std::bind and rvalue reference - c++

Let's consider the following piece of code:
class Widget{
};
int main(){
Widget w;
auto lambda = bind([](Widget&& ref){ return; }, std::move(w));
return 0;
}
and it triggers error
no match for call to ‘(std::_Bind<main()::<lambda(Widget&&)>(Widget)>) ()’
lambda();
And my question is: Why the error has appeared? After all, I do an explicit cast to rvalue reference – I mean std::move(w) and I take argument by rvalue reference – I mean Widget&& ref.
What's up?
Moreover the the below code works, what makes me worried the more:
class Widget{
};
int main(){
Widget w;
auto lambda = bind([](Widget& ref){ return; }, std::move(w));
return 0;
}

It might become clearer if you write down what std::bind schematically does.
// C++14, you'll have to write a lot of boilerplate code for C++11
template <typename FuncT, typename ArgT>
auto
bind(FuncT&& func, ArgT&& arg)
{
return
[
f = std::forward<FuncT>(func),
a = std::forward<ArgT>(arg)
]() mutable { return f(a); }; // NB: a is an lvalue here
}
Since you can call the function object std::bind gives you multiple times, it cannot “use up” the captured argument so it will be passed as an lvalue reference. The fact that you pass bind itself an rvalue only means that there is no copy made on the line where a is initialized.
If you try to compile your example with the schematic bind shown above, you'll also get a more helpful error message from your compiler.
main.cxx: In instantiation of ‘bind(FuncT&&, ArgT&&)::<lambda()> mutable [with FuncT = main()::<lambda(Widget&&)>; ArgT = Widget]’:
main.cxx:10:33: required from ‘struct bind(FuncT&&, ArgT&&) [with FuncT = main()::<lambda(Widget&&)>; ArgT = Widget]::<lambda()>’
main.cxx:11:31: required from ‘auto bind(FuncT&&, ArgT&&) [with FuncT = main()::<lambda(Widget&&)>; ArgT = Widget]’
main.cxx:18:59: required from here
main.cxx:11:26: error: no match for call to ‘(main()::<lambda(Widget&&)>) (Widget&)’
]() mutable { return f(a); }; // NB: a is an lvalue here
^
main.cxx:11:26: note: candidate: void (*)(Widget&&) <conversion>
main.cxx:11:26: note: conversion of argument 2 would be ill-formed:
main.cxx:11:26: error: cannot bind ‘Widget’ lvalue to ‘Widget&&’
main.cxx:18:33: note: candidate: main()::<lambda(Widget&&)> <near match>
auto lambda = bind([](Widget&&){ return; }, std::move(w));
^
main.cxx:18:33: note: conversion of argument 1 would be ill-formed:
main.cxx:11:26: error: cannot bind ‘Widget’ lvalue to ‘Widget&&’
]() mutable { return f(a); }; // NB: a is an lvalue here

To make it work you need to write it like this:
#include <functional>
#include <iostream>
class Widget{};
int main()
{
Widget a;
auto lf = [](Widget&& par){ };
auto f = std::bind
(
lf,
std::bind
(
std::move<Widget&>, a
)
);
f();
return 0;
}
My compiler is gcc version 4.9.2 20141101 (Red Hat 4.9.2-1) (GCC)

Related

invoke_result_t<> not matching lambda with a reference parameter

Using a function that accepts templated functions works great when the type is either an rvalue reference or has no reference, but as soon as I make it an lvalue reference it breaks.
Note that V is currently unused here, but it still fails to compile anyways regardless of whether it's used or not.
using namespace std;
template <typename F, typename V = std::invoke_result_t<F, string>>
void func(F f) {
std::vector<string> v = { "a", "b", "c" };
std::for_each(v.begin(), v.end(), f);
}
int main() {
func([](string s) { return s.length(); }); // Good
// func([](string& s) { return s.length(); }); // Bad
func([](const string& s) { return s.length(); }); // Good
}
main.cpp: In function 'int main()':
main.cpp:18:46: error: no matching function for call to 'func(main()::)'
func([](string& s) { return s.length(); });
^
main.cpp:11:6: note: candidate: 'template void func(F)'
void func(F f) {
^~~~
main.cpp:11:6: note: template argument deduction/substitution failed:
I can't do something like
std::invoke_result_t<F, string&>
and I couldn't do something like
std::invoke_result_t<F, std::add_lvalue_reference_t<string>>
The last one was a shot in the dark. My template knowledge is not that great. I've been searching around on here and on various blogs/google/etc, haven't had much success.
std::invoke_result_t<F, string>
this means passing F a string rvalue. And you cannot if F takes an lvalue reference.
I can't do something like
std::invoke_result_t<F, string&>
well yes you can. Do that if you want to know what the result of calling it with a non-const lvalue is.
At your point of use in your sample code, you pass it an lvalue. The string&& overload does not work.

