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Can template deduction guides call constexpr functions?
I have my own fixed-size array type I want to be constexpr constructible from an std::initializer_list without having to explicitly define the size template argument.
I thought I'd be able to use a template deduction guide but it looks like it's not treating std::initializer_list::size() as a constexpr function for it.
Here's an example of trying to make a deduction guide for std::array (which is similar to my type and has the same problem):
namespace std
{
template<typename T> array(initializer_list<T> initialiserList) -> array<T, initialiserList.size()>;
}
static constexpr std::array myArray = {1,2,3};
static constexpr std::array myArray2 = {{1,2,3}};
I've tried on MSVC and Clang, both give roughly the same errors:
myArray has an error complaining about too many arguments to the function.
myArray2 says "substitution failure [with T = int]: non-type template argument is not a constant expression"
I tried putting constexpr in front of the deduction guide or the function argument but neither appears to be allowed, so it appears that the deduction guide is invalid even though it should work fine in a constexpr context.
Is there a way to make this work without going down the make_array() route?
c++ c++17 constexpr template-deduction
edited Jan 6 at 23:17
max66
35.8k74165
asked Jan 6 at 22:53
Blake PrestonBlake Preston
985
add a comment |
I have my own fixed-size array type I want to be constexpr constructible from an std::initializer_list without having to explicitly define the size template argument.
I thought I'd be able to use a template deduction guide but it looks like it's not treating std::initializer_list::size() as a constexpr function for it.
Here's an example of trying to make a deduction guide for std::array (which is similar to my type and has the same problem):
namespace std
{
template<typename T> array(initializer_list<T> initialiserList) -> array<T, initialiserList.size()>;
}
static constexpr std::array myArray = {1,2,3};
static constexpr std::array myArray2 = {{1,2,3}};
I've tried on MSVC and Clang, both give roughly the same errors:
myArray has an error complaining about too many arguments to the function.
myArray2 says "substitution failure [with T = int]: non-type template argument is not a constant expression"
I tried putting constexpr in front of the deduction guide or the function argument but neither appears to be allowed, so it appears that the deduction guide is invalid even though it should work fine in a constexpr context.
Is there a way to make this work without going down the make_array() route?
c++ c++17 constexpr template-deduction
edited Jan 6 at 23:17
max66
35.8k74165
asked Jan 6 at 22:53
Blake PrestonBlake Preston
985
add a comment |
I have my own fixed-size array type I want to be constexpr constructible from an std::initializer_list without having to explicitly define the size template argument.
I thought I'd be able to use a template deduction guide but it looks like it's not treating std::initializer_list::size() as a constexpr function for it.
Here's an example of trying to make a deduction guide for std::array (which is similar to my type and has the same problem):
namespace std
{
template<typename T> array(initializer_list<T> initialiserList) -> array<T, initialiserList.size()>;
}
static constexpr std::array myArray = {1,2,3};
static constexpr std::array myArray2 = {{1,2,3}};
I've tried on MSVC and Clang, both give roughly the same errors:
myArray has an error complaining about too many arguments to the function.
myArray2 says "substitution failure [with T = int]: non-type template argument is not a constant expression"
I tried putting constexpr in front of the deduction guide or the function argument but neither appears to be allowed, so it appears that the deduction guide is invalid even though it should work fine in a constexpr context.
Is there a way to make this work without going down the make_array() route?
c++ c++17 constexpr template-deduction
edited Jan 6 at 23:17
max66
35.8k74165
asked Jan 6 at 22:53
Blake PrestonBlake Preston
985
I have my own fixed-size array type I want to be constexpr constructible from an std::initializer_list without having to explicitly define the size template argument.
I thought I'd be able to use a template deduction guide but it looks like it's not treating std::initializer_list::size() as a constexpr function for it.
Here's an example of trying to make a deduction guide for std::array (which is similar to my type and has the same problem):
namespace std
{
template<typename T> array(initializer_list<T> initialiserList) -> array<T, initialiserList.size()>;
}
static constexpr std::array myArray = {1,2,3};
static constexpr std::array myArray2 = {{1,2,3}};
I've tried on MSVC and Clang, both give roughly the same errors:
myArray has an error complaining about too many arguments to the function.
myArray2 says "substitution failure [with T = int]: non-type template argument is not a constant expression"
I tried putting constexpr in front of the deduction guide or the function argument but neither appears to be allowed, so it appears that the deduction guide is invalid even though it should work fine in a constexpr context.
