Introduction template<int F(), int N = F()> void func (); In this answer we will go through the relevant sections of the International Standard, step by step, to prove that the above snippet is well-formed. What does the International Standard (N3337) say? The Standardese 14.1p9 Template parameters [temp.param] A default template-argument is a template-argument (14.3) … Read more
Update: The situation got a bit more complicated after the resolution of CWG DR 1684 removed the requirement quoted below. Some more information can be found in this discussion on the std-discussion mailing list and in the related question Why isn’t `std::initializer_list` defined as a literal type? [decl.constexpr]/8: A constexpr specifier for a non-static member … Read more
A constexpr function does NOT have to be defined before its first use, however the result of any call made prior to definition is not a constant expression. Source: C++ Standard draft n4296, section 5.20: A conditional-expression e is a core constant expression unless the evaluation of e, following the rules of the abstract machine, … Read more
Unfortunately, even x86-64 implementations can differ – from Intel’s instruction set reference,BSF and BSR, with a source operand value of (0), leaves the destination undefined, and sets the ZF (zero flag). So the behaviour may not be consistent between micro-architectures or, say, AMD and Intel. (I believe AMD leaves the destination unmodified.) The newer LZCNT … Read more
Aren’t they basically the same? No. Absolutely not. Not even close. Apart from the fact your macro is an int and your constexpr unsigned is an unsigned, there are important differences and macros only have one advantage. Scope A macro is defined by the preprocessor and is simply substituted into the code every time it … Read more
Is there a way to cause a compile-time error with a constexpr function, but not do anything at run time? You can use the exact same trick, but instead of using a throw-expression, use an expression that is not a constant expression but does what you want at runtime. For instance: int runtime_fallback(int x) { … Read more
Constant variables of integral types with constant initializers are integral constant expressions (de facto implicitely constexpr; see expr.const in ISO C++). float is not an integral type and does not meet the requirements for constant expression without the use of constexpr. (A similar case is why int can be but float cannot be a template … Read more
The standard doesn’t say much about the discarded statement of an if constexpr. There are essentially two statements in [stmt.if] about these: In an enclosing template discarded statements are not instantiated. Names referenced from a discarded statement are not required ODR to be defined. Neither of these applies to your use: the compilers are correct … Read more
We can have our cake and eat it to, too. Let’s just consider the simplest specialization of copy, the one that copies chars. In C++17, that might look like: char* copy(char const* first, char const* last, char* d) { memcpy(d, first, last – first); return d + (last – first); } Of course, we can’t … Read more
At compile time, the compiler doesn’t know the address of sqrt. However, you cannot do anything at compile time with a constexpr function pointer that would allow you to access that pointer’s address. Therefore, a function pointer at compile time can be treated as an opaque value. And since you can’t change a constexpr variable … Read more