I'm trying to pass a list of pairs of types constructed from boost::mp11::mp_product to a function that takes the pairs along with an additional function argument via boost::mp11::mp_for_each.
The docs I've found for mp_for_each are limited to use with generic lambdas or pure functions, so I can't seem to figure out if the use of std::bind is the way to go; and if it is, what I'm doing wrong yielding the following compiler error:
error: no matching function for call to 'bind'
std::bind(inject_foo, m, std::placeholders::_1));
^~~~~~~~~
/Applications/Xcode.app/Contents/Developer/Toolchains/XcodeDefault.xctoolchain/usr/bin/../include/c++/v1/functional:2953:1: note: candidate template ignored: couldn't infer template argument '_Fp'
bind(_Fp&& __f, _BoundArgs&&... __bound_args)
^
/Applications/Xcode.app/Contents/Developer/Toolchains/XcodeDefault.xctoolchain/usr/bin/../include/c++/v1/functional:2962:1: note: candidate template ignored: couldn't infer template argument '_Rp'
bind(_Fp&& __f, _BoundArgs&&... __bound_args)
Code I'm using:
#include <pybind11/numpy.h>
#include <boost/mp11.hpp>
#include <functional>
using boost::mp11::mp_product;
using boost::mp11::mp_for_each;
template <typename...> struct type_list {};
// all possible types
using my_type_list = type_list<
double, float, py::ssize_t, int, unsigned int, unsigned long>;
// construct all possible pairs of types with help from boost::mp11
using my_type_pairs = mp_product<
type_list, my_type_list, my_type_list>;
// the C++ function that we bind to a python module in the next function.
template <typename Tx, typename Ty>
py::array<py::ssize_t> foo(p::array_t<Tx> x, py::array_t<Ty>) {
py::array_t<py::ssize_t> z;
// do something with x and y
return z;
}
// bind foo<Tx, Ty> function to py::module m
template <typename Tx, typename Ty>
void inject_foo(py::module_& m, const type_list<Tx, Ty>&) {
m.def("_foo", &foo<Tx, Ty>, py::arg("x").noconvert(), py::arg("y").noconvert());
}
PYBIND11_MODULE(_backend, m) {
// these function calls work as expected:
// inject_foo(m, type_list<double, double>{});
// inject_foo(m, type_list<double, float>{});
// inject_foo(m, type_list<double, int>{});
// .....
// trying to make my life easier with the loop
// over all possible types of pairs is not working
mp_for_each(pg_type_pairs{}, std::bind(inject_foo, m, std::placeholders::_1));
}
PiotrNycz's comment led to an answer:
mp_for_each<pg_type_pairs>([&](const auto& x) { inject_foo(m, x); });
Provides the desired behavior (In the question I was even using the mp_for_each API incorrectly; switching to a lambda helped find that error).
Related
I was playing around with toy tuple implementations and eventually stuck with how get function works.
Consider this simple example
#include <iostream>
#include <utility>
template <size_t Tag, typename ValueType>
struct TagedValue { ValueType value; };
struct Test : TagedValue<0, int>, TagedValue<1, std::string>, TagedValue<2, double> {};
template <size_t Idx, typename T>
auto& get(Test& test) {
((TagedValue<Idx, T>&)(test)).value;
}
template <size_t Idx, typename T>
auto& get_impl(TagedValue<Idx, T>& tagged_value) {
return tagged_value.value;
}
template <size_t Idx>
auto& get_2(Test& test) {
return get_impl<Idx>(test);
}
int main()
{
Test test;
get_2<0>(test);
get<0>(test);
}
I get this error:
<source>: In function 'int main()':
<source>:29:16: error: no matching function for call to 'get<0>(Test&)'
29 | get<0>(test);
| ^
<source>:10:7: note: candidate: 'template<long unsigned int Idx, class T> auto& get(Test&)'
10 | auto& get(Test& test) {
| ^~~
<source>:10:7: note: template argument deduction/substitution failed:
<source>:29:16: note: couldn't deduce template parameter 'T'
29 | get<0>(test);
| ^
I do have couple of questions:
Basically why get_2 works and get doesn't compile. To me it looks like get_2 does exactly what I'm trying to do inside get
Does deducing T for get_2 take O(1) time, if yes how is it possible? Does compiler store some kind of map internally?
When you call a function template then all template arguments must either be specified explicitly or be deduced from the function arguments. When you call
get<0>(test);
Then Idx is 0, but there is no way for the compiler to know what T is supposed to be. The parameter is just Test, and T cannot be deduced from that.
Sorry for the generic title, but I'm unable to focus the problem.