Cannot capture reference in lambda from universal reference?

I am trying to figure out why the captured variable cannot be passed to the foo function by reference, the program shown below does not compile with the following error:
error: invalid initialization of reference of type
'Object&' from expression of type 'const Object'
Is there some way to perfect forward the universal reference into the lambda so that it is captured by reference for later use? Types T might also be R-Values or a combination of R-Values and L-Values, so they cannot always be captured by reference.
struct Object
{
};
void foo(Object& a)
{
}
template<typename... T>
auto bar(T&&... values)
{
return [values...](){
foo(values...);
};
}
int main ()
{
Object obj;
auto lambda = bar(obj);
return 0;
}
You're by capturing by values which stores values as const members inside the closure, and they cannot bind to non-const lvalue references. For capture by-ref you've forgot to put '&' before values:
return [&values...](){
foo(values...);
};
A small example.
The only diff between the two files:
diff omg.cpp nyan.cpp
6c6
< return [values...]{ foo(values...); };
---
> return [&values...]{ foo(values...); };
Compile the one with by-value capture:
$ g++ -std=c++14 omg.cpp && ./a.out
omg.cpp: In instantiation of ‘bar(T&& ...)::<lambda()> [with T = {int&}]’:
omg.cpp:6:16: required from ‘struct bar(T&& ...) [with T = {int&}]::<lambda()>’
omg.cpp:6:38: required from ‘auto bar(T&& ...) [with T = {int&}]’
omg.cpp:11:16: required from here
omg.cpp:6:25: error: binding reference of type ‘int&’ to ‘const int’ discards qualifiers
return [values...]{ foo(values...); };
~~~^~~~~~~~~~~
omg.cpp:3:6: note: initializing argument 1 of ‘void foo(int&)’
void foo(int &a) { std::cout << a << std::endl; }
^~~
Compile the other:
$ g++ -std=c++14 nyan.cpp && ./a.out
42
314

std::bind on a generic lambda - auto type deduction

Consider the following code:
#include <iostream>
#include <functional>
int main() {
auto run = [](auto&& f, auto&& arg) {
f(std::forward<decltype(arg)>(arg));
};
auto foo = [](int &x) {};
int var;
auto run_foo = std::bind(run, foo, var);
run_foo();
return 0;
}
Which gives the following compilation error when compiled with clang:
$ clang++ -std=c++14 my_test.cpp
my_test.cpp:6:9: error: no matching function for call to object of type 'const (lambda at my_test.cpp:8:16)'
f(std::forward<decltype(arg)>(arg));
^
/usr/bin/../lib64/gcc/x86_64-pc-linux-gnu/6.3.1/../../../../include/c++/6.3.1/functional:998:14: note: in instantiation of function template specialization 'main()::(anonymous class)::operator()<const (lambda at my_test.cpp:8:16) &, const int &>' requested here
= decltype( std::declval<typename enable_if<(sizeof...(_Args) >= 0),
^
/usr/bin/../lib64/gcc/x86_64-pc-linux-gnu/6.3.1/../../../../include/c++/6.3.1/functional:1003:2: note: in instantiation of default argument for 'operator()<>' required here
operator()(_Args&&... __args) const
^~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
my_test.cpp:11:12: note: while substituting deduced template arguments into function template 'operator()' [with _Args = <>, _Result = (no value)]
run_foo();
^
my_test.cpp:8:16: note: candidate function not viable: 1st argument ('const int') would lose const qualifier
auto foo = [](int &x) {};
^
my_test.cpp:8:16: note: conversion candidate of type 'void (*)(int &)'
1 error generated.
Why is arg deduced to be const int& instead of just int&?
std::bind documentation says:
Given an object g obtained from an earlier call to bind, when it is
invoked in a function call expression g(u1, u2, ... uM), an invocation
of the stored object takes place, as if by std::invoke(fd,
std::forward(v1), std::forward(v2), ...,
std::forward(vN)), where fd is a value of type std::decay_t the
values and types of the bound arguments v1, v2, ..., vN are determined
as specified below.
...
Otherwise, the
ordinary stored argument arg is passed to the invokable object as
lvalue argument: the argument vn in the std::invoke call above is
simply arg and the corresponding type Vn is T cv &, where cv is the
same cv-qualification as that of g.
But in this case, run_foo is cv-unqualified. What am I missing?
MWE:
#include <functional>
int main() {
int i;
std::bind([] (auto& x) {x = 1;}, i)();
