Consider the following piece of code:
struct foo {
static constexpr const void* ptr = reinterpret_cast<const void*>(0x1);
};
auto main() -> int {
return 0;
}
The above example compiles fine in g++ v4.9 (Live Demo), while it fails to compile in clang v3.4 (Live Demo) and generates the following error:
error: constexpr variable 'ptr' must be initialized by a constant expression
Questions:
Which of the two compilers is right according to the standard?
What's the proper way of declaring an expression of such kind?
A constexpr variable must be initialized at compile time. All constexpr variables are const . A variable can be declared with constexpr , when it has a literal type and is initialized. If the initialization is performed by a constructor, the constructor must be declared as constexpr .
const applies for variables, and prevents them from being modified in your code. constexpr tells the compiler that this expression results in a compile time constant value, so it can be used in places like array lengths, assigning to const variables, etc.
A static constexpr variable has to be set at compilation, because its lifetime is the the whole program. Without the static keyword, the compiler isn't bound to set the value at compilation, and could decide to set it later. So, what does constexpr mean?
A constructor that is declared with a constexpr specifier is a constexpr constructor. Previously, only expressions of built-in types could be valid constant expressions. With constexpr constructors, objects of user-defined types can be included in valid constant expressions.
TL;DR
clang
is correct, this is known gcc
bug. You can either use intptr_t
instead and cast when you need to use the value or if that is not workable then both gcc
and clang
support a little documented work-around that should allow your particular use case.
Details
So clang
is correct on this one if we go to the draft C++11 standard section 5.19
Constant expressions paragraph 2 says:
A conditional-expression is a core constant expression unless it involves one of the following as a potentially evaluated subexpression [...]
and includes the following bullet:
— a reinterpret_cast (5.2.10);
One simple solution would be to use intptr_t:
static constexpr intptr_t ptr = 0x1;
and then cast later on when you need to use it:
reinterpret_cast<void*>(foo::ptr) ;
It may be tempting to leave it at that but this story gets more interesting though. This is know and still open gcc
bug see Bug 49171: [C++0x][constexpr] Constant expressions support reinterpret_cast. It is clear from the discussion that gcc
devs have some clear use cases for this:
I believe I found a conforming usage of reinterpret_cast in constant expressions useable in C++03:
//---------------- struct X { X* operator&(); }; X x[2]; const bool p = (reinterpret_cast<X*>(&reinterpret_cast<char&>(x[1])) - reinterpret_cast<X*>(&reinterpret_cast<char&>(x[0]))) == sizeof(X); enum E { e = p }; // e should have a value equal to 1 //----------------
Basically this program demonstrates the technique, the C++11 library function addressof is based on and thus excluding reinterpret_cast unconditionally from constant expressions in the core language would render this useful program invalid and would make it impossible to declare addressof as a constexpr function.
but were not able to get an exception carved for these use cases, see closed issues 1384:
Although reinterpret_cast was permitted in address constant expressions in C++03, this restriction has been implemented in some compilers and has not proved to break significant amounts of code. CWG deemed that the complications of dealing with pointers whose tpes changed (pointer arithmetic and dereference could not be permitted on such pointers) outweighed the possible utility of relaxing the current restriction.
BUT apparently gcc
and clang
support a little documented extension that allows constant folding of non-constant expressions using __builtin_constant_p (exp) and so the following expressions is accepted by both gcc
and clang
:
static constexpr const void* ptr =
__builtin_constant_p( reinterpret_cast<const void*>(0x1) ) ?
reinterpret_cast<const void*>(0x1) : reinterpret_cast<const void*>(0x1) ;
Finding documentation for this is near impossible but this llvm commit is informative with the following snippets provide for some interesting reading:
support the gcc __builtin_constant_p() ? ... : ... folding hack in C++11
and:
// __builtin_constant_p ? : is magical, and is always a potential constant.
and:
// This macro forces its argument to be constant-folded, even if it's not // otherwise a constant expression. #define fold(x) (__builtin_constant_p(x) ? (x) : (x))
We can find a more formal explanation of this feature in the gcc-patches email: C constant expressions, VLAs etc. fixes which says:
Furthermore, the rules for __builtin_constant_p calls as conditional expression condition in the implementation are more relaxed than those in the formal model: the selected half of the conditional expression is fully folded without regard to whether it is formally a constant expression, since __builtin_constant_p tests a fully folded argument itself.
Clang is right. The result of a reinterpret-cast is never a constant expression (cf. C++11 5.19/2).
