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;
}
Related
While exploring templates in C++, I stumbled upon the example in the following code:
#include <iostream>
#include <functional>
template <typename T>
void call(std::function<void(T)> f, T v)
{
f(v);
}
int main(int argc, char const *argv[])
{
auto foo = [](int i) {
std::cout << i << std::endl;
};
call(foo, 1);
return 0;
}
To compile this program, I am using the GNU C++ Compiler g++:
$ g++ --version // g++ (Ubuntu 6.5.0-1ubuntu1~16.04) 6.5.0 20181026
After compiling for C++11, I get the following error:
$ g++ -std=c++11 template_example_1.cpp -Wall
template_example_1.cpp: In function ‘int main(int, const char**)’:
template_example_1.cpp:15:16: error: no matching function for call to ‘call(main(int, const char**)::<lambda(int)>&, int)’
call(foo, 1);
^
template_example_1.cpp:5:6: note: candidate: template<class T> void call(std::function<void(T)>, T)
void call(std::function<void(T)> f, T v)
^~~~
template_example_1.cpp:5:6: note: template argument deduction/substitution failed:
template_example_1.cpp:15:16: note: ‘main(int, const char**)::<lambda(int)>’ is not derived from ‘std::function<void(T)>’
call(foo, 1);
^
(same for C++14 and C++17)
From the compiler error and notes I understand that the compiler failed to deduce the type of the lambda, since it cannot be matched against std::function.
Looking at previous questions (1, 2, 3, and 4) regarding this error, I am still confused about it.
As pointed out in answers from questions 3 and 4, this error can be fixed by explicitly specifying the template argument, like so:
int main(int argc, char const *argv[])
{
...
call<int>(foo, 1); // <-- specify template argument type
// call<double>(foo, 1) // <-- works! Why?
return 0;
}
However, when I use other types instead of int, like double, float, char, or bool, it works as well, which got me more confused.
So, my questions are as follow:
Why does it work when I explicitly specify int (and others) as the template argument?
Is there a more general way to solve this?
A std::function is not a lambda, and a lambda is not a std::function.
A lambda is an anonymous type with an operator() and some other minor utility. Your:
auto foo = [](int i) {
std::cout << i << std::endl;
};
is shorthand for
struct __anonymous__type__you__cannot__name__ {
void operator()(int i) {
std::cout << i << std::endl;
}
};
__anonymous__type__you__cannot__name__ foo;
very roughly (there are actual convert-to-function pointer and some other noise I won't cover).
But, note that it does not inherit from std::function<void(int)>.
A lambda won't deduce the template parameters of a std::function because they are unrelated types. Template type deduction is exact pattern matching against types of arguments passed and their base classes. It does not attempt to use conversion of any kind.
A std::function<R(Args...)> is a type that can store anything copyable that can be invoked with values compatible with Args... and returns something compatible with R.
So std::function<void(char)> can store anything that can be invoked with a char. As int functions can be invoked with a char, that works.
Try it:
void some_func( int x ) {
std::cout << x << "\n";
}
int main() {
some_func('a');
some_func(3.14);
}
std::function does that some conversion from its signature to the callable stored within it.
The simplest solution is:
template <class F, class T>
void call(F f, T v) {
f(v);
}
now, in extremely rare cases, you actually need the signature. You can do this in c++17:
template<class T>
void call(std::function<void(T)> f, T v) {
f(v);
}
template<class F, class T>
void call(F f_in, T v) {
std::function f = std::forward<F>(f_in);
call(std::move(f), std::forward<T>(v));
}
Finally, your call is a crippled version of std::invoke from c++17. Consider using it; if not, use backported versions.
I wanted to implement a overload for operator<< that allowed me to call a given function and output the result.
I therefore wrote an overload, but the conversion to bool is selected and when writing a function myself, it would not compile.
EDIT: Know that I do not want to call the lambda,
but instead pass it to the function where it should be called with a default constructed parameter list.