}
[func.bind]/(10.4) states that the cv-qualifiers of the argument passed to the lambda are those of the argument to bind, augmented by the cv-qualifiers of the call wrapper; but there are none, and thus a non-const int should be passed in.
Both libc++ and libstdc++ fail to resolve the call. For libc++, reported as #32856, libstdc++ as #80564. The main problem is that both libraries infer the return type in the signature somehow, looking like this for libstdc++:
// Call as const
template<typename... _Args, typename _Result
= decltype( std::declval<typename enable_if<(sizeof...(_Args) >= 0),
typename add_const<_Functor>::type&>::type>()(
_Mu<_Bound_args>()( std::declval<const _Bound_args&>(),
std::declval<tuple<_Args...>&>() )... ) )>
_Result operator()(_Args&&... __args) const
During template argument deduction as necessitated by overload resolution, the default template argument will be instantiated, which causes a hard error due to our ill-formed assignment inside the closure.
This can be fixed by perhaps a deduced placeholder: remove _Result and its default argument entirely, and declare the return type as decltype(auto). This way, we also get rid of SFINAE which influences overload resolution and thereby induces incorrect behaviour:
#include <functional>
#include <type_traits>
struct A {
template <typename T>
std::enable_if_t<std::is_const<T>{}> operator()(T&) const;
};
int main() {
int i;
std::bind(A{}, i)();
}
This should not compile—as explained above, the argument passed to A::operator() should be non-const because i and the forwarding call wrapper are. However, again, this compiles under libc++ and libstdc++, because their operator()s fall back on const versions after the non-const ones fail under SFINAE.

How to use bind correctly here?

I can not figure out the correct syntax to bind member functions.
If I have a function that takes a function with a single argument,
how do I pass an object to it?
In the following example, what would be the correct syntax of passing the function?
#include <iostream>
#include <functional>
void caller(std::function<void(int)> f)
{
f(42);
}
class foo
{
public:
void f(int x)
{
std::cout<<x<<std::endl;
}
};
int main()
{
foo obj;
caller(std::bind(&foo::f,obj));
//^Wrong
}
Error was:
a.cpp: In function ‘int main()’:
a.cpp:18:34: error: could not convert ‘std::bind(_Func&&, _BoundArgs&& ...) [with _Func = void (foo::*)(int); _BoundArgs = {foo&}; typename std::_Bind_helper<std::__or_<std::is_integral<typename std::decay<_Tp>::type>, std::is_enum<typename std::decay<_Tp>::type> >::value, _Func, _BoundArgs ...>::type = std::_Bind<std::_Mem_fn<void (foo::*)(int)>(foo)>]((* & obj))’ from ‘std::_Bind_helper<false, void (foo::*)(int), foo&>::type {aka std::_Bind<std::_Mem_fn<void (foo::*)(int)>(foo)>}’ to ‘std::function<void(int)>’
caller(std::bind(&foo::f,obj));
Placeholders create a "space" for the actual arguments to be bound later:
int main()
{
foo obj;
caller(std::bind(&foo::f, &obj, std::placeholders::_1));
// ^ placeholder
// ^ address or even foo()
}
These placeholders are needed to correctly generate an appropriate signature for the std::bind result to be bound to the std::function<void(int)>.
You may also want to use the address of your object, or std::ref (so it won't be copied); this will vary on what you want the semantics to be.
There is an implicit first argument to member functions, which is the this point. You need to send it as a pointer; you also need a placeholder for the int argument. So:
caller(std::bind(&foo::f, &obj, std::placeholders::_1));
// ^ ^^^^^^^^^^^^^^^^^^^^^^^
You need to specify that the created function object takes a parameter using a placeholder:
std::bind(&foo::f,obj, std::placeholders::_1)

C++1y no viable conversion from std::bind to std::function

I am trying to store a forward function into std::function. If I use std::bind, I get error message like no viable conversion from .... If I use lambda, it compile okay.
Here is sample code
#include <functional>
template<typename Handler>void func1(int a, Handler&& handler) {}
template<typename Handler>void func2(Handler&& handler)
{
// this line compile fine
std::function<void ()> funcA = [handler = std::move(handler)]() { func1(1, std::move(handler)); };
// this line got compile error
