auto bind2nd = [] < auto func_object,auto second_arg>(){
return [=](auto&& first_arg){
return func_object(first_arg,second_arg);
};
};
auto h =bind2nd.template operator()<std::greater<int>(),5>();
compiler result:
<source>:9:60: error: no matching function for call to '<lambda()>::operator()<std::greater<int>(), 5>()'
9 | auto x =bind2nd.template operator()<std::greater<int>(),5>();
| ^
<source>:3:16: note: candidate: 'template<auto func_object, auto second_arg> <lambda()>'
3 | auto bind2nd = [] < auto func_object,auto second_arg>(){
| ^
<source>:3:16: note: template argument deduction/substitution failed:
<source>:9:60: error: type/value mismatch at argument 1 in template parameter list for 'template<auto func_object, auto second_arg> <lambda()>'
9 | auto x =bind2nd.template operator()<std::greater<int>(),5>();
| ^
<source>:9:60: note: expected a constant of type 'auto', got 'std::greater<int>()'
<source>:9:60: note: ambiguous template argument for non-type template parameter is treated as function type
I want to use lambda with template and it does not work.
But I can run that :
auto x =[]<auto t>(){
return t;
};
auto test = []<auto func_object,auto second_arg>(){
return [=](auto&& first_arg){
return func_object.template operator()<second_arg>();
};
};
auto z =test.template operator()<x,5>();
int main(){
std::cout<<z(5);
}
It will print 5.
What's the right way to use lambda with template and how to fix this problem?
As a template argument std::greater<int>() is parsed as a function type (a function that takes no arguments and returns the class type std::greater<int>). This is where you can use curly brackets to help the compiler differentiate:
auto h = bind2nd.template operator()<std::greater<int>{}, 5>();
Related
Compare the following case when I have a class object that takes a vector. The non-deduced parameter T can be substituted fine with the default template argument:
#include <vector>
template <typename T = int>
struct container
{
container(std::vector<T> vec) {}
};
int main()
{
container C = std::vector{1,2,3,4,5};
}
This is not the case for my class which is a bit more complicated (CompilerExplorer):
#include <cstdio>
#include <initializer_list>
#include <variant>
template <size_t> struct obj;
template<size_t Vs>
using val = std::variant<std::monostate, int, struct obj<Vs>>;
template <size_t Vs = 0>
struct obj
{
obj() = default;
obj(std::initializer_list<val<Vs>> init) {
printf("initializer of object called, Vs = %d\n", Vs);
}
};
template <size_t Vs = 0>
struct container : public obj<Vs>
{
container(obj<0> init) {}
};
int main()
{
container<5> some_container = obj{1,2,5,2,obj{1,2,33},2,2};
}
This fails with the following error:
<source>: In function 'int main()':
<source>:29:57: error: class template argument deduction failed:
29 | container<5> some_container = obj{1,2,5,2,obj{1,2,33},2,2};
| ^
<source>:29:57: error: no matching function for call to 'obj(int, int, int)'
<source>:14:5: note: candidate: 'template<long unsigned int Vs> obj(std::initializer_list<std::variant<std::monostate, int, obj<Vs> > >)-> obj<<anonymous> >'
14 | obj(std::initializer_list<val<Vs>> init) {
| ^~~
But it works when I supplement the template specialization obj<0> in the instantiation of the container (in main). Any ideas why this doesn't work for my class and how I can fix it? I don't want to force the user to specify the template each time.
This problem already exists in the simpler case of just
auto o = obj{1,2,33};
which yields this error:
<source>:29:24: error: class template argument deduction failed:
29 | auto o = obj{1,2,33};
| ^
<source>:29:24: error: no matching function for call to 'obj(int, int, int)'
<source>:14:5: note: candidate: 'template<long unsigned int Vs> obj(std::initializer_list<std::variant<std::monostate, int, obj<Vs> > >)-> obj<<anonymous> >'
14 | obj(std::initializer_list<val<Vs>> init) {
| ^~~
<source>:14:5: note: template argument deduction/substitution failed:
<source>:29:24: note: mismatched types 'std::initializer_list<std::variant<std::monostate, int, obj<Vs> > >' and 'int'
29 | auto o = obj{1,2,33};
So, the compiler is unable to deduce, that the three ints should be an initializer list. If you add extra braces around them, the compiler recognizes that this should actually be a single list argument instead of three separate ones and it works:
auto o = obj{{1,2,33}};
This also carries over to the more complicated case:
container some_container = obj{{1,2,5,2,obj{{1,2,33}},2,2}};
I'm writing a template wrapper function that can be applied to a functions with different number/types of arguments.
