pp">format!</span>(<span class="st">"Rustaceans"</span>);
<span class="kw">let</span> str3 = strcat(str1, str2);
<span class="pp">println!</span>(<span class="st">"{}"</span>, str3);
}
<span class="co">/// Concatenate `suffix` onto the end of `prefix`.</span>
<span class="kw">fn</span> strcat(<span class="kw">mut</span> prefix: <span class="dt">String</span>, suffix: <span class="dt">String</span>) -> <span class="dt">String</span> {
<span class="kw">for</span> ch <span class="kw">in</span> suffix.chars() {
prefix.push(ch);
}
prefix
}
<span class="co">// Challenge: Convert `strcat` to use borrowing, not ownership.</span>
<span class="co">// Question: Now that you've converted `strcat`, what happens if you</span>
<span class="co">// call `strcat` using the same string for `prefix` and `suffix`?</span>
<span class="co">// Why?</span></code></pre></div>
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