I have appended my code:
#include <iostream>
template<typename T>
void test(T *) {
std::cout << "ptr" << std::endl;
}
template<typename T>
void test(bool) {
std::cout << "bool" << std::endl;
}
template<typename Ret, typename ...Args>
void test(Ret(*el)(Args...)) {
std::cout << "function ptr\n" << el(Args()...) << std::endl;
}
template<typename Char_T, typename Char_Traits, typename Ret, typename ...Args>
std::basic_ostream<Char_T, Char_Traits>& operator<<(
std::basic_ostream<Char_T, Char_Traits> &str, Ret(*el)(Args...)) {
return str << el(Args()...);
}
int main() {
std::boolalpha(std::cout);
std::cout << []{return 5;} << std::endl; // true is outputted
test([]{return 5;}); // will not compile
}
I use gcc 7.3.1 with the version flag -std=c++14.
EDIT: Error message:
main.cc: In function ‘int main()’:
main.cc:25:23: error: no matching function for call to ‘test(main()::<lambda()>)’
test([]{return 5;});
^
main.cc:5:6: note: candidate: template<class T> void test(T*)
void test(T *) {
^~~~
main.cc:5:6: note: template argument deduction/substitution failed:
main.cc:25:23: note: mismatched types ‘T*’ and ‘main()::<lambda()>’
test([]{return 5;});
^
main.cc:9:6: note: candidate: template<class T> void test(bool)
void test(bool) {
^~~~
main.cc:9:6: note: template argument deduction/substitution failed:
main.cc:25:23: note: couldn't deduce template parameter ‘T’
test([]{return 5;});
^
main.cc:13:6: note: candidate: template<class Ret, class ... Args> void test(Ret (*)(Args ...))
void test(Ret(*el)(Args...)) {
^~~~
main.cc:13:6: note: template argument deduction/substitution failed:
main.cc:25:23: note: mismatched types ‘Ret (*)(Args ...)’ and ‘main()::<lambda()>’
test([]{return 5;});
Your problem here is that Template Argument Deduction is only done on the actual argument passed to test. It's not done on all possible types that the argument could possibly converted to. That might be an infinite set, so that's clearly a no-go.
So, Template Argument Deduction is done on the actual lambda object, which has an unspeakable class type. So the deduction for test(T*) fails as the lambda object is not a pointer. T can't be deduced from test(bool), obviously. Finally, the deduction fails for test(Ret(*el)(Args...)) as the lambda object is not a pointer-to-function either.
There are a few options. You might not even need a template, you could accept a std::function<void(void)> and rely on the fact that it has a templated constructor. Or you could just take a test(T t) argument and call it as t(). T will now deduce to the actual lambda type. The most fancy solution is probably using std::invoke, and accepting a template vararg list.
Even though non-capturing lambdas have an implicit conversion to function pointers, function templates must match exactly for deduction to succeed, no conversions will be performed.
Therefore the easiest fix is to force the conversion with a +
int main() {
std::boolalpha(std::cout);
std::cout << []{return 5;} << std::endl; // true is outputted
test(+[]{return 5;});
// ^
}
template<typename T>
void test(bool) {
std::cout << "bool" << std::endl;
}
Template is not needed. In fact you overload functions, not templates. Replace it with
void test(bool) {
std::cout << "bool" << std::endl;
}
Now your sample will compile.
I'm trying to write a member function that can instantiate an object of a custom type (templatized), initializing its const& member to a local object of the function.
This is consistent since the lifetime of the custom type object is the same as the local_object.
The objective is caching some metadata of the local object because they don't change during its lifetime. The operator() (or any member function) computes some values, then used later in func, and the objective is offering a hook to change the behaviour of func.
Please no polymorphic solutions (currently used) due to (profiled) slowness.
