Preprocessor and template arguments or conditional compilation of piece of code

Question

How I can compile template function with pre-processor condition? Like that (but it is not working):

template <bool var>
void f()
{
    #if (var == true)
    // ...
    #endif
}

score 9 · Answer 1 · answered Nov 14 '12 at 11:35

9

You can't. The preprocessor, as this names indicates, processes the source file before the compiler. It has therefore no knowledge of the values of your template arguments.

answered Nov 14 '12 at 11:35

slaphappy

6,806
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score 7 · Answer 2 · answered Nov 14 '12 at 12:05

You can't do that with the preprocessor. All you can do is delegate the code to a separate template, something like this:

template <bool var>
void only_if_true()
{}

template <>
void only_if_true<true>()
{
  your_special_code_here();
}


template <bool var>
void f()
{
  some_code_always_used();
  only_if_true<var>();
  some_code_always_used();
}

Of course, if you need information shared between f() and only_if_true() (which is likely), you have to pass it as parameters. Or make only_if_true a class and store the shared data in it.

Milo Yip · Answer 3 · 2012-11-14T12:16:47.717

3

If you need to generate different code paths with template parameter, you can just simply use if or other C++ statement:

template <bool var>
void f()
{
    if (var == true) ｛
        // ...
    ｝
｝

Compiler can optimize it and generate code that doesn't contain such branches.

A little drawback is that some compiler (e.g. Msvc) will generate warnings for conditions which is always constant.

edited Nov 14 '12 at 12:16

answered Nov 14 '12 at 12:04

Milo Yip

4,742
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Unfortunately, this only works if both branches are syntactically and semantically valid. – Angew is no longer proud of SO Nov 14 '12 at 12:06
But this way is useful when the code path depending on local variables in `f()`. Using template specialization will be troublesome in this case. – Milo Yip Nov 14 '12 at 12:18
@MiloYip, your answer is right, but when I launch program in debug mode all optimizations (except inline functions) not work. – 4Bytes Nov 16 '12 at 17:29
Interested readers might like to read the answers to this question asking about these two different approaches ('deterministic `if`' vs SFINAE): http://stackoverflow.com/q/16178756/2757035 – underscore_d Jun 28 '16 at 17:35

score 1 · Answer 4 · answered Aug 25 '19 at 07:03

With C++17's introduction of if constexpr you can discard branches inside a template, much like conditional compilation allows.

template <bool var>
void f()
{
    if constexpr (var == true) {
    // ...
    }
}

The code inside the branch has to be syntactically correct, but doesn't have to be well-formed when var is false, because it will be discarded entirely.

Preprocessor and template arguments or conditional compilation of piece of code

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