4 ms·

Is there no way to indicate to GCC when you want something fully evaluated at compile-time? Maybe some pragmas? 98% of the time I probably don't care whether GC

Is there no way to indicate to GCC when you want something fully evaluated at compile-time? Maybe some pragmas? 98% of the time I probably don't care whether GCC does a good job optimizing this properly, but for the times I do care: what should I do?

lultimouomo 10y ago
c++ has constexpr for this. a constexpr expression will be evaluated at compile time if the result is used in a constant expression (i.e. assignd to an enum constant or a constexpr variable), or an error will be generated if it is not possible to do so (the expression does not fulfill the requirements to be constexpr).
- krona 10y ago
  working example: https://godbolt.org/g/EXdQx1 https://godbolt.org/g/EXdQx1
  - TillE 10y ago
    Your example still gets calculated at compile time if you remove the constexpr. However, if you change -O3 to -O2, then there's a difference.
matt_d 10y ago
A `constexpr` function storing result in a `constexpr` variable (you need both for the compile-time evaluation guarantee): http://en.cppreference.com/w/cpp/language/constexpr http://en.cppreference.com/w/cpp/language/constexpr An extra complication in this example is that `std::accumulate` currently lacks `constexpr` support: http://stackoverflow.com/questions/32395408/why-arent-stdalgorithms-constexpr-and-which-could-be http://stackoverflow.com/questions/32395408/why-arent-stdalg... There's been a proposal to relax this limitation: http://www.open-std.org/jtc1/sc22/wg21/docs/papers/2016/p0202r0.html http://www.open-std.org/jtc1/sc22/wg21/docs/papers/2016/p020... For now you could use a workaround in form of a range-based for loop: https://godbolt.org/g/aVEqlm https://godbolt.org/g/aVEqlm or even https://godbolt.org/g/GDrFr5 https://godbolt.org/g/GDrFr5 (homogeneous types). A fold expression is much cleaner, though (also supporting heterogeneous types): https://godbolt.org/g/Hrnkc1 https://godbolt.org/g/Hrnkc1
andrepd 10y ago
Use constexpr: http://en.cppreference.com/w/cpp/language/constexpr http://en.cppreference.com/w/cpp/language/constexpr You can't make sum constexpr in this case, though, since it calls a non-constexpr function, accumulate.

lultimouomo 10y ago

c++ has constexpr for this. a constexpr expression will be evaluated at compile time if the result is used in a constant expression (i.e. assignd to an enum constant or a constexpr variable), or an error will be generated if it is not possible to do so (the expression does not fulfill the requirements to be constexpr).

krona 10y ago
working example: https://godbolt.org/g/EXdQx1 https://godbolt.org/g/EXdQx1
- TillE 10y ago
  Your example still gets calculated at compile time if you remove the constexpr. However, if you change -O3 to -O2, then there's a difference.

matt_d 10y ago

A `constexpr` function storing result in a `constexpr` variable (you need both for the compile-time evaluation guarantee): http://en.cppreference.com/w/cpp/language/constexpr http://en.cppreference.com/w/cpp/language/constexpr An extra complication in this example is that `std::accumulate` currently lacks `constexpr` support: http://stackoverflow.com/questions/32395408/why-arent-stdalgorithms-constexpr-and-which-could-be http://stackoverflow.com/questions/32395408/why-arent-stdalg... There's been a proposal to relax this limitation: http://www.open-std.org/jtc1/sc22/wg21/docs/papers/2016/p0202r0.html http://www.open-std.org/jtc1/sc22/wg21/docs/papers/2016/p020... For now you could use a workaround in form of a range-based for loop: https://godbolt.org/g/aVEqlm https://godbolt.org/g/aVEqlm or even https://godbolt.org/g/GDrFr5 https://godbolt.org/g/GDrFr5 (homogeneous types). A fold expression is much cleaner, though (also supporting heterogeneous types): https://godbolt.org/g/Hrnkc1 https://godbolt.org/g/Hrnkc1

andrepd 10y ago

Use constexpr: http://en.cppreference.com/w/cpp/language/constexpr http://en.cppreference.com/w/cpp/language/constexpr You can't make sum constexpr in this case, though, since it calls a non-constexpr function, accumulate.