Why constexpr is not evaluated at compile time (MSVC 2015)? - c++11

recently I've tried to take advantage of C++0x constexpr under MSVC 2015 and my objective was to achieve compile-time hash strings. I wrote a simple FNV-1a hash algorithm as a constexpr function using, as required, a single return statement (ternary operator) and calling only constexpr functions, here it is:
template <size_t N>
constexpr U32 StringID_FNV1a_32(const char(&str)[N], I32 charIndex = 0, U32 hash = 2166136261U)
{
return charIndex < N-1 ? StringID_FNV1a_32(str, charIndex +1, (hash ^ str[charIndex]) * 16777619U) : hash;
}
I also made a little macro to be able to change the algorithm under the hood without any trouble:
#define STRING_ID(str) core::utility::StringID_FNV1a_32(str)
then I used this macro in my code, carefully checking if any breakpoint was hit and, also, the generated assembly code. Here's the little scenario:
//1. normal variable test
U32 hash1 = STRING_ID("abc");
//2. enum test
enum {
hash2 = STRING_ID("abc")
};
//3. constexpr variable test
constexpr U32 hash3 = STRING_ID("abc");
And here the facts:
first test was called at run time
second test was performed at compile time
third test was called at run time
As you can imagine I'm a little confused about the first and the third attempt.
Why in the third scenario is the compiler allowed to call the function at runtime? even though the msdn says clearly "The primary difference between const and constexpr variables is that the initialization of a const variable can be deferred until run time whereas a constexpr variable must be initialized at compile time." [https://msdn.microsoft.com/it-it/library/dn956974.aspx#Anchor_3]
Can be related to the fact that I'm in debug mode with all the optimizations turned off? and what about the first test?, is there any way to force the compiler to perform the hash at compile time?

MSVC's behavior can be quite strange, however it is possible to force it to make constexpr functions run at compile time.
#define COMPILE_TIME(value) ((decltype(value))CompileTime<decltype(value), value>::ValueHolder::VALUE)
template<typename T, T Value>
struct CompileTime
{
enum class ValueHolder : T
{
VALUE = Value
};
};
This forces the value to be passed as a template argument + an enumeration value, thus making it strictly compile-time only. Also please note that this works only for integer types.
You can use it simply by putting the call to the constexpr function as a parameter to the COMPILE_TIME macro:
constexpr U32 hash = COMPILE_TIME(STRING_ID("abc"));

Related

Remove class member type part from decltype

I ran into I case I had not seen before, while using decltype on a member of a templated class. I wanted to make a nicer make_unique so that changing type on the member does not cause fixing the make_unique calls. I wanted to avoid this using decltype(member)::element_type as the type for make_unique but got an error. Here is a simple snippet that shows the error (and I understand why it is shown):
#include <memory>
template<typename T>
struct foo
{
foo()
{
// g++ gives:
// dependent-name 'decltype (((foo<T>*)this)->foo<T>::p_)::element_type' is parsed as a non-type, but instantiation yields a type
// say 'typename decltype (((foo<T>*)this)->foo<T>::p_)::element_type' if a type is meant
//
// How can I atleast remove the class name from the type?
p_ = std::make_unique<decltype(p_)::element_type>();
// g++ gives:
// dependent-name 'decltype (p)::element_type' is parsed as a non-type, but instantiation yields a type
// say 'typename decltype (p)::element_type' if a type is meant
//
// makes sense since p here is dependent on T
std::unique_ptr<T> p = std::make_unique<decltype(p)::element_type>();
// This one is fine, makes sense, since the type is known
std::unique_ptr<int> p2 = std::make_unique<decltype(p2)::element_type>();
}
std::unique_ptr<T> p_;
};
int main()
{
foo<int> f;
return 0;
}
My question is, is there a nice/pretty way to remove the 'is a member of' ((foo<T>*)this)->foo<T>::p_))part from the decltype value, so that at least I could use the same fix and simply provide typename on the member variable p_ ? The long fix suggested by g++ seems kind of ugly.
5 minutes after posting I had an idea that I could do
p_ = std::make_unique<decltype(std::remove_reference(*p_)::type)>();
but that seems to give a parse error.
You can simply place a typename before decltype().
I mean
p_ = std::make_unique<typename decltype(p_)::element_type>();

Optional throw in constexpr?