I have some code that works but I'm trying to change more arguments into template parameters.
The working code:
#include <iostream>
int func0(bool b) { return b ? 1 : 2; }
//There is a few more funcX...
template<typename ...ARGS>
int wrapper(int (*func)(ARGS...), ARGS... args) { return (*func)(args...) * 10; }
int wrappedFunc0(bool b) { return wrapper<bool>(func0, b); }
int main()
{
std::cout << wrappedFunc0(true) << std::endl;
return 0;
}
Now I want int (*func)(ARGS...) to also be a template parameter. (It's for performance reasons. I want the pointer to be backed into the wrapper, because the way I'm using it prevents the compiler from optimizing it out.)
Here is what I came up with (The only difference is I've changed the one argument into a template parameter.):
#include <iostream>
int func0(bool b) { return b ? 1 : 2; }
//There is a few more funcX...
template<typename ...ARGS, int (*FUNC)(ARGS...)>
int wrapper(ARGS... args) { return (*FUNC)(args...) * 10; }
int wrappedFunc0(bool b) { return wrapper<bool, func0>(b); }
int main()
{
std::cout << wrappedFunc0(true) << std::endl;
return 0;
}
This doesn't compile. It shows:
<source>: In function 'int wrappedFunc0(bool)':
<source>:9:55: error: no matching function for call to 'wrapper<bool, func0>(bool&)'
9 | int wrappedFunc0(bool b) { return wrapper<bool, func0>(b); }
| ~~~~~~~~~~~~~~~~~~~~^~~
<source>:7:5: note: candidate: 'template<class ... ARGS, int (* FUNC)(ARGS ...)> int wrapper(ARGS ...)'
7 | int wrapper(ARGS... args) { return (*FUNC)(args...) * 10; }
| ^~~~~~~
<source>:7:5: note: template argument deduction/substitution failed:
<source>:9:55: error: type/value mismatch at argument 1 in template parameter list for 'template<class ... ARGS, int (* FUNC)(ARGS ...)> int wrapper(ARGS ...)'
9 | int wrappedFunc0(bool b) { return wrapper<bool, func0>(b); }
| ~~~~~~~~~~~~~~~~~~~~^~~
<source>:9:55: note: expected a type, got 'func0'
ASM generation compiler returned: 1
<source>: In function 'int wrappedFunc0(bool)':
<source>:9:55: error: no matching function for call to 'wrapper<bool, func0>(bool&)'
9 | int wrappedFunc0(bool b) { return wrapper<bool, func0>(b); }
| ~~~~~~~~~~~~~~~~~~~~^~~
<source>:7:5: note: candidate: 'template<class ... ARGS, int (* FUNC)(ARGS ...)> int wrapper(ARGS ...)'
7 | int wrapper(ARGS... args) { return (*FUNC)(args...) * 10; }
| ^~~~~~~
<source>:7:5: note: template argument deduction/substitution failed:
<source>:9:55: error: type/value mismatch at argument 1 in template parameter list for 'template<class ... ARGS, int (* FUNC)(ARGS ...)> int wrapper(ARGS ...)'
9 | int wrappedFunc0(bool b) { return wrapper<bool, func0>(b); }
| ~~~~~~~~~~~~~~~~~~~~^~~
<source>:9:55: note: expected a type, got 'func0'
Execution build compiler returned: 1
(link to the compiler explorer)
It looks like a problem with the compiler to me, but GCC and Clang agree on it so maybe it isn't.
Anyway, how can I make this template compile correctly with templated pointer to a function?
EDIT:
Addressing the duplicate flag Compilation issue with instantiating function template
I think the core of the problem in that question is the same as in mine, however, it lacks a solution that allows passing the pointer to function (not only its type) as a template parameter.
