In C++17 we got inline variables and I have assumed that global constexpr variables are implicitly inline.
But apparently this is true only for static member variables.
What is the logic/technical limitation behind this?
source:
A static member variable (but not a namespace-scope variable) declared constexpr is implicitly an inline variable.
The reason why constexpr static data members were made implicitly inline was to solve a common problem in C++: when defining a class-scoped constant, one was previously forced to emit the definition in exactly one translation unit, lest the variable be ODR-used:
// foo.h
struct foo {
static constexpr int kAnswer = 42;
};
// foo.cpp
// a linker error will occur if this definition is omitted before C++17
#include "foo.h"
constexpr int foo::kAnswer;
// main.cpp
#include "foo.h"
#include <vector>
int main() {
std::vector<int> bar;
bar.push_back(foo::kAnswer); // ODR-use of 42
}
In such cases, we usually care only about the value of the constant, not its address; and it's convenient for the compiler to synthesize a unique location for the constant in case it really is ODR-used, but we don't care where that location is.
Thus, C++17 changed the rules so that the out-of-line definition is no longer required. In order to do so, it makes the declaration of foo::kAnswer an inline definition, so that it can appear in multiple translation units without clashing, just like inline functions.
For namespace-scope constexpr variables (which are implicitly static, and therefore have internal linkage, unless declared extern) there is no similar issue. Each translation unit has its own copy. inline, as it's currently specified, would have no effect on such variables. And changing the existing behaviour would break existing programs.
The point here is that constexpr int x = 1; at namespace scope has internal linkage in C++14.
If you make it implicitly inline without changing the internal linkage part, the change would have no effect, because the internal linkage means that it can't be defined in other translation units anyway. And it harms teachability, because we want things like inline constexpr int x = 1; to get external linkage by default (the whole point of inline, after all, is to permit the same variable to be defined in multiple translation units).
If you make it implicitly inline with external linkage, then you break existing code:
// TU1
constexpr int x = 1;
// TU2
constexpr int x = 2;
This perfectly valid C++14 would become an ODR violation.
Related
This question already has answers here:
constexpr variable at namespace scope with and without explicit inline definition
(2 answers)
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I know global constexpr variables have internal linkage. so how is it that inline constexpr are introduced with having external linkage? does adding inline just converts internal linakges to external linkages in all cases?
There seems to be a little bit of confusion about what "linkage" and "inline" actually means. They are independent (orthogonal) properties of a variable, but nevertheless coupled together.
To inline a variable one declares it inline. Declaring a constexpr variable at namescope does not imply inline [1]. To declare a variable to have internal linkage one declares it static or more preferrable puts it into an anonymous namespace [2],[3]. For const and constexpr (which implies const) variables there is a special rule, which gives them internal linkage as long as they are non-inline [4].
Because constexpr variables require an immediate definition [5], you typically want them to be be inline which allows multiple (equivalent) definitions in multiple translation units:
\\ c.hpp
inline constexpr int c = 0; // define it in header
\\ a.cpp
#include "c.hpp" // c is defined in a.cpp
int a = c; // use it
\\ b.cpp
#include "c.hpp" // c is re-defined in b.cpp
int b = c; // use it
The linkage of c in that example above is external, because the special rule for const variables only applies to non-inline variables.
Note that when ommiting the inline specifier in the example makes each source file get an independent non-inline definition of c with internal linkage. It will still compile but you have to be careful to not use c in any inline functions [6].
You can put inline constexpr variables into an anonymous namespace or declare it static to make its linkage internal. If we changed the example above into
\\ c.hpp
namespace {
inline constexpr int c = 0;
};
\\ a.cpp
...
the effects would be almost the same as if ommitin the inline in the original example. Each translation unit gets its own version of the (now inlined) variable and you have to make sure that you don't use c in an inline function.
inline variable or function make compiler merge* multiple definition into one.
for the same reason, multiple inline constexpr with same name would has only one instance after link.
then you're accessing the variable in other TU, it's effectively has external linkage.
* it's undefined behavior if the definition are not the same though.
** you cannot declare extern constexpr, btw
I know global constexpr variables have internal linkage
You are missing a few qualifiers (emphasis mine):
internal linkage
Any of the following names declared at namespace scope have internal
linkage:
...
non-volatile non-template (since C++14) non-inline (since C++17) non-exported (since C++20) const-qualified variables (including
constexpr) (since C++11) that aren't declared extern and aren't
previously declared to have external linkage;
...
If I define a function in my program.cpp:
constexpr bool isThree(const int number)
{
return number == 3;
}
is that any different from declaring it static?
static constexpr bool isThree(const int number)
{
return number == 3;
}
It seems that those should be equivalent, since constexpr means the function is inline and therefore not shared among compilation units.
