I want to make functor to generic function, but I get compiler error.
Here is the code:
template <class T>
struct Creator
{
template <typename...Ts>
static std::shared_ptr<T> create(Ts&&... vs)
{
std::shared_ptr<T> t(new T(std::forward<Ts>(vs)...));
return t;
}
};
class Car:
public Creator<Car>
{
private:
friend class Creator<Car>;
Car()
{
}
};
int main()
{
auto car=Car::create();
std::function< std::shared_ptr<Car> () > createFn=&Car::create;
return 0;
}
I get the following error in GCC 4.6.3 on the second statement(the first is OK):
error: conversion from ‘<unresolved overloaded function type>’
to non-scalar type ‘std::function<std::shared_ptr<Car>()>’ requested
Any hint appreciated.
If the pointer of a template function is needed, the template must be instantiated first.
std::function<std::shared_ptr<Car>()> createFn = &Car::create<>;
This will make it compile on clang++ 3.1, but g++ 4.8 still refuses to compile, which I believe is a bug.
You could provide a lambda function instead:
std::function<std::shared_ptr<Car>()> createFn = []{ return Car::create(); };
Related
I am trying to pass a pointer to the predicate function into the Foo and Bar functions.
The Bar function works correctly, but the Foo function raises a compile-time error:
error: no matching function for call to Foo<int>(bool (&)(int))
Why does the compiler raise an error?
Is there any difference between Foo's and Bar's template arguments types after Args' unpacking?
#include <functional>
bool predicate(int a) {
return (a > 5);
}
// sizeof...(Args) == 1 and I suppose it is int
template<typename... Args>
void Foo(std::function<bool(Args...)> predicate) {
// clang: note: candidate template ignored:
// could not match 'function<bool (int, type-parameter-0-0...)>'
// against 'bool (*)(int)'
}
template<typename Args>
void Bar(std::function<bool(Args)> predicate) {
}
int main(int argc, char const *argv[]) {
// gcc: error: no matching function for call to
// 'Foo<int>(bool (&)(int))'
Foo<int>(predicate);
Bar<int>(predicate);
return 0;
}
See Compiler Explorer for a live example.
I also tried to change the Foo function a little and it works somehow:
template<typename... Args>
void Foo(bool(*predicate)(Args...)) {
std::function<bool(Args...)> func(predicate);
}
I want to have std::function type argument in the Foo function, but I don't know how to do it
The error is because the exact type of std::function is not same as predicate. To get around this, you can explicitly call the constructor of std::function:
int main() {
Foo<int>( std::function<bool(int){predicate} );
//OR
Foo<int>( {predicate} );
return 0;
}
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 wrote some code that retrieves the types of the non-auto parameters when given a generic lambda function. As you can see in the code below, the idea is to call the connect function with a generic lambda and provide arguments for the auto parameters (which will always be at the front in my use case). So in the code below my goal was to detect that the second parameter is of type float.
The code works fine with clang 3.8 but it doesn't compile with gcc 6.1.1, so I was wondering whether this was a bug in gcc or if this is just not valid c++ code? Can I assume that a generic lambda is implemented with a templated operator() function or is this compiler-specific?
template <typename Functor, typename... AllArgs, typename... ProvidedArgs>
void findArgTypes(void(Functor::*)(AllArgs...) const, Functor, ProvidedArgs...)
{
// AllArgs == int, float
// ProvidedArgs == int
}
template <typename Func, typename... ProvidedArgs>
void connect(Func func, ProvidedArgs... providedArgs)
{
findArgTypes(&Func::template operator()<ProvidedArgs...>, func, providedArgs...);
}
int main()
{
int tmp = 0;
connect([&](auto, float){ ++tmp; }, 0);
}
The error that gcc gives is this:
main.cpp: In instantiation of ‘void connect(Func, ProvidedArgs ...) [with Func = main()::<lambda(auto:1, float)>; ProvidedArgs = {int}]’:
main.cpp:16:33: required from here
main.cpp:11:17: error: no matches converting function ‘operator()’ to type ‘void (struct main()::<lambda(auto:1, float)>::*)() const’
findArgTypes(&Func::template operator()<ProvidedArgs...>, func, providedArgs...);
~~~~~~~~~~~~^~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
main.cpp:16:27: note: candidate is: template<class auto:1> main()::<lambda(auto:1, float)>
connect([](auto, float){}, 0);
^
Removing the const in findArgTypes gives the same result.
Using the following code works with both compilers:
struct Foo
{
template <typename T>
void operator()(T, float) const {}
};
int main()
{
Foo f;
connect(f, 0);
}
You have error because you are expecting functor (object) but lambda with empty capture is convertible to free function:
int main() {
using function = void (*)(int, float);
function a = [](auto, float){};
}
See lambda from cppreference:
For the newest version of your question that implementation satisfies both compilers:
template <typename Func, typename... ProvidedArgs>
void connect(Func func, ProvidedArgs... providedArgs)
{
auto mf = &Func::template operator()<ProvidedArgs...>;
findArgTypes(mf, func, providedArgs...);
}
I think this is gcc compiler bug that gcc needs this auto local variable to work correctly...
