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POC++ linking and template specializations
 

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  1. POC++ linking and template specializations
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    <p>I'm studying the behavior of the C++ linker with respect to template specializations. I'm using Microsoft Visual C++ 2010 for these experiments. I don't know if the behavior is the same with other toolchains (e.g. gcc).</p> <p>Here's a first code snippet:</p> <pre class="lang-cpp prettyprint-override"><code>// bar.cpp template &lt;typename T&gt; int foo() { return 1; } int bar() { return foo&lt;double&gt;(); } // main.cpp template &lt;typename T&gt; int foo() { return 1; } template &lt;&gt; int foo&lt;double&gt;() { return 2; } int bar(); int main() { const int x = bar(); const int y = foo&lt;double&gt;(); // doesn't link } </code></pre> <p>Expectedly, this code doesn't link because <code>foo&lt;double&gt;()</code> has multiple definitions as it gets instantiated once in <em>bar.cpp</em> and once in <em>main.cpp</em> (via specialization). We would then expect, <em>if this program would link</em>, that <code>bar()</code> and <code>main()</code> would use distinct instantiations of <code>foo()</code> such that at the end we would have x&nbsp;==&nbsp;1 and y&nbsp;==&nbsp;2.</p> <p>Let's fix the link error by declaring the specialization of <code>foo&lt;double&gt;()</code> as <code>static</code>:</p> <pre class="lang-cpp prettyprint-override"><code>// bar.cpp template &lt;typename T&gt; int foo() { return 1; } int bar() { return foo&lt;double&gt;(); } // main.cpp template &lt;typename T&gt; int foo() { return 1; } template &lt;&gt; static int foo&lt;double&gt;() { return 2; } // note: static int bar(); int main() { const int x = bar(); // x == 1 const int y = foo&lt;double&gt;(); // y == 2 } </code></pre> <p>We now have x&nbsp;==&nbsp;1 and y&nbsp;==&nbsp;2, as we expected. (Note: we <em>must</em> use the <code>static</code> keyword here: an anonymous namespace won't do since we can't specialize a template function in a different namespace than its declaration.)</p> <p>Now, the use of the <code>static</code> keyword is rather unintuitive. Typically, the specialization <code>foo&lt;double&gt;()</code> would reside somewhere in a header file, and thus would be marked as inline, like in the following snippet:</p> <pre class="lang-cpp prettyprint-override"><code>// bar.cpp template &lt;typename T&gt; int foo() { return 1; } int bar() { return foo&lt;double&gt;(); } // main.cpp template &lt;typename T&gt; int foo() { return 1; } template &lt;&gt; inline int foo&lt;double&gt;() { return 2; } // note: inline int bar(); int main() { const int x = bar(); // x == 2 const int y = foo&lt;double&gt;(); // y == 2 } </code></pre> <p>This code now links properly and when we run it we get x&nbsp;==&nbsp;2 and y&nbsp;==&nbsp;2. <strong>This is the bit I find surprising: why is there a single definition of <code>foo&lt;double&gt;()</code>? What is the meaning of <code>inline</code> in this code?</strong></p> <p>A last snippet:</p> <pre class="lang-cpp prettyprint-override"><code>// bar.cpp template &lt;typename T&gt; int foo() { return 1; } int bar() { return foo&lt;double&gt;(); } // main.cpp template &lt;typename T&gt; int foo() { return 1; } template &lt;&gt; inline int foo&lt;double&gt;() { return 2; } // note: inline int bar(); int main() { const int x = bar(); // x == 1 // const int y = foo&lt;double&gt;(); // note: commented out } </code></pre> <p>This case is actually not surprising: the specialization of <code>foo&lt;double&gt;()</code> is no longer instantiated in <em>main.cpp</em> (although the declaration is still there), so the only instantiation remaining is the one in <em>bar.cpp</em>.</p>
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    1. COIn your third paragraph, the call in bar.cpp is not "implicit specialization", but rather "instantiation", since you only *call* `foo<double>()` (you're not specializing anything). Also, next sentence, "explicit specialization" should probably just be "specialization" or "total specialization", to avoid confusion with the similar sounding "explicit instantiation" (which you don't have).
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      1. USKerrek SB
    2. COI think if you `inline` a function, you promise that it has the same definition in all translation units. If that's true, then you're incurring undefined behaviour.
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    3. CO@Kerrek Actually that's not true. Inline functions are allowed to have different implementations in different translation units. They are an exception from the One Definition Rule.
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      1. USLet_Me_Be
 

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