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POIs there a way to get the C++ virtual member function address
 

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  1. POIs there a way to get the C++ virtual member function address
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    <p>I searched this article: <a href="https://stackoverflow.com/questions/4998318/c-getting-function-virtual-address-with-member-function-pointer">C++ : Getting function virtual &#39;address&#39; with member function pointer</a></p> <p>In order to test if the virtual member function is usually at the beginning address of the object, I wrote the code as following:</p> <pre><code>#include &lt;pwd.h&gt; #include &lt;string.h&gt; #include &lt;stdio.h&gt; class Base { public: int mBase1; char mBase2; virtual void foo() { fprintf(stderr,"Base foo called\n"); } }; class Child: public Base { public: int cc; virtual void foo() { fprintf(stderr,"Child foo called"); } }; int main(int argc, char *argv[]) { Base bb; fprintf(stderr,"[mBase1 %p][mBase2 %p] [foo %p]\n",&amp;bb.mBase1,&amp;bb.mBase2,&amp;Base::foo); return 0; } </code></pre> <p>when compiling, I got a warning:</p> <pre><code>test.cpp:30:88: warning ‘%p’ expects argument of type ‘void*’, but argument 5 has type ‘void (Base::*)()’ [-Wformat] </code></pre> <p>The output is:</p> <pre><code>[mBase1 0xbfc2ca38][mBase2 0xbfc2ca3c] [foo 0x1] </code></pre> <p>I consider it's wired. </p> <ol> <li><p>Is there any "pretty method" to get the member function (not static member function address?). Besides the following method, is there any other elegant way?</p> <pre><code>typedef void (Base::*Foo)(); Foo f = &amp;Base::foo(); </code></pre></li> <li><p>Why doesn't <code>&amp;Base::foo</code> get the correct C++ member address?</p></li> </ol>
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    1. COThe function itself will *not* be at the beginning of the object; a pointer to the v-table will (generally) be at the beginning of the object, and within the pointed to table you will find a pointer to the function (but not the function itself).
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    2. CO"Normal" pointers, including function pointers, are usually represented as machine addresses; member pointers aren't. Consider that `Derived d; (d.*f)();` *works*, but calls `Derived::Foo` as one would expect. You couldn't do that if `f` were *one* specific address.
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