BTW, one question - one bug in clang, one in gcc - I really advice you to find simpler way to achieve your goals - maybe consider to just use std::function instead of quite fresh generic-lambda?
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.
I like to use local classes in template classes to perform constructions like "static if". But I've faced with the problem that gcc 4.8 does not want to compile my code. However 4.7 does.
This sample:
#include <type_traits>
#include <iostream>
#include <string>
using namespace std;
struct A {
void printA() {
cout << "I am A" << endl;
}
};
struct B {
void printB() {
cout << "I am B" << endl;
}
};
template <typename T>
struct Test {
void print() {
struct IfA {
constexpr IfA(T &value) : value(value) {
}
T &value;
void print() {
value.printA();
}
};
struct IfB {
constexpr IfB(T &value) : value(value) {
}
T &value;
void print() {
value.printB();
}
};
struct Else {
constexpr Else(...) {}
void print() {
}
};
typename conditional<is_same<T, A>::value, IfA, Else>::type(value).print();
typename conditional<is_same<T, B>::value, IfB, Else>::type(value).print();
}
T value;
};
int main() {
Test<A>().print();
Test<B>().print();
}
Options:
g++ --std=c++11 main.cc -o local-sfinae
Task:
Given classes A and B with different interfaces for printing.
Write a generic class Test that can print both A and B.
Do not pollute either any namespace or class scope.
Description of the code:
This is only a clean example.
I use an approach like this, because I want to generalize the construction "static if". See, that I pass the arguments to IfA and IfB classes via their fields, not directly to the print() function.
I use such constructions a lot.
I've found that these constructions should not be in (pollute) class scope. I mean they should be placed in a method scope.
So the question.
This code can not be compiled with GCC 4.8. Because it checks ALL classes, even if they are never used. But it has not instantiate them in binary (I've commented the lines that cause errors and compiled it with gcc 4.8). Proof:
$ nm local-sfinae |c++filt |grep "::If.*print"
0000000000400724 W Test<A>::print()::IfA::print()
00000000004007fe W Test<B>::print()::IfB::print()
See, there is no Test::print()::IfB::print(). (See later: 'void Test::print()::IfB::print() [with T = A]')
The errors if I compile aforementioned code with gcc 4.8:
g++ --std=c++11 main.cc -o local-sfinae
main.cc: In instantiation of 'void Test<T>::print()::IfB::print() [with T = A]':
main.cc:36:9: required from 'void Test<T>::print() [with T = A]'
main.cc:49:21: required from here
main.cc:34:17: error: 'struct A' has no member named 'printB'
value.printB();
^
main.cc: In instantiation of 'void Test<T>::print()::IfA::print() [with T = B]':
main.cc:28:9: required from 'void Test<T>::print() [with T = B]'
main.cc:50:21: required from here
main.cc:26:17: error: 'struct B' has no member named 'printA'
value.printA();
^
Is it a GCC 4.8 bug?
Or is it GCC 4.7 bug? Maybe the code should not be compiled.
Or it is a my bug, and I should not rely on the compiler behavior/should not use such approach to implement "static if".
Additional info:
This simple code compiles on 4.7, but not on 4.8. I shortened it.
struct A {
void exist() {
}
};
template <typename T>
struct Test {
void print() {
struct LocalClass {
constexpr LocalClass(T &value) : value(value) {
}
T &value;
void print() {
value.notExist();
}
};
}
T value;
};
int main() {
Test<A>().print();
}
Errors:
main.cc: In instantiation of 'void Test<T>::print()::LocalClass::print() [with T = A]':
main.cc:16:9: required from 'void Test<T>::print() [with T = A]'
main.cc:22:21: required from here
main.cc:14:17: error: 'struct A' has no member named 'notExist'
value.notExist();
^
Have tested two GCC 4.8 versions: 2012.10 and 2013.02. Hope it is GCC 4.8 bug and it can be fixed.
LocalClass is not a template. The "not instantiated if not used" rule is only applicable to member functions of class templates.
That is, when Test::print() is instantiated, everything that is inside is brought to life, including the unused member of its local class.
There is no SFINAE in your code.
SFINAE applies during template argument deduction and argument substitution (the 'S' in SFINAE stands for substitution) but the only substitution in your program happens when substituting A for T in the template parameter list of Test, which doesn't fail.
You then call print() which instantiates Test<A>::print(), which doesn't involve any substitution, and you get an error because value.notExist(); is not valid.
SFINAE has to be used in substitution contexts, such as template argument deduction caused by a function call or when deducing template parameters with default arguments.