I was researching how to implement a compile time strlen for a toy project in this repo https://github.com/elbeno/constexpr when I noticed an unusual pattern of throws in some of the code. It seemingly does nothing, why would you do this?
namespace err {
extern const char * strlen_runtime_error;
}
constexpr int strlen(const char * str) {
return true ? constexpr_func() : throw strlen_runtime_error;
}
I got curious if it has any use what so ever but couldn’t find anything useful on my own. The extern err is undefined.
It seems it is trying to enforce the function being used only in compile time, according to this comment in another of the functions of the library:
// convenience function for inferring the string size and ensuring no
// accidental runtime encryption
template <uint64_t S, size_t N>
constexpr encrypted_string<S, N> make_encrypted_string(const char(&s)[N])
{
return true ? encrypted_string<S, N>(s) :
throw err::strenc_runtime_error;
}
However, as you point out, it is not doing anything here. Typically, that trick with the ternary operator in constexpr functions is used to trigger compile-time errors given a condition -- not to ensure all calls to the function are constant expressions. See constexpr error at compile-time, but no overhead at run-time for an explanation of that pattern.
If you need to ensure that the result was found out during compile time, you can easily assign the result to a constexpr variable:
constexpr int result = strlen("asd");

C++11, Is it possible to force an instance to be extern but also a constant expression of a non-type template parameter?

Using C++11, g++ (GCC) 4.4.7 20120313 (Red Hat 4.4.7-18).
Lets pretend I have a templated function (pardon my terminology if it isn't quite right).
I want to perform a "general" algorithm based on what was supposed to be compile-time instances of "field". Where the only things that really changed are these constants which I moved into trait classes (only added one here but imagine there are more). Originally I was declaring it as
constexpr field FIELD1{1};
However in C++11, non-type template params need to have external linkage (unlike C++14 which can have internal and external linkage?). So because not's in the same translation unit I needed to use extern in order to give it external linkage (sorry if I butchered that explanation also). But by defining it extern I can't define it using constexpr and it seems that losing that constexpr constructor this field is no longer a valid constant expression to qualify as a non-type template param.
Any suggestions if there is some way I can get around this? Open to a new method of doing things. Below is a simplified (incomplete, and non-compiling version to get the gist of the organization).
So the error I am seeing is along the lines of
error: the value of ‘FIELD1’ is not usable in a constant expression
note: ‘FIELD1’ was not declared ‘constexpr’
extern const field FIELD1;
Not quite sure what could be a best alternative.
I can get rid of the second error by removing the constexpr from the constructor. But then I don't know how to approach the constant expression issue.
field.H
struct field
{
int thingone;
constexpr field(int i):thingone(i){}
};
extern const field FIELD1;
field.C
#include "field.H"
const field FIELD1{0};
field_traits.H
#include "field.H"
template< const field& T >
class fieldTraits;
template< >
class fieldTraits<FIELD1>
{
public:
// Let's say I have common field names
// with different constants that I want to plug
// into the "function_name" algorithm
static constexpr size_t field_val = 1;
};
function.H
#include "field.H"
template< const field& T, typename TT = fieldTraits<T> >
void function_name()
{
// Let's pretend I'm doing something useful with that data
std::cout << T.thingone << std::endl;
std::cout << TT::field_val << std::endl;
}
So because not's in the same translation unit I needed to use extern in order to give it external linkage (sorry if I butchered that explanation also). But by defining it extern I can't define it using constexpr [...]
Per my comment, you can. It wouldn't work for you, but it's a step that helps in coming up with something that would work:
extern constexpr int i = 10;
This is perfectly valid, gives i external linkage, and makes i usable in constant expressions.
But it doesn't allow multiple definitions, so it can't work in a header file which is included in multiple translation units.
Ordinarily, the way around that is with inline:
extern inline constexpr int i = 10;
But variables cannot be declared inline in C++11.
Except... when they don't need to be declared inline because the effect has already been achieved implicitly:
struct S {
static constexpr int i = 10;
};
Now, S::i has external linkage and is usable in constant expressions!
You may not even need to define your own class for this, depending on the constant's type: consider std::integral_constant. You can write
using i = std::integral_constant<int, 10>;
and now i::value will do exactly what you want.

Is it possible to write a generic function that distiguishes between an rvalue and an lvalue?

