C++ obfuscated typedef declaration

Question

What is the meaning of this declaration? (was given during an interview) :

typedef void * (A:: *B)(char *);

Pointer to a method on class A returning a pointer to void and taking a pointer to char. — Richard J. Ross III, Jan 21 '14 at 11:39
All your answers are clear and concise, I think I was blinded by the pointer to member function here. — Laurent, Jan 21 '14 at 11:48

score 10 · Accepted Answer · answered Jan 21 '14 at 11:39

10

Define B as a pointer to member-function of class A, which receives char* and returns void*.

answered Jan 21 '14 at 11:39

ForEveR

@RichardJ.RossIII: C++11 uses `using B = void* (A::*)(char *);`. It is no so *old*. – Jarod42 Jan 21 '14 at 12:22

score 5 · Answer 2 · edited May 23 '17 at 12:28

5

Declare B as the type of signature of member functions of class A getting a char* argument and returning a void* pointer.

I don't feel it is obfuscated. It permits much more readable code. See this answer (for C, but you could adapt it for C++).

edited May 23 '17 at 12:28

Community

answered Jan 21 '14 at 11:39

Ok maybe it is not the correct word, let say it is not natural for some people to read it. – Laurent Jan 21 '14 at 11:42
Well, avoiding such a `typedef` makes less readable code IMHO. – Basile Starynkevitch Jan 21 '14 at 11:43
sure, I didnt meant the typedef was adding complexity to reading – Laurent Jan 21 '14 at 11:46

2 Answers2