I have a templatized class method that accept an argument pack and provides a new type in return, to hide the details of the implementation. More specifically, the class handles SQLite queries, and the method calls sqlite3_prepare() to prepare the statement before executing the query.
class Table {
...
template <typename ...Ts>
class PreparedStatement { ... };
template <typename ...Ts>
PreparedStatement<Ts...> prepare(std::tuple<Ts...> tuple) {
// do something
return PreparedStatement<Ts...> ( ... );
}
That works well with "normal" types, but the problem occurs when the arguments are declared const:
const Field<int> fld = createField<int>("name");
...
PreparedStatement<decltype(fld)> s = prepare(make_tuple(fld));
The error is the following:
no match for 'operator =' (operand types are PreparedStatenent<const Field<int>> and PreparedStatement<Field<int>>
I suspect the issue is in my declaration of the function, is there a way to fix this issue and make the function more "elegant" ?
NOTE: I know I can fix the issue by manually declare the s variable, but my doubts are on how the method was implemented.
As Many Asked, here's an example:
#include <tuple>
template <typename T>
struct Field {
};
class Table {
public:
template <typename ...Ts>
class PreparedStatement {
public:
PreparedStatement() {};
};
template <typename ...Ts>
PreparedStatement<Ts...> prepare(std::tuple<Ts...> tuple) {
// do something
return PreparedStatement<Ts...> ( );
}
};
int main()
{
Field<int> fld;
Table t;
Table::PreparedStatement<decltype(fld)> p;
p = t.prepare(std::make_tuple(fld));
// here comes the problem
const Field<int> f2;
Table::PreparedStatement<decltype(f2)> p2;
p2 = t.prepare(std::make_tuple(f2));
return 0;
}
and here's the compiler output
main.cpp: In function 'int main()': main.cpp:35:39: error: no match
for 'operator=' (operand types are 'Table::PreparedStatement >' and 'Table::PreparedStatement >')
p2 = t.prepare(std::make_tuple(f2));
^ main.cpp:10:10: note: candidate: constexpr Table::PreparedStatement >&
Table::PreparedStatement >::operator=(const
Table::PreparedStatement >&)
class PreparedStatement {
^~~~~~~~~~~~~~~~~ main.cpp:10:10: note: no known conversion for argument 1 from 'Table::PreparedStatement >'
to 'const Table::PreparedStatement >&'
main.cpp:10:10: note: candidate: constexpr
Table::PreparedStatement >&
Table::PreparedStatement
::operator=(Table::PreparedStatement >&&) main.cpp:10:10: note: no known conversion for argument 1 from
'Table::PreparedStatement >' to
'Table::PreparedStatement >&&'
UPDATE
As many noted, I could use auto to deduce the type, but in some condition auto cannot practically be used. One is, for example, if I need to declare the statement in the Class Context.
So suppose auto is forbidden for some reason. Isn't any other solution available? See the updated code above.
cppreference.com for make_tuple tells us:
template< class... Types >
tuple<VTypes...> make_tuple( Types&&... args );
For each Ti in Types..., the corresponding type Vi in Vtypes... is
std::decay<Ti>::type unless application of std::decay results in
std::reference_wrapper<X> for some type X, in which case the deduced
type is X&.
While std::decay, among other things, removes cv-qualifiers. So your type will be no PreparedStatement<const Field<int>>, but PreparedStatement<Field<int>>.
You can use auto, as manni66 proposed, to avoid such problems.
auto s = prepare(make_tuple(fld));
I could use auto to deduce the type, but in some condition auto cannot practically be used. One is, for example, if I need to declare the statement in the Class Context. So suppose auto is forbidden for some reason. Isn't any other solution available? See the updated code above.
Instead of auto, you can use a decltype expression that take in count the value returned by prepare.
I mean... instead of
Table::PreparedStatement<decltype(f2)> p2;
you can try with
decltype(t.prepare(std::make_tuple(f2))) p2;
or
decltype(std::declval<Table>().prepare(
std::make_tuple(std::declval<Field<int>>()))) p2;
I suppose you can use a similar decltype() also to declare members of your classes.
The problem came from here - I wanted to create an approach solving a little bit more general problem. Consider an example:
#include <utility>
template<class T, std::size_t>
using deduced = T;
template<std::size_t N, class = std::make_index_sequence<N>>
struct Foo;
template<std::size_t N, std::size_t... Is>
struct Foo<N, std::index_sequence<Is...>>{
template <class... Args>
void Bar(deduced<Args, Is>...)
{ }
};
int main() {
Foo<3> myfoo;
myfoo.Bar(true, 2.0f, 3); // OK
//myfoo.Bar(1, 2, 3, 4); // error
}
clang has no problem with compiling the code, gcc on the other hand shows following errors:
prog.cc: In function 'int main()':
prog.cc:18:27: error: no matching function for call to 'Foo<3ul>::Bar(bool, float, int)'
myfoo.Bar(true, 2.0f, 3); // valid
^
prog.cc:12:10: note: candidate: template<class ... Args> void Foo<N, std::integer_sequence<long unsigned int, Is ...> >::Bar(deduced<Args, Is>...) [with Args = {Args ...}; long unsigned int N = 3ul; long unsigned int ...Is = {0ul, 1ul, 2ul}]
void Bar(deduced<Args, Is>...)