This doesn't work because a pack parameter (the one including ...) consumes all remaining arguments. All arguments following it can't be specified explicitly and must be deduced.
Normally you write such wrappers like this:
template <typename F, typename ...P>
int wrapper(F &&func, P &&... params)
{
return std::forward<F>(func)(std::forward<P>(params)...) * 10;
}
(And if the function is called more than once inside of the wrapper, all calls except the last can't use std::forward.)
This will pass the function by reference, which should be exactly the same as using a function pointer, but I have no reasons to believe that it would stop the compiler from optimizing it.
You can force the function to be encoded in the template argument by passing std::integral_constant<decltype(&func0), func0>{} instead of func0, but again, I don't think it's going to change anything.
The 2nd snippet is not valid because:
a type parameter pack cannot be expanded in its own parameter clause.
As from [temp.param]/17:
If a template-parameter is a type-parameter with an ellipsis prior to its optional identifier or is a parameter-declaration that declares a pack ([dcl.fct]), then the template-parameter is a template parameter pack. A template parameter pack that is a parameter-declaration whose type contains one or more unexpanded packs is a pack expansion. ... A template parameter pack that is a pack expansion shall not expand a template parameter pack declared in the same template-parameter-list.
So consider the following invalid example:
template<typename... Ts, Ts... vals> struct mytuple {}; //invalid
The above example is invalid because the template type parameter pack Ts cannot be expanded in its own parameter list.
For the same reason, your code example is invalid. For example, a simplified version of your 2nd snippet doesn't compile in msvc.
I would like to use a function and pass a constexpr lambda. However, it only compiles successfully if I let the type be deduced through auto. Explicitly giving the type through -> std::array<event, l()> seems to fail (the first instance). Why is this?
template <typename Lambda_T>
constexpr static auto foo(Lambda_T l) -> std::array<event, l()> {
return {};
} // error
template <typename Lambda_T>
constexpr static auto foo(Lambda_T l) {
return std::array<event, (l())>{};
} // OK
template <typename Lambda_T>
constexpr static auto foo(Lambda_T l) -> decltype(l()) { return {}; }
// OK
Note that, the lambda returns a size_t.
gcc errors on this without a call (clang accepts it):
prog.cc:9:63: error: template argument 2 is invalid
9 | constexpr static auto foo(Lambda_T l) -> std::array<event, l()>
| ^
prog.cc:9:63: error: template argument 2 is invalid
prog.cc:9:63: error: template argument 2 is invalid
prog.cc:9:63: error: template argument 2 is invalid
prog.cc:9:42: error: invalid template-id
9 | constexpr static auto foo(Lambda_T l) -> std::array<event, l()>
| ^~~
prog.cc:9:61: error: use of parameter outside function body before '(' token
9 | constexpr static auto foo(Lambda_T l) -> std::array<event, l()>
| ^
prog.cc:9:23: error: deduced class type 'array' in function return type
9 | constexpr static auto foo(Lambda_T l) -> std::array<event, l()>
| ^~~
In file included from prog.cc:4:
/opt/wandbox/gcc-head/include/c++/9.0.1/array:94:12: note:
'template<class _Tp, long unsigned int _Nm> struct std::array' declared here
94 | struct array
| ^~~~~
prog.cc: In function 'int main()':
prog.cc:14:5: error: 'foo' was not declared in this scope
14 | foo([]() {return 3; });
| ^~~
Parameters to constexpr functions are not themselves constexpr objects - so you cannot use them in constant expressions. Both of your examples returning arrays are ill-formed because there is no valid call to them.
To understand why, consider this nonsense example:
struct Z { int i; constexpr int operator()() const { return i; }; };
template <int V> struct X { };
template <typename F> constexpr auto foo(F f) -> X<f()> { return {}; }
constexpr auto a = foo(Z{2});
constexpr auto b = foo(Z{3});
Z has a constexpr call operator, and this is well-formed:
constexpr auto c = Z{3}();
static_assert(c == 3);
But if the earlier usage were allowed, we'd have two calls to foo<Z> that would have to return different types. This could only fly if the actual value f were the template parameter.