Are constexpr global functions implicitly static?
constexpr functions are implicitly inline.
inline is a linking feature. An inline function with definitions in different compilation units is not an error; if their definitions vary, your program is ill-formed no diagnostic required, but if they have the same definition then all but one version is discarded and that version is used.
static, on a non-method function, is also a linking feature. A static definition is not shared outside of its compilation unit; the compilation unit does not 'advertise' that it has a definition for isThree.
static on a method function has nothing to do with linking. In that case, it just means that this is not implicitly passed to the function. A method with/without this it doesn't work has differences, but they are mostly unrelated to them being constexpr. Note that in at least c++14 a constexpr method that doesn't use this can still be constant evaluated. Some versions of c++ make constexpr methods implicitly const; c++17 does not.
&isThree in one compilation unit and &isThree in another can (and usually do) vary when static (barring aggressive ICF, which is a matter for a different question). When inline they may not vary.
inline functions are shared between compilation units. Their full definition is also often visible in all compilation units aware of it, so it makes compiler "inlining" (as opposed to the keyword) your code easier. static are not. constexpr functions are implicitly inline, but not implicitly static.
Note that constexpr functions can be evaluated in a runtime context sometimes. When evaluated in a compile time context, their inline vs static or linkage state really doesn't matter.
constexpr means other things as well, but you wanted to know the difference between two different constexpr declarations, and none of those meanings change.
constexpr functions are not implicitly static. They have the same linkage as non-constexpr functions:
// external linkage
constexpr int f1(int x) { /* ... */ }
// internal linkage
static constexpr int f2(int x) { /* ... */ }
// internal linkage
namespace {
constexpr int f3(int x) { /* ... */ }
}
// no linkage
void enclosing() {
struct S {
constexpr int f4(int x) { /* ... */ }
};
}
When a constexpr function has external linkage, it has the same address in all translation units. When it has internal linkage, there is a different copy in each translation unit, and those copies have different addresses. However, I believe the result of calling a constexpr function should not depend on whether it has internal or external linkage (since constexpr functions may not contain static variables).
Consider the following header and assume it is used in several TUs:
static int x = 0;
struct A {
A() {
++x;
printf("%d\n", x);
}
};
As this question explains, this is an ODR violation and, therefore, UB.
Now, there is no ODR violation if our inline function refers to a non-volatile const object and we do not odr-use it within that function (plus the other provisions), so this still works fine in a header:
constexpr int x = 1;
struct A {
A() {
printf("%d\n", x);
}
};
But if we do happen to odr-use it, we are back at square one with UB:
constexpr int x = 1;
struct A {
A() {
printf("%p\n", &x);
}
};
Thus, given we have now inline variables, should not the guideline be to mark all namespace-scoped variables as inline in headers to avoid all problems?
constexpr inline int x = 1;
struct A {
A() {
printf("%p\n", &x);
}
};
This also seems easier to teach, because we can simply say "inline-everything in headers" (i.e. both function and variable definitions), as well as "never static in headers".
Is this reasoning correct? If yes, are there any disadvantages whatsoever of always marking const and constexpr variables in headers as inline?
As you have pointed out, examples one and third does indeed violate ODR as per [basic.def.odr]/12.2.1
[..] in each definition of D, corresponding names, looked up according to [basic.lookup], shall refer to an entity defined within the definition of D, or shall refer to the same entity, after overload resolution and after matching of partial template specialization, except that a name can refer to
a non-volatile const object with internal or no linkage if the object
is not odr-used in any definition of D, [..]
Is this reasoning correct?
Yes, inline variables with external linkage are guaranteed to refer to the same entity even when they are odr-used as long all the definitions are the same:
[dcl.inline]/6
An inline function or variable shall be defined in every translation unit in which it is odr-used and shall have exactly the same definition in every case ([basic.def.odr]). [..] An inline function or variable with external linkage shall have the same address in all translation units.
The last example is OK because it meets and don't violate the bold part of the above.
are there any disadvantages whatsoever of always marking const and constexpr variables in headers as inline?
I can't think of any, because if we keep the promise of having the exact same definition of an inline variable with external linkage through TU's, the compiler is free to pick any of them to refer to the variable, this will be the same, technically, as having just one TU and have a global variable declared in the header with appropriate header guards
Why does cppreference define type_traits xxx_v shortcuts as inline constexpr and not just constexpr?
For example, see is_integral_v:
template< class T >
inline constexpr bool is_integral_v = is_integral<T>::value;
Is this just a matter of style or is there some difference in behavior? As far as I know constexpr variables are implicitly inline.
Edit: Looking at the draft of the latest standard, it also uses inline constexpr. The question actually applies to the standard, then.
[dcl.constexpr]/9
A constexpr specifier used in an object declaration declares the object as const.
[basic.link]/3.2
A name having namespace scope has internal linkage if it is the name of
-a non-inline variable of non-volatile const-qualified type that is neither explicitly declared extern nor previously declared to have external linkage
Without inline specifier, is_integral_v would have internal linkage. This could be problematic if you compared two pointers to this same variable name taken in different translation unit.