^~~
prog.cc:12:10: note: template argument deduction/substitution failed:
prog.cc:18: confused by earlier errors, bailing out
[live demo]
What confuses me gcc does not have a problem with deduction when using the same alias but outer parameter pack isn't involved, e.g.:
void Bar(deduced<Args, 0>...)
So the question is - is it legal to combine parameter packs from outer and inner class with alias of this form to make compiler deduce one of the template parameters or is it gcc bug?
Edit (based on bogdan's comment):
The code causes trouble also to MSVC (2017 RC), but works in this form with EDG compiler, icc (in version 16 and 17) also seem to deal well with a code. It is also worth noting that similar code with class instead of alias (also considered as deduced context in some places example by bogdan) causes even more trouble to compilers - only clang seems to deal well with this version(?)
I am trying to pass a callback function as function parameter. But getting template substitution failure errors in following code. Not sure why template substitution is failing.
#include<iostream>
#include <map>
#include <tuple>
#include <functional>
template<typename A,typename B>
void myfun(std::map<A,B> & mm, std::function<std::tuple<A,B>(void)> fn)
{
A key;
B val;
std::tie(key,val) = fn();
mm[key] = val;
}
std::tuple<std::string,int> fun()
{
return std::make_tuple(std::string("hi"),1);
}
int main()
{
std::map<std::string,int> gg;
#if 0
//fixed version
std::function<std::tuple<std::string,int>(void)> yy = fun;//fixed
myfun(gg,yy);//fixed
#else
// error causing code
myfun(gg,fun);
#endif
}
And error is as following
main.cpp:8:6: note: template argument deduction/substitution failed:
main.cpp:25:17: note: mismatched types 'std::function<std::tuple<_T1, _T2>()>' and 'std::tuple<std::__cxx11::basic_string<char, std::char_traits<char>, std::allocator<char> >, int> (*)()'
myfun(gg,fun);
The compiler can't both cast to a std::function and deduce the template arguments. It doesn't understand the mapping between an arbitrary function pointer and a std::function.
There are a few ways round this.
You could explicitly create a std::function at the call site:
myfun(gg,std::function<std::tuple<std::string,int>(void)>{fun});`
You could write a make_function function to deduce the types for you. You can find discussions and implementations of this online, such as here, here and here.
myfun(gg,make_function(fun));
You could just forget about std::function and deduce the entire function type. This is the approach I would take:
template<typename A,typename B, typename Fun>
void myfun(std::map<A,B> & mm, Fun fn)
{
A key;
B val;
std::tie(key,val) = fn();
mm[key] = val;
}
I have a compile error when using std::function in a templated member function, the following code is a simple example:
#include <functional>
#include <memory>
using std::function;
using std::bind;
using std::shared_ptr;
class Test {
public:
template <typename T>
void setCallback(function<void (T, int)> cb);
};
template <typename T>
void Test::setCallback(function<void (T, int)> cb)
{
// do nothing
}
class TestA {
public:
void testa(int a, int b) { }
};
int main()
{
TestA testA;
Test test;
test.setCallback(bind(&TestA::testa, &testA, std::placeholders::_1, std::placeholders::_2));
return 0;
}
And come with the following compile error:
testtemplate.cpp: In function ‘int main()’:
testtemplate.cpp:29:92: error: no matching function for call to
‘Test::setCallback(std::_Bind_helper)(int, int),
TestA, const std::_Placeholder<1>&, const
std::_Placeholder<2>&>::type)’
testtemplate.cpp:29:92: note: candidate is: testtemplate.cpp:10:7:
note: template void Test::setCallback(std::function)
testtemplate.cpp:10:7: note: template argument
deduction/substitution failed:
testtemplate.cpp:29:92: note: ‘std::_Bind(TestA*, std::_Placeholder<1>,
std::_Placeholder<2>)>’ is not derived from ‘std::function’
I'm using C++11 and g++ 4.7
To figure out the problem let separate statements:
auto f = bind(&TestA::testa, &testA, _1, _2); // OK
test.setCallback(f); // <<--- Error is here
setCallback needs to know type of T and it can't deduce it from f, so give it a type
test.setCallback<TYPE>(f); // TYPE: int, float, a class, ...
You can make type deduction work with some variant of:
template<typename CALLBACK>
void setCallback(CALLBACK cb) {
typedef CALLBACK::first_argument_type T;
static_assert(is_same_type<CALLBACK,function<void(T,int)>>::value);
...
}
This way CALLBACK can be determined by looking at the argument. It might get into trouble if bind doesn't actually return a std::function but rather something that can be cast as one. I'm not sure.