Note that clang compiling the declaration is not, in of itself, a compiler error. This is a class of situations that are ill-formed, no diagnostic required.
While writing some template code, I ran into <unresolved overloaded function type> errors which can be reduced to the following.
template <int N>
auto bar()
{
return N;
}
int main(int, char* [])
{
auto foo = [] (auto func) {
return func();
};
foo(bar<3>);
}
With the errors being:
unresolved_overload.cpp: In function 'int main(int, char**)':
unresolved_overload.cpp:26:28: error: no match for call to '(main(int, char**)::<lambda(auto:1)>) (<unresolved overloaded function type>)'
std::cout << foo(bar<3>) << std::endl;
^
unresolved_overload.cpp:21:29: note: candidate: template<class auto:1> constexpr main(int, char**)::<lambda(auto:1)>::operator decltype (((const main(int, char**)::<lambda(auto:1)>*)((const main(int, char**)::<lambda(auto:1)>* const)0))->operator()(static_cast<auto:1&&>(<anonymous>))) (*)(auto:1)() const
auto foo = [] (auto func) {
^
unresolved_overload.cpp:21:29: note: template argument deduction/substitution failed:
unresolved_overload.cpp:26:28: note: couldn't deduce template parameter 'auto:1'
std::cout << foo(bar<3>) << std::endl;
^
unresolved_overload.cpp:21:29: note: candidate: template<class auto:1> main(int, char**)::<lambda(auto:1)>
auto foo = [] (auto func) {
^
unresolved_overload.cpp:21:29: note: template argument deduction/substitution failed:
unresolved_overload.cpp:26:28: note: couldn't deduce template parameter 'auto:1'
std::cout << foo(bar<3>) << std::endl;
If we replace the auto-return with the explicit return type, int, the example will compile fine.
Why does auto-return run into these issues? I looked into template argument deduction and substitution but the search was largely unfruitful. I thought it might have something to do with the order of template instantiation / etc but couldn't make too much sense of it...
Per AndyG's suggestion, I found the same issue on GCC's bug list. Bug 64194. First reported in 2014. Thus the conclusion seems to be that this is a GCC bug and thankfully not another special case for templates.
Working around this just requires having something else to trigger the instantiation (e.g. assign to a variable, a using declaration).
Try this:
template <typename func>
auto bar(func&& f)->decltype(f())
{
return f();
}
int main()
{
int i = 100;
auto f = [=]()
{
return i;
};
bar(f);
return 0;
}
Nothing clearer than an old good MCVE:
struct X {
auto get(int) const -> int { return {}; }
auto get(int) -> int { return {}; }
};
template <class R> auto f(auto (X::*)(int) const -> R) {}
// ^~~~ ~~~~
// trailing return type
int main() {
f(&X::get);
}
This fails in g++ (4.9.2 & 5.1.0). However if the old return type is used:
template <class R> auto f(R (X::*)(int) const) {}
// ^
// old return type
it works.
On clang (3.5.0) both variants work.
I know that trailing return type changes when the return type is inferred and the scope of it, so I wouldn't be quick to cast it as a gcc bug. So what does the standard says? Which compiler is right?
The most significant message in the error I think is
couldn't deduce template parameter ‘R’`
g++ full message:
main2.cpp: In function ‘int main()’:
main2.cpp:21:12: error: no matching function for call to ‘f(<unresolved overloaded function type>)’
f(&X::get);
^
main2.cpp:18:25: note: candidate: template<class R, class auto:1> auto f(auto:1 (X::*)(int) const)
template <class R> auto f(auto (X::*)(int) const -> R) {}
^
main2.cpp:18:25: note: template argument deduction/substitution failed:
main2.cpp:21:12: note: types ‘auto:1 (X::)(int) const’ and ‘int (X::)(int)’ have incompatible cv-qualifiers
f(&X::get);
^
main2.cpp:21:12: note: couldn't deduce template parameter ‘R’
<builtin>: recipe for target 'main2' failed
make: *** [main2] Error 1
As pointed in the question this is a gcc bug which was beed fixed in version 6
gcc.gnu.org/bugzilla/show_bug.cgi?id=69139