This is a M(N)WE:
#include <vector>
class cls {
public:
template <typename Custom> int func() {
std::vector<int> local_object{0, 14, 32};
Custom c(local_object, 42);
return c();
}
};
template<typename AType> class One {
public:
One(const AType& obj, const int n): objref(obj), param(n), member_that_should_depend_on_objref(obj.size()) {}
int operator()() { return 42; }
private:
const AType& objref;
const int param;
float member_that_should_depend_on_objref;
};
template<typename AType> class Two {
public:
Two(const AType& obj, const int n): objref(obj), param(n), other_member_that_should_depend_on_objref(obj.empty()), other_dependent_member(obj.back()) {}
int operator()() { return 24; }
private:
const AType& objref;
const int param;
bool other_member_that_should_depend_on_objref;
int other_dependent_member;
};
int main() {
cls myobj;
auto a = myobj.func<One>();
auto b = (myobj.func<Two>)();
}
G++ 5.3.0 says
tmp.cpp: In function 'int main()':
tmp.cpp:34:30: error: no matching function for call to 'cls::func()'
auto a = myobj.func<One>();
^
tmp.cpp:4:36: note: candidate: template<class Custom> int cls::func()
template <typename Custom> int func() {
^
tmp.cpp:4:36: note: template argument deduction/substitution failed:
tmp.cpp:35:32: error: no matching function for call to 'cls::func()'
auto b = (myobj.func<Two>)();
^
tmp.cpp:4:36: note: candidate: template<class Custom> int cls::func()
template <typename Custom> int func() {
^
tmp.cpp:4:36: note: template argument deduction/substitution failed:
Clang++ 3.7.1 says:
tmp.cpp:34:20: error: no matching member function for call to 'func'
auto a = myobj.func<One>();
~~~~~~^~~~~~~~~
tmp.cpp:4:36: note: candidate template ignored: invalid explicitly-specified argument for template
parameter 'Custom'
template <typename Custom> int func() {
^
tmp.cpp:35:21: error: no matching member function for call to 'func'
auto b = (myobj.func<Two>)();
~~~~~~~^~~~~~~~~~
tmp.cpp:4:36: note: candidate template ignored: invalid explicitly-specified argument for template
parameter 'Custom'
template <typename Custom> int func() {
^
2 errors generated.
auto a = myobj.func<One>();
is wrong since One is a class template, not a class. Use
auto a = myobj.func<One<SomeType>>();
It's not clear from your code what SomeType should be.
Update
If you want to use:
auto a = myobj.func<One>();
you need to change func to use a template template parameter:
class cls {
public:
template <template <class> class Custom > int func() {
std::vector<int> local_object{0, 14, 32};
Custom<std::vector<int>> c(local_object, 42);
return c();
}
};
Perhaps that was your intention.
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
The following code does not compile with G++ (although I believe it should):
#include <iostream>
template <unsigned N>
struct foo_traits {
typedef const char ArrayArg[N];
typedef int Function (ArrayArg *);
};
template <unsigned N>
int foo (typename foo_traits<N>::Function *ptr) {
return ptr(&"good");
}
int bar (const char (*x)[5]) {
std::cout << *x << "\n";
return 0;
}
int main ()
{
return foo(bar);
}
I checked this with GCC 4.4 through 4.7, and I get a template argument deduction failure. With 4.7.1:
prog.cpp: In function ‘int main()’:
prog.cpp:21:19: error: no matching function for call to ‘foo(int (&)(const char (*)[5]))’
prog.cpp:21:19: note: candidate is:
prog.cpp:10:5: note: template<unsigned int N> int foo(typename foo_traits<N>::Function*)
prog.cpp:10:5: note: template argument deduction/substitution failed:
prog.cpp:21:19: note: couldn't deduce template parameter ‘N’
If I use an explicit template argument (i.e., foo<5>(bar)), it compiles fine. If I use a version of the code without the typedefs, it compiles fine:
#include <iostream>
template <unsigned N>
int fixfoo (int (*ptr) (const char (*)[N])) {
return ptr(&"good");
}
int bar (const char (*x)[5]) {
std::cout << *x << "\n";
return 0;
}
int main ()
{
return fixfoo(bar);
}
Is the failing code supposed to compile (i.e., did I make a silly mistake)?
int foo(typename foo_traits<N>::Function *ptr);
The signature makes it a non-deductible context, so you must include the template arguments so that the value N is known and so consequentially the type of the pointer ptr be known as well.
Your second example compiles because the type of the signature through bar can be deduced.