std::function<void ()> funcB = std::bind(func1<Handler>, 1, std::move(handler));
}
int main()
{
func2(&main); // this just a sample, I am using functor as argument in real code
}
Trying both g++ --std=c++1y (v4.9.0) and clang++ --std=c++1y (v3.4.1) yield the same result
edit: clang++ error message
main.cpp:8:28: error: no viable conversion from 'typename _Bind_helper<__is_socketlike<void (*)(int, int (*&&)())>::value, void (*)(int, int
(*&&)()), int, int (*)()>::type' (aka '_Bind<__func_type (typename decay<int>::type, typename decay<int (*)()>::type)>') to
'std::function<void ()>'
std::function<void ()> funcB = std::bind(&func1<Handler>, 1, std::move(handler));
^ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
main.cpp:14:5: note: in instantiation of function template specialization 'func2<int (*)()>' requested here
func2(&main);
^
/usr/bin/../lib64/gcc/x86_64-unknown-linux-gnu/4.9.0/../../../../include/c++/4.9.0/functional:2181:7: note: candidate constructor not viable: no
known conversion from 'typename _Bind_helper<__is_socketlike<void (*)(int, int (*&&)())>::value, void (*)(int, int (*&&)()), int, int
(*)()>::type' (aka '_Bind<__func_type (typename decay<int>::type, typename decay<int (*)()>::type)>') to 'nullptr_t' for 1st argument
function(nullptr_t) noexcept
^
/usr/bin/../lib64/gcc/x86_64-unknown-linux-gnu/4.9.0/../../../../include/c++/4.9.0/functional:2192:7: note: candidate constructor not viable: no
known conversion from 'typename _Bind_helper<__is_socketlike<void (*)(int, int (*&&)())>::value, void (*)(int, int (*&&)()), int, int
(*)()>::type' (aka '_Bind<__func_type (typename decay<int>::type, typename decay<int (*)()>::type)>') to 'const std::function<void ()> &'
for 1st argument
function(const function& __x);
^
/usr/bin/../lib64/gcc/x86_64-unknown-linux-gnu/4.9.0/../../../../include/c++/4.9.0/functional:2201:7: note: candidate constructor not viable: no
known conversion from 'typename _Bind_helper<__is_socketlike<void (*)(int, int (*&&)())>::value, void (*)(int, int (*&&)()), int, int
(*)()>::type' (aka '_Bind<__func_type (typename decay<int>::type, typename decay<int (*)()>::type)>') to 'std::function<void ()> &&' for
1st argument
function(function&& __x) : _Function_base()
^
/usr/bin/../lib64/gcc/x86_64-unknown-linux-gnu/4.9.0/../../../../include/c++/4.9.0/functional:2226:2: note: candidate template ignored:
substitution failure [with _Functor = std::_Bind<void (*(int, int (*)()))(int, int (*&&)())>]: no matching function for call to object of
type 'std::_Bind<void (*(int, int (*)()))(int, int (*&&)())>'
function(_Functor);
^
1 error generated.
INTRODUCTION
std::bind will try to call func1<Handler> with an lvalue-reference, but your instantiation of func1 will make it only accept rvalues.
EXPLANATION
Here we have reduced your testcase to the bare minimum to show what is going on, the snippet below is ill-formed and an explanation will follow to why that is.
#include <functional>
template<class T>
void foobar (T&& val);
int main() {
std::function<void()> f = std::bind (&foobar<int>, std::move (123));
}
In the above we will instantiate foobar with T = int, which makes the type of argument val to be an rvalue-reference to int (int&&).
std::move(123) will move-construct our value to be stored inside the object created by std::bind, but the Standard says that when std::bind later invokes the stored function, all arguments are passed as TiD cv &; ie. as lvalues.
This behavior is mandated by the Standard (n3797), as stated in section [func.bind.bind]p10.
By changing the previous ill-formed snippet into the following, no error will be raised, since foobar<int> now accepts an lvalue-reference; suitable to be bound to the lvalue passed to our function by the function-object returned by std::bind.
std::function<void()> f = std::bind (&foobar<int&>, std::move (123));
???
#include <functional>
#include <type_traits>
#include <iostream>
int main() {
auto is_lvalue = [](auto&& x) {
return std::is_lvalue_reference<decltype(x)> { };
};
auto check = std::bind (is_lvalue, std::move (123));
bool res = check (); // res = true
}
in short: function has to be copyable. bind with rvalue returns non-copyable object. Workaround is to capture/bind with shared_ptr containing abovementioned value