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    <p>It's not possible. The usual way you do it is this:</p> <pre><code>template&lt;int N&gt; struct foo { static const int value = N; }; </code></pre> <p>and for types</p> <pre><code>template&lt;typename T&gt; struct foo { typedef T type; }; </code></pre> <p>You can access it then as <code>foo&lt;39&gt;::value</code> or <code>foo&lt;int&gt;::type</code>. </p> <p>If you have a particular type, you can use partial template specialization:</p> <pre><code>template&lt;typename&gt; struct steal_it; template&lt;std::size_t N&gt; struct steal_it&lt; std::bitset&lt;N&gt; &gt; { static const std::size_t value = N; }; </code></pre> <p>The same principle is possible for type parameters too, indeed. Now you can pass any bitset to it, like <code>steal_it&lt; std::bitset&lt;16&gt; &gt;::value</code> (note to use size_t, not int!). Because we have no variadic many template paramters yet, we have to limit ourself to a particular parameter count, and repeat the steal_it template specializations for count from 1 up to N. Another difficulty is to scan types that have mixed parameters (types and non-types parameters). This is probably nontrivial to solve. </p> <p>If you have not the type, but only an object of it, you can use a trick, to still get a value at compile time:</p> <pre><code>template&lt;typename T&gt; char (&amp; getN(T const &amp;) )[steal_it&lt;T&gt;::value]; int main() { std::bitset&lt;16&gt; b; sizeof getN(b); // assuming you don't know the type, you can use the object } </code></pre> <p>The trick is to make the function template auto-deduce the type, and then return a reference to a character array. The function doesn't need to be defined, the only thing needed is its type.</p>
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    1. POExtract C++ template parameters
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    1. USJohannes Schaub - litb
    1. USJohannes Schaub - litb
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    1. POExtract C++ template parameters
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    1. CONice trick. One thing though, you start by saying how it's not possible, but then proceed to show that it is! :)
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      1. USRichard Corden
    2. CO@Richard Corden: I think he was saying it's not possible to do using RTTI
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      1. UScdleary
    3. CORichart Corden. I meant it's generally not possible to pick arbitrary template parameters. consider this: template<template<class, int> class, class, size_t> class foo; that trick won't get their types :)
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      1. USJohannes Schaub - litb
 

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