Is there a way to make this work without going down the make_array() route?
c++ c++17 constexpr template-deduction
c++ c++17 constexpr template-deduction
edited Jan 6 at 23:17
max66
35.8k74165
asked Jan 6 at 22:53
Blake PrestonBlake Preston
985
edited Jan 6 at 23:17
max66
35.8k74165
asked Jan 6 at 22:53
Blake PrestonBlake Preston
985
edited Jan 6 at 23:17
max66
35.8k74165
edited Jan 6 at 23:17
max66
35.8k74165
edited Jan 6 at 23:17
max66
35.8k74165
35.8k74165
asked Jan 6 at 22:53
Blake PrestonBlake Preston
985
asked Jan 6 at 22:53
Blake PrestonBlake Preston
985
asked Jan 6 at 22:53
Blake PrestonBlake Preston
985
985
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
You can do:
template <class T, class... U>
array(T, U...) -> array<T, 1 + sizeof...(U)>;
The problem is not that you cannot call constexpr functions in deduction guides. You can. This example is ridiculous, but works:
constexpr size_t plus_one(size_t i) { return i + 1; }
template <class T, class... U>
array(T, U...) -> array<T, plus_one(sizeof...(U))>;
The problem is that function parameters are not constexpr objects, so you cannot invoke constexpr member functions on them if those member functions read kind of local state.
answered Jan 6 at 23:22
BarryBarry
180k19313573
add a comment |
Is there a way to make this work without going down the
make_array()route?
Why don't you try with the following deduction guide ?
template <typename T, std::size_t N>
array(T const (&)[N]) -> array<T, N>;
This way, the argument in myArray2 = {{1,2,3}} isn't interpreted as a std::initializer_list (that as argument can't be considered constexpr, so it's size() can't be used for a template argument) but as a C-style array.
So can be deduced, as template arguments, type and size (T and N) and also the size (N) can be used as template argument.
answered Jan 6 at 23:10
max66max66
35.8k74165
add a comment |
Parameter/argument values are not constexpr.
You might use variadic template to know size at compile time, or type with know size (std::array or C-array reference).
answered Jan 6 at 23:07
Jarod42Jarod42
115k12101182
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
You can do:
template <class T, class... U>
array(T, U...) -> array<T, 1 + sizeof...(U)>;
The problem is not that you cannot call constexpr functions in deduction guides. You can. This example is ridiculous, but works:
constexpr size_t plus_one(size_t i) { return i + 1; }
template <class T, class... U>
array(T, U...) -> array<T, plus_one(sizeof...(U))>;
The problem is that function parameters are not constexpr objects, so you cannot invoke constexpr member functions on them if those member functions read kind of local state.
answered Jan 6 at 23:22
BarryBarry
180k19313573
add a comment |
You can do:
template <class T, class... U>
array(T, U...) -> array<T, 1 + sizeof...(U)>;
The problem is not that you cannot call constexpr functions in deduction guides. You can. This example is ridiculous, but works:
constexpr size_t plus_one(size_t i) { return i + 1; }
template <class T, class... U>
array(T, U...) -> array<T, plus_one(sizeof...(U))>;
The problem is that function parameters are not constexpr objects, so you cannot invoke constexpr member functions on them if those member functions read kind of local state.
answered Jan 6 at 23:22
BarryBarry
180k19313573
add a comment |
You can do:
template <class T, class... U>
array(T, U...) -> array<T, 1 + sizeof...(U)>;
The problem is not that you cannot call constexpr functions in deduction guides. You can. This example is ridiculous, but works:
constexpr size_t plus_one(size_t i) { return i + 1; }
template <class T, class... U>
array(T, U...) -> array<T, plus_one(sizeof...(U))>;
The problem is that function parameters are not constexpr objects, so you cannot invoke constexpr member functions on them if those member functions read kind of local state.
