C++ 11, how to use a const inside an #if

Question

I use a const variable in my code. And would like to tell my pre-processor to use it. Namely:

const double x = 1.2;
const bool xIsZero = x==0;

#if xIsZero
...
#endif

But that doesn't work. In C++ 17, if constexpr doest the trick. But I'm stuck with C++11 for now.

So instead I could use a workaround like:

#define X 1.2
#define xIsZero (x==0)
const double x = X;

#if xIsZero
...
#endif

But I just dislike giving x to a #define, I'd like to give it directly to a const. Is there a way to do that?

Answer 1

If the condition is known at compile-time, you can use overloading to mimic if constexpr in C++11:

void foo(std::true_type) {   // if (xIsZero)
}

void foo(std::false_type) {  // if (!xIsZero)
}

constexpr bool xIsZero = ...;
foo(std::integral_constant<bool, xIsZero>{});

---

As LoPiTaL noted in comments, this is not fully equivalent to if constexpr, because both foos have to compile. This technique becomes important when we work with templates. With plain if both branches are compiled with the same set of template parameters. With overloading we can effectively discard branches.

For example, this code fails:

template<unsigned int i>
void foo(std::integral_constant<unsigned int, i>) {
    if (i > 0) {
        consume_positive_i(i);
        foo(std::integral_constant<unsigned int, i - 1>{});
    } else
        consume_zero_i(i);
}

In C++17 it can easily be fixed with if constexpr:

template<unsigned int i>
void foo(std::integral_constant<unsigned int, i>) {
    if constexpr (i > 0) {
        consume_positive_i(i);
        foo(std::integral_constant<unsigned int, i - 1>{});
    } else
        consume_zero_i(i);
}

The workaround in C++11 is overloading:

void foo(std::integral_constant<unsigned int, 0>) {   
    consume_zero_i(i);
}

template<unsigned int i>
void foo(std::integral_constant<unsigned int, i>) {
    consume_positive_i(i);
    foo(std::integral_constant<unsigned int, i - 1>{});
}