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    <p>For non-member functions, a type such as <code>typedef void(Function)()</code> has several uses, but for member functions the only application is to declare a variable which holds a function pointer. Hence, other than a stylistic preference, there's no strict need to allow this syntax and it has been omitted from the standard.</p> <h3>Background</h3> <p>The <code>::</code> is a scope resolution operator, and the syntax <code>X::Y</code> is reserved for <code>static</code> member access if <code>X</code> is a class type. So <code>X::*Z</code> was another syntax invented to define <em>pointer-to-member</em>.</p> <p>Forget <em>member-function</em> for a while, just think about <em>member-data</em>, and see this code:</p> <pre><code>struct X { int a; }; int X::*pa = &amp;X::a; //pointer-to-member X x = {100}; //a = 100 cout &lt;&lt; (x.*pa) &lt;&lt; endl; </code></pre> <p>It defines a <em>pointer-to-member-data</em>, and the <code>cout</code> uses it to print the value of <code>a</code> of object <code>x</code>, and it prints:</p> <pre><code>100 </code></pre> <p>Demo : <a href="http://www.ideone.com/De2H1" rel="nofollow">http://www.ideone.com/De2H1</a></p> <p>Now think, if <code>X::pa</code> (as opposed to <code>X::*pa</code>) were allowed to do that, then you've written the above as:</p> <pre><code>int X::pa = X::a; //not &amp;X::a </code></pre> <p>Seeing this syntax, how would you tell if <code>X::a</code> is a <code>static</code> member or non-static member? That is one reason why the Standard came up with <em>pointer-to-member</em> syntax, and uniformly applies it to <em>non-static member-data</em> as well as <em>non-static member-function</em>.</p> <p>In fact, you <strong>cannot</strong> write <code>X::a</code>, you've to write <code>&amp;X::a</code>. The syntax <code>X::a</code> would result in compilation error (see <a href="http://www.ideone.com/FfGao" rel="nofollow">this</a>).</p> <hr> <p>Now extend this argument of <em>member-data</em> to <em>member-function</em>. Suppose you've a typedef defined as:</p> <pre><code>typedef void fun(); </code></pre> <p>then what do you think the following code does?</p> <pre><code>struct X { fun a; }; </code></pre> <p>Well, it defines member <code>a</code> of type <code>fun</code> (which is function taking no argument, and returning void), and is equivalent to this:</p> <pre><code>struct X { void a(); }; </code></pre> <p>Surprised? Read on.</p> <pre><code>struct X { fun a; //equivalent to this: void a(); }; void X::a() //yes, you can do this! { cout &lt;&lt; "haha" &lt;&lt; endl; } </code></pre> <p>We can use exactly the same syntax to refer to <code>a</code> which is now a member-function:</p> <pre><code>X x; x.a(); //normal function call void (X::*pa)() = &amp;X::a; //pointer-to-member (x.*pa)(); //using pointer-to-member </code></pre> <p>The similarity is the synatax on the right hand side : <code>&amp;X::a</code>. Whether <code>a</code> refers to a member-function or member-data, the syntax is same.</p> <p>Demo : <a href="http://www.ideone.com/Y80Mf" rel="nofollow">http://www.ideone.com/Y80Mf</a></p> <p><strong>Conclusion:</strong></p> <p>As we know that we cannot write <code>X::a</code> on the RHS, no matter if <code>a</code> is a member-data or member-function. The only syntax which is allowed is <code>&amp;X::f</code> which makes it necessary that the target type (on LHS) must be <em>pointer</em> as well, which in turn makes the syntax <code>void (X::*pa)()</code> absolutely necessary and fundamental, as it fits in with other syntax in the language.</p>
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    1. USNawaz
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    1. COArgument for `static` member access is quite **reasonable**. However, one thing is still to be addressed that, for member function pointer syntax we enclose it in `()` which will distinguish it with the `static` members.
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    2. CO@iammilind: I don't quite follow your comment... in the case of a member function, as in the case of a non-member function, you have to enclose with `()` which is funnily interpreted as enclosing *everything else other than what is inside the parenthesis* (niceties of the C declaration syntax) in the declaration, to distinguish the pointer to function from a function returning a pointer. I don't see where `static` members have anything to do with that set of parenthesis, or maybe I misunderstood your comment?
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      1. USDavid Rodríguez - dribeas
    3. CO@David, according to Nawaz's answer, `::` is reserved for `static` member access (when used with `class` at LHS). So it's difficult for compiler to interpret that `::` is meant for class member function pointer declaration. I agree with that. ..... However, it can be argued that member function pointer can be still distinguished by extra `()` we put across. E.g. `typedef void (Base :: Callback) ();` ... in this case we can still say that `Callback` should be meant for class members and not `static` members; because it's enclosed in `()`.
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