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When I declare a class static method, is it possible to refer current class using decltype (or in any other similar style)? For example,

class
    static AAA const make();
I am trying to make something like this.
class
    static decltype(*this) const make();  // Not working because there's no `this`.
The *this is used to describe what I want to do. I want to know some decltype() expression which can be resolved to AAA.
If it's possible how can I do that?
                I don't know why you want to trade the clear and concise form of static AAA const make() for the indirect static decltype(*this) const make(). Readability is king, and being explicit about your types helps your readers a lot.
– cmaster - reinstate monica
                Nov 25, 2013 at 21:46
                Is the idea behind this to add a make function to a whole hierarchy of classes in a particular "smart" way? That'd be a case for the curiously recurring template parameter pattern. If this is only for one class, I don't understand the need to derive the return type, AAA would do fine (but then, I don't understand what's wrong with calling the constructor, either).
– Damon
                Nov 26, 2013 at 0:23
                I don't have specific instance of use-case. I suddenly am curious on this, and if this is possible, I think maybe I can figure out some interesting applications later. I don't claim any goodness of badness of idea in this question. I also thought this wouldn't that much useful, but this question was bugging me for a long time, so I decided to ask this.
– eonil
                Nov 26, 2013 at 15:00
This is not legal in C++11 as you need to specify a return type for make().  In C++98/03/11 you can:
class
public:
    typedef AAA Self;
    static Self make()
        return AAA();
That is low-tech, but very readable.
<aside>
You should avoid returning const-qualified types by value.  This inhibits efficient move semantics.  If you want to avoid assigning to rvalues, then create an assignment operator qualified with &.
</aside>
   AAA(int i): _i(i) {}
   static auto make(int i) -> typename std::remove_reference<decltype(*this)>::type
      return {i};
   void print()
      std::cout << _i << std::endl;
int main()
    AAA aaa = AAA::make(1);
    aaa.print();
    return 0;
It compiles on GCC 4.7.2 at least :) 
EDIT 26/11-13: The above code is not legal c++, even though it compiles with gcc, it does not with clang or icpc. My apologies.
                User hvd already made a similar suggestion (now deleted), but it's not allowed: [expr.prim.general]/3 about the expression this: "It shall not appear before the optional cv-qualifier-seq and it shall not appear within the declaration of a static member function"
– dyp
                Nov 26, 2013 at 11:34
Just invented a way using member pointers, and it seems to work:
https://godbolt.org/z/v-g5z0
    #define self_test(name) \
    void dummy__##name(void) {} \
    template  static T type__##name( void (T::*func)(void) ) { return T(); } \
    typedef decltype(type__##name(&dummy__##name)) self__##name; \
    static void test( void ) { self__##name::print(); }
    struct T1 {
      self_test(X);
      static void print() { printf( "this is T1\n" ); }
    struct T2 {
      self_test(X);
      static void print() { printf( "this is T2\n" ); }
    int main() {
      T1::test();
      T2::test();
This is not perfect either, it requires -fpermissive to compile with gcc,
but at least gcc/VS/Intel all compile it and it works.
Also in fact gcc/mingw doesn't even require -fpermissive for this.
        
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