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    <p>Why using a template?</p> <pre><code>#include &lt;cstdio&gt; struct Foo { int a; int b; } foo = {2, 3}; int const (Foo::*mp) = &amp;Foo::b; int main() { printf("%d\n", foo.*mp); return 0; } </code></pre> <p>The following compiles <code>mp</code> to this on gcc-4.4.1 (I don't have access to MSVC at the moment):</p> <pre><code>.globl mp .align 4 .type mp, @object .size mp, 4 mp: .long 4 </code></pre> <p>It is just an offset to the member, which looks pretty compile-time to me.</p> <p>With template, you need to specify the definition outside of the class:</p> <pre><code>#include &lt;cstdio&gt; struct Foo { int m; int r; } foo = {2, 3}; template&lt;int Foo::*Mem&gt; struct B { static int Foo::* const mp; }; template&lt;int Foo::*Mem&gt; int Foo::* const B&lt;Mem&gt;::mp = Mem; int main() { typedef B&lt;&amp;Foo::m&gt; Bm; typedef B&lt;&amp;Foo::r&gt; Br; printf("%d, %d\n", foo.*(Bm::mp), foo.*(Br::mp)); } </code></pre> <p>Which compiles to:</p> <pre><code>g++ -O2 -S -o- b.cc | c++filt ... .weak B&lt;&amp;(Foo::r)&gt;::mp .section .rodata._ZN1BIXadL_ZN3Foo1rEEEE2mpE,"aG",@progbits,B&lt;&amp;(Foo::r)&gt;::mp,comdat .align 4 .type B&lt;&amp;(Foo::r)&gt;::mp, @object .size B&lt;&amp;(Foo::r)&gt;::mp, 4 B&lt;&amp;(Foo::r)&gt;::mp: .long 4 .weak B&lt;&amp;(Foo::m)&gt;::mp .section .rodata._ZN1BIXadL_ZN3Foo1mEEEE2mpE,"aG",@progbits,B&lt;&amp;(Foo::m)&gt;::mp,comdat .align 4 .type B&lt;&amp;(Foo::m)&gt;::mp, @object .size B&lt;&amp;(Foo::m)&gt;::mp, 4 B&lt;&amp;(Foo::m)&gt;::mp: .zero 4 </code></pre> <p>However this all smacks of standard library features reimplementation (see <code>std::tr1::mem_fn</code>).</p>
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    1. POHow to save pointer to member in compile time?
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    1. COYou just do not understand my question.
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    2. COThen please clarify your question, a lot of people seem to be in the same boat.
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    3. CODo not be offended. I want to save pointer to member in template class as a type or an integer. I need *a pointer to member traits* as type traits.
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