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    <p>It's probably not the latest, but my C++0x draft standard from June 2008 says you can do the following:</p> <pre><code>auto x = 5; // OK: x has type int const auto *v = &amp;x, u = 6; // OK: v has type const int*, u has type const int </code></pre> <p>So unless something has changed from June this is (or will be) permitted in a limited form with a pretty intuitive interpretation.</p> <p>The limitation is that if you do want to string multiple auto declarations like this (using the example above), it works because the inferred type of <code>v</code> and <code>u</code> have the same 'base type' (int in this case) to use an inexact term.</p> <p>If you want the precise rule, The draft standard says this:</p> <blockquote> <p>If the list of declarators contains more than one declarator, the type of each declared variable is determined as described above. If the type deduced for the template parameter U is not the same in each deduction, the program is ill-formed.</p> </blockquote> <p>where the "deduced template parameter U" is determined by:</p> <blockquote> <p>the deduced type of the parameter u in the call f(expr) of the following invented function template:</p> <pre><code> `template &lt;class U&gt; void f(const U&amp; u);` </code></pre> </blockquote> <p>Why they've come up with this rule instead of saying something like:</p> <pre><code>auto a = 10, b = 3.f , * c = new Class(); </code></pre> <p>is equivalent to:</p> <pre><code>auto a = 10; auto b = 3.f; auto * c = new Class(); </code></pre> <p>I don't know. But I don't write compilers. Probably something to do with once you've figured out the the <code>auto</code> keyword replaces, you can't change it in the same statement.</p> <p>Take for example:</p> <pre><code>int x = 5; CFoo * c = new CFoo(); auto a1 = x, b1 = c; // why should this be permitted if int a2 = x, CFoo* b2 = c; // this is not? </code></pre> <p>In any case, I'm not a fan of putting multiple declarations on the same statement anyway.</p>
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