answered Jan 6 at 23:22
BarryBarry
180k19313573
You can do:
template <class T, class... U>
array(T, U...) -> array<T, 1 + sizeof...(U)>;
The problem is not that you cannot call constexpr functions in deduction guides. You can. This example is ridiculous, but works:
constexpr size_t plus_one(size_t i) { return i + 1; }
template <class T, class... U>
array(T, U...) -> array<T, plus_one(sizeof...(U))>;
The problem is that function parameters are not constexpr objects, so you cannot invoke constexpr member functions on them if those member functions read kind of local state.
answered Jan 6 at 23:22
BarryBarry
180k19313573
answered Jan 6 at 23:22
BarryBarry
180k19313573
answered Jan 6 at 23:22
BarryBarry
180k19313573
answered Jan 6 at 23:22
BarryBarry
180k19313573
180k19313573
add a comment |
add a comment |
Is there a way to make this work without going down the
make_array()route?
Why don't you try with the following deduction guide ?
template <typename T, std::size_t N>
array(T const (&)[N]) -> array<T, N>;
This way, the argument in myArray2 = {{1,2,3}} isn't interpreted as a std::initializer_list (that as argument can't be considered constexpr, so it's size() can't be used for a template argument) but as a C-style array.
So can be deduced, as template arguments, type and size (T and N) and also the size (N) can be used as template argument.
answered Jan 6 at 23:10
max66max66
35.8k74165
add a comment |
Is there a way to make this work without going down the
make_array()route?
Why don't you try with the following deduction guide ?
template <typename T, std::size_t N>
array(T const (&)[N]) -> array<T, N>;
This way, the argument in myArray2 = {{1,2,3}} isn't interpreted as a std::initializer_list (that as argument can't be considered constexpr, so it's size() can't be used for a template argument) but as a C-style array.
So can be deduced, as template arguments, type and size (T and N) and also the size (N) can be used as template argument.
answered Jan 6 at 23:10
max66max66
35.8k74165
add a comment |
Is there a way to make this work without going down the
make_array()route?
Why don't you try with the following deduction guide ?
template <typename T, std::size_t N>
array(T const (&)[N]) -> array<T, N>;
This way, the argument in myArray2 = {{1,2,3}} isn't interpreted as a std::initializer_list (that as argument can't be considered constexpr, so it's size() can't be used for a template argument) but as a C-style array.
So can be deduced, as template arguments, type and size (T and N) and also the size (N) can be used as template argument.
answered Jan 6 at 23:10
max66max66
35.8k74165
Is there a way to make this work without going down the
make_array()route?
Why don't you try with the following deduction guide ?
template <typename T, std::size_t N>
array(T const (&)[N]) -> array<T, N>;
This way, the argument in myArray2 = {{1,2,3}} isn't interpreted as a std::initializer_list (that as argument can't be considered constexpr, so it's size() can't be used for a template argument) but as a C-style array.
So can be deduced, as template arguments, type and size (T and N) and also the size (N) can be used as template argument.
answered Jan 6 at 23:10
max66max66
35.8k74165
edited Jan 7 at 1:41
answered Jan 6 at 23:10
max66max66
35.8k74165
answered Jan 6 at 23:10
max66max66
35.8k74165
answered Jan 6 at 23:10
max66max66
35.8k74165
35.8k74165
add a comment |
add a comment |
Parameter/argument values are not constexpr.
You might use variadic template to know size at compile time, or type with know size (std::array or C-array reference).
answered Jan 6 at 23:07
Jarod42Jarod42
115k12101182
add a comment |
Parameter/argument values are not constexpr.
You might use variadic template to know size at compile time, or type with know size (std::array or C-array reference).
answered Jan 6 at 23:07
Jarod42Jarod42
115k12101182
add a comment |
Parameter/argument values are not constexpr.
You might use variadic template to know size at compile time, or type with know size (std::array or C-array reference).
answered Jan 6 at 23:07
Jarod42Jarod42
115k12101182
Parameter/argument values are not constexpr.
You might use variadic template to know size at compile time, or type with know size (std::array or C-array reference).
answered Jan 6 at 23:07
Jarod42Jarod42
115k12101182
answered Jan 6 at 23:07
Jarod42Jarod42
115k12101182
answered Jan 6 at 23:07
Jarod42Jarod42
115k12101182
answered Jan 6 at 23:07
Jarod42Jarod42
115k12101182
115k12101182
add a comment |
add a comment |
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