I am trying to learn rvalue references, as an exercise I tried to do answer the following.
Is it possible to write a function that can tell (at least at runtime, better if at compile time) if the passed value is a value (non reference), a rvalue or an lvalue? for a generic type? I want to extract as much information about the type as possible.
An alternative statement of the problem could be:
Can I have a typeid-like function that can tell as much as possible about the calling expression?, for example (and ideally) if the type is T, T&, T const&, or T&&.
Currently, for example, typeid drops some information about the type and one can do better (as in the example the const and non-const reference are distiguished). But how much better than typeid can one possibly do?
This is my best attempt so far. It can't distinguish between a rvalue and a "constant". First and second case in the example).
Maybe distiguishing case 1 and 2 is not possible in any circumstance? since both are ultimately rvalue? the the question is Even if both are rvalues can the two cases trigger different behavior?
In any case, it seems I overcomplicated the solution as I needed to resort to rvalue conditional casts, and ended up with this nasty code and not even 100% there.
#include<iostream>
#include<typeinfo>
template<class T>
void qualified_generic(T&& t){
std::clog << __PRETTY_FUNCTION__ << std::endl;
std::clog
<< typeid(t).name() // ok, it drops any qualification
<< (std::is_const<typename std::remove_reference<decltype(std::forward<T>(t))>::type>::value?" const":"") // seems to detect constness rigth
<< (std::is_lvalue_reference<decltype(std::forward<T>(t))>::value?"&":"")
<< (std::is_rvalue_reference<decltype(std::forward<T>(t))>::value?"&&":"") // cannot distiguish between passing a constant and an rvalue expression
<< std::endl
;
}
using namespace std;
int main(){
int a = 5;
int const b = 5;
qualified_generic(5); // prints "int&&", would plain "int" be more appropriate?
qualified_generic(a+1); // prints "int&&" ok
qualified_generic(a); // print "int&", ok
qualified_generic(b); // print "int const&", ok
}
Maybe the ultimate solution to distiguish between the cases will involve detecting a constexpr.
UPDATE: I found this talk by Scott Meyers where he claims that "The Standard sometimes requires typeid to give the 'wrong' answer". http://vimeo.com/97344493 about minute 44. I wonder if this is one of the cases.
UPDATE 2015: I revisited the problem using Boost TypeIndex and the result is still the same. For example using:
template<class T>
std::string qualified_generic(T&& t){
return boost::typeindex::type_id_with_cvr<decltype(t)>().pretty_name();
// or return boost::typeindex::type_id_with_cvr<T>().pretty_name();
// or return boost::typeindex::type_id_with_cvr<T&&>().pretty_name();
// or return boost::typeindex::type_id_with_cvr<T&>().pretty_name();
}
Still it is not possible to distinguish the type of 5 and a+1 in the above example.

Why declare int a constexpr

While watching a C++11 tutorial video linked on isocpp.org I noticed something:
constexpr int windowWidth{800}, windowHeight{600};
What is the point in declaring these int variables as constexpr, and not just const?
Nice video Vittorio!
Here is a summary of the difference between declaring an int const and constexpr:
int get_int(); // Some run time function that returns int
template <int N> // example use requiring a compile time int
struct test {};
const int w = get_int(); // initialized at run time
const int x = 5; // initialized at compile time
constexpr int y = get_int(); // error, can not initialize at compile time
constexpr int z = 6; // initialized at compile time
int
main()
{
test<w> tw; // error, w is not a compile time constant
test<x> tx; // ok, x is a compile time constant
test<y> ty; // error, there is no compile time constant named y
test<z> tz; // ok, z is a compile time constant
}
When you use constexpr, you require that the initialization happens at compile time, lest you will get a compile time error. When you use const, you allow the initialization to happen at run time, though it will still happen at compile time if the initializer is itself a compile time constant.
If you have a const int, the code reviewer must look at the initialization (following back to the original if this is a copy of a const int) to know if that const int is a compile time constant, or a run time constant.
If you have a constexpr int, the code reviewer can immediately assume that it is a compile time constant without analyzing how it was initialized. If that assumption turns out to be false, the compiler will flag it as an error.
<Disclaimer>
In the comments below Kerrek SB correctly points out that I've played "fast and loose" with the terminology in this answer. I've done so in an effort to make the answer short and understandable. What I'm calling "initialized at compile time" and "compile time constant" are what the standard in section 5.19 "Constant expressions" [expr.const] refers to as integral constant expressions.
Integral constant expressions are initialized with what is called constant initialization, which together with zero-initialization is referred to as static initialization ([basic.start.init]/p2).
Any deviations between what I write here and what appears in the standard are accidental, and what is in the standard is correct.
</Disclaimer>
I'm the author of the video.
Intent.
constexpr clearly expresses the intent of a compile-time immutable value. const doesn't really mean compile-time immutable value.
Both modifiers can be casted away, but that results in undefined behavior. Check DyP's comment for more information.
When using C++11, in my opinion, the first keyword that should come to mind when dealing with compile-time values is not const but constexpr.
The code would behave exactly the same without constexpr, or with const in place of constexpr.
But when you take a look at the code, and see constexpr int windowWidth; you can be 100%
sure that's an immutable constant that will never change during run-time.
In my second tutorial video, there's an addendum on constexpr in the first three minutes, that shows constexpr functions and more constexpr examples/explanations.

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