The purpose of constant expressions is that they can be reasoned about as values, and values have to be valid. What you're writing is not provably a valid pointer (since it's not the address of an object, or related to the address of an object by pointer arithmetic), so you're not allowed to use it as a constant expression. If you just want to store the number 1
, store it as a uintptr_t
and do the reinterpret cast at the use site.
As an aside, to elaborate a bit on the notion of "valid pointers", consider the following constexpr
pointers:
constexpr int const a[10] = { 1 };
constexpr int * p1 = a + 5;
constexpr int const b[10] = { 2 };
constexpr int const * p2 = b + 10;
// constexpr int const * p3 = b + 11; // Error, not a constant expression
static_assert(*p1 == 0, ""); // OK
// static_assert(p1[5] == 0, ""); // Error, not a constant expression
static_assert(p2[-2] == 0, ""); // OK
// static_assert(p2[1] == 0, ""); // Error, "p2[1]" would have UB
static_assert(p2 != nullptr, ""); // OK
// static_assert(p2 + 1 != nullptr, ""); // Error, "p2 + 1" would have UB
Both p1
and p2
are constant expressions. But whether the result of pointer arithmetic is a constant expression depends on whether it is not UB! This kind of reasoning would be essentially impossible if you allowed the values of reinterpret_casts to be constant expressions.
I have also been running into this problem when programming for AVR microcontrollers. Avr-libc has header files (included through <avr/io.h>
that make available the register layout for each microcontroller by defining macros such as:
#define TCNT1 (*(volatile uint16_t *)(0x84))
This allows using TCNT1
as if it were a normal variable and any reads and writes are directed to memory address 0x84 automatically. However, it also includes an (implicit) reinterpret_cast
, which prevents using the address of this "variable" in a constant expression. And since this macro is defined by avr-libc, changing it to remove the cast is not really an option (and redefining such macros yourself works, but then requires defining them for all the different AVR chips, duplicating the info from avr-libc).
Since the folding hack suggested by Shafik here seems to no longer work in gcc 7 and above, I have been looking for another solution.
Looking at the avr-libc header files more closely, it turns out they have two modes:
- Normally, they define variable-like macros as shown above.
- When used inside the assembler (or when included with _SFR_ASM_COMPAT
defined), they define macros that just contain the address, e.g.:
#define TCNT1 (0x84)
At first glance the latter seems useful, since you could then set _SFR_ASM_COMPAT
before include <avr/io.h>
and simply use intptr_t
constants and use the address directly, rather than through a pointer. However, since you can include the avr-libc header only once (iow, only have TCNT1
as either a variable-like-macro, or an address), this trick only works inside a source file that does not include any other files that would need the variable-like-macros. In practice, this seems unlikely (though maybe you could have constexpr (class?) variables that are declared in a .h file and assigned a value in a .cpp file that includes nothing else?).
In any case, I found another trick by Krister Walfridsson, that defines these registers as external variables in a C++ header file and then defines them and locates them at a fixed location by using an assembler .S file. Then you can simply take the address of these global symbols, which is valid in a constexpr expressions. To make this work, this global symbol must have a different name as the original register macro, to prevent a conflict between both.
E.g. in your C++ code, you would have:
extern volatile uint16_t TCNT1_SYMBOL;
struct foo {
static constexpr volatile uint16_t* ptr = &TCNT1_SYMBOL;
};
And then you include a .S file in your project that contains:
#include <avr/io.h>
.global TCNT1_SYMBOL
TCNT1_SYMBOL = TCNT1
While writing this, I realized the above is not limited to the AVR-libc case, but can also be applied to the more generic question asked here. In that case, you could get a C++ file that looks like:
extern char MY_PTR_SYMBOL;
struct foo {
static constexpr const void* ptr = &MY_PTR_SYMBOL;
};
auto main() -> int {
return 0;
}
And a .S file that looks like:
.global MY_PTR_SYMBOL
MY_PTR_SYMBOL = 0x1
Here's how this looks: https://godbolt.org/z/vAfaS6 (I could not figure out how to get the compiler explorer to link both the cpp and .S file together, though
This approach has quite a bit more boilerplate, but does seem to work reliably across gcc and clang versions. Note that this approach looks like a similar approach using linker commandline options or linker scripts to place symbols at a certain memory address, but that approach is highly non-portable and tricky to integrate in a build process, while the approach suggested above is more portable and just a matter of adding a .S file into the build.
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