Conditionally defined variable (static if)

Question

In multiple cases I would want to use something like

template<bool condition>
struct S
{
  int value;
  if constexpr(condition) /*#if condition*/
    double my_extra_member_variable;
  /*#endif*/
}

or

if constexpr(sizeof...(Ts) != 0)
  // define something extra ( a tuple or so )

This is possible with preprocessor flags, but we want to be the "cool kids" that don't use preprocessor flags, but rather metaprogramming

It could also work with specilizations in some cases, but assume you have multiple conditions, and multiple member variables that need to be conditionally activated, it could get nasty quite fast.

I tried

constexpr in_series_production_mode = false;
struct dummy{ dummy(auto& x){}; operator=(auto& x){return *this;} }; /*1 byte*/
struct debug_data_t { /* lots of parameters with big size*/};

template <typename T>
using maybe_empty = typename std::conditional<in_series_production_mode ,T,dummy>::type; 

maybe_empty<debug_data_t> my_debugging_variable;

But with that you still get 1 byte for the unused dummy variable. While if you used #if of something similar you would have needed 0 bytes.

Does anybody know a better practice for that?

i dont understand the question. `if constexpr (...) T t;` is discarded when the condition is false. You dont have to be a "cool kid" to use it — 463035818_is_not_an_ai, Mar 09 '22 at 11:57
If you try to compile that ```template struct S { int a; if contstexpr(x>0) double b;};``` probably won't compile .. or ? — Ahmed Mahrous, Mar 09 '22 at 11:58
no. You didnt mention that it is for a member. You can specialize `S` such that `S` does not have certain members while `S` does have them — 463035818_is_not_an_ai, Mar 09 '22 at 12:00
yup, it would work fine for one specilization but if you have multiple conditions like ```if(x>2) && (x<4) ...``` and so on or you have multiple member variables/fcns ```if(x>3) double c;``` as well, it could messy quite fast — Ahmed Mahrous, Mar 09 '22 at 12:04
why would it be messy? `S` has the first member not the second but the third. Can you show how you would use the struct and its member? — 463035818_is_not_an_ai, Mar 09 '22 at 12:06
related / dupe: https://stackoverflow.com/q/15473643/4117728 — 463035818_is_not_an_ai, Mar 09 '22 at 12:07

score 4 · Accepted Answer · answered Mar 09 '22 at 12:06

4

In C++20, the "good enough" solution is with the [[no_unique_address]] attribute

struct empty_t {};

template<bool condition>
struct S {
    [[no_unique_address]]] std::conditional_t<condition, double, empty_t> var;
};

It's not perfect because var is always defined, but it will not take any space. Note that if there are multiple variables, you will need them to be different types, e.g.

template<int>
struct empty_t {};

template<bool condition>
struct S {
    [[no_unique_address]]] std::conditional_t<condition, double, empty_t<0>> var; 
    [[no_unique_address]]] std::conditional_t<condition, double, empty_t<1>> rav;
};

answered Mar 09 '22 at 12:06

Passer By

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Great idea, thanks ...^^ .. Does it have to be `empty_t` ? Would it work with `empty` with no templates as well? – Ahmed Mahrous Mar 09 '22 at 13:26
"*if there are multiple variables, you will need them to be different types*" - why? – Remy Lebeau Mar 09 '22 at 17:00
@RemyLebeau It's a simplification. `no_unique_address` actually means "maybe no unique address if the types involved are different". Two objects of the same type having the same address is the same object. That can't happen, so they have different addresses. – Passer By Mar 10 '22 at 04:47
@PasserBy [according to this](https://en.cppreference.com/w/cpp/language/attributes/no_unique_address), the types don't matter at all, only the names. – Remy Lebeau Mar 10 '22 at 16:00
@RemyLebeau See the comments in the example. Also notice how it's worded "need not", as in the attribute isn't a strict requirement. The compiler can in fact ignore the attribute completely. – Passer By Mar 10 '22 at 16:44

Conditionally defined variable (static if)

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