I do not understand why neither the lambda nor the function are recognized as an std::invocable compliant types in the following code:
#include <concepts>
#include <iostream>
void f( std::invocable auto callback)
{
callback(47);
}
void function_callback(int i)
{
std::cout << i << std::endl;
}
auto lambda_callback = [](int i )
{
std::cout << i << std::endl;
};
int main(int)
{
f(&function_callback);
f(lambda_callback);
}
I am using GCC trunk with -std=c++2a flag enabled.
If you look at the definition of invocable (or in the standard):
template< class F, class... Args >
concept invocable =
requires(F&& f, Args&&... args) {
std::invoke(std::forward<F>(f), std::forward<Args>(args)...);
/* not required to be equality preserving */
};
What this means:
void f( std::invocable auto callback)
And it might be clearer if we write it long-form:
template <typename F>
requires std::invocable<F>
void f(F callback);
Is that F is invocable with no arguments - that it's a nullary function (Args... is an empty pack here). Neither your function nor your lambda are nullary functions - they're both unary, so the constraint correctly rejects those.
What you probably want is:
void f( std::invocable<int> auto callback)
which checks if callback is invocable with a single argument of type int.
Related
I am trying to use a Concept to constrain a template parameter to only allow invocables that are noexcept.
E.g.:
template<NoExceptFunc Fn>
void foo(Fn invocable) noexcept {
invocable();
}
The difficulty I have is that the invocable should be any type of invocable (free function, lambda, ...) and allow any number of parameters.
EDIT: Thanks to How can unspecified types be used in C++20 'requires' expressions?, I understand this is not possible to do in the general case, you need to know what invocable actually is, and then can constraint that.
Still, even when I know what invocable actually is, I do not know how to express the contrain, see below.
The following
template <typename T, typename... Args>
concept NoExceptFunc = requires(T &&f, Args &&...args) { requires noexcept(f(args...)); };
works, but requires that the arguments types be explicitly given, e.g.:
template<NoExceptFunc<int, int> Fn>
void foo(Fn invocable) noexcept {
invocable(0, 1);
}
The problem is that I do not know how to specify the parameters in the following case:
template <NoExceptFunc<??> Fn>
void foo(Fn f) noexcept {
f(1);
f("hello", "world");
}
int main() {
auto lambda = [](auto&& a, auto&&... rest) {
std::cout << a;
if constexpr(sizeof...(rest) != 0) {
std::cout << (rest << ...);
}
std::cout << std::endl;
};
foo(lambda);
}
How can I say that the parameters can be <int> OR <const char*, const char*>?
Based on: How do I expand a tuple into variadic template function's arguments?
#include <string>
#include <iostream>
#include <tuple>
template <typename... Args>
void print_all(const Args &... args) {
((std::cout << " " << args), ...) << std::endl;
}
int main()
{
// Create a tuple
auto values = std::make_tuple(1, 2, 3.4f, 4.5, "bob");
// Need to pass the tuple through the lambda for template type deduction and to pass param to template function?
std::apply([](auto &&... args) { print_all(args...); }, values);
// This does not work - other then there is no parameter I can't see how this does not work
// and how the lambda does work as it has the same (roughly) param list
std::apply(print_all(), values);
return 0;
}
can someone explain why one works and the other does not?
Behind the scences, this lambda expression [](auto &&... args) { print_all(args...); } is roughly of type:
struct [unnamed] {
template <typename ...Args>
void operator()(Args&&...args) { ... };
};
It is a type with a templated operator(), ie overload resolution and template argument dedcution only take place once the operator() is actually called. print_all on the other hand is a template, hence you cannot pass it to std::apply.
In other words, no matter what Args... is, the lambda is always of same type, but print_all isn't. You would need to instantiate it before you can get a pointer to the function. As mentioned by Scheff, this is fine:
std::apply(&print_all<int, int, float, double, const char*>, values);
What I'm trying to achieve is creating a struct which stores any kind of method. I can later call struct_object.run() to run the method I've stored.
This method can return any kind of value and, most importantly, use any amount of parameters; however, I can't get around the "any amount of parameters" issue.
Mind you, the following code doesn't even build, mostly because I have no clue on what the correct syntax would be like.
ApplicationPair.h
template<typename T, typename... Args>
struct ApplicationPair
{
ApplicationPair(boost::function<T()> func, Args... arguments )
{