Nota Bene: the inline specifier is redundant with constexpr only if the variable is a class static data member.
Following an exemple of easy violation of the one definition rule that could happen if is_integral_v where not inline.
bad_type_trait.h
template<class T>
constexpr auto bad_is_integral_v=std::is_integral<T>::value;
my_header.h
#include "bad_type_trait.h"
void f(const bool& x);
inline void g()
{
f(bad_is_integral_v<int>);
//g ODR use the static variable bad_is_integral_v.
//"ODR use" approximate definition is:
// the variable is refered by its memory address.
}
source1.cpp
#include "my_header.h"
void my_func1(){
g(); //the definition of g in this translation unit.
}
source2.cpp
#include "my_header.h"
void my_func2(){
g(); //is not the same as the definition of g in this translation unit.
}
In the two translation units, the variable bad_is_integral_v is instantiated as separate static variables. The inline function g() is defined in these two translation units. Inside the definition of g(), the variable bad_is_integral_v is ODR used, so the two definitions of g() are different, which is a violation of the one definition rule.
[basic.link]/3.2 applies to names of "a non-inline variable of non-volatile const-qualified type". A variable template isn't a variable, just like a class template isn't a class, a function template isn't a function, and a cookie cutter isn't a cookie. So that rule doesn't apply to the variable template is_integral_v itself.
A variable template specialization is a variable, however, so there are some questions about whether that rule gives it internal linkage. This is core issue 1713, the direction of which is that the rule is not applicable.
Core issue 1713, however, wasn't resolved in time for C++17. So LWG decided to simply plaster inline all over the variable templates just to be safe, because they don't hurt, either.
In the C++11 standard, what is the difference between constexpr and static constexpr global variables when defined in a header? More specifically, when multiple translation units include the same header, which declaration (if any) is guaranteed to define the same variable across the translation units?
e.g.,
cexpr.h:
#ifndef CEXPR_H
#define CEXPR_H
constexpr int cint = 1;
static constexpr int scint = 1;
#endif
a.cpp:
#include "cexpr.h"
b.cpp:
#include "cexpr.h"
In your current example there is no difference: On variable declarations, constexpr implies const, and a const variable at namespace scope has internal linkage by default (so adding static does not change anything).
In C++14, you cannot declare a variable as constexpr and have it have external linkage unless you only ever do this in one single translation unit. The reason is that constexpr variables require an initializer, and a declaration with initializer is a definition, and you must only have a single definition.
However, what you can do is use a normal integral constant, which you can declare (not define) as extern, and in the translation unit where it is defined it can even be used as a constant expression:
lib.h:
extern const int a;
lib.cpp:
#include "lib.h"
const int a = 10;
int b[a] = {1, 2, 3}; // OK in this translation unit
In C++17, there is a new feature "inline variables" which lets you say:
inline constexpr int a = 10;
And this is an "inline definition" that can appear repeatedly, and each definition defines the same entity (just like all the other "inline" entities in the language).
I think this article explains more clear. 6.8 — Global constants and inline variables
Because const globals have internal linkage, each .cpp file gets an independent version of the global variable that the linker can’t see. In most cases, because these are const, the compiler will simply optimize the variables away.
The term “optimizing away” refers to any process where the compiler optimizes the performance of your program by removing things in a way that doesn’t affect the output of your program. For example, lets say you have some const variable x that’s initialized to value 4. Wherever your code references variable x, the compiler can just replace x with 4 (since x is const, we know it won’t ever change to a different value) and avoid having to create and initialize a variable altogether.
So, the "cint " and "scint" are all internal linkage variables.
The best practice to define global variable after C++ 17:
inline constexpr double pi = 0;
Working Mechanism:
C++17 introduced a new concept called inline variables. In C++, the term inline has evolved to mean “multiple definitions are allowed”. Thus, an inline variable is one that is allowed to be defined in multiple files without violating the one definition rule. Inline global variables have external linkage by default.
Inline variables have two primary restrictions that must be obeyed:
1) All definitions of the inline variable must be identical (otherwise, undefined behavior will result).
2) The inline variable definition (not a forward declaration) must be present in any file that uses the variable.
The compiler will consolidate all inline definitions into a single variable definition. This allows us to define variables in a header file and have them treated as if there was only one definition in a .cpp file somewhere. These variables also retain their constexpr-ness in all files in which they are included.
If you can, prefer the static constexpr because with the constexpr it depends on the toolchain how likely it will get done on compile-time. Gcc is most aggressive, MSVS least aggressive and clang is in between.
Instead of leaving some values up to optimizer to decide it will do it at compile-time be more explicit and force it.
Reference:
https://www.youtube.com/watch?v=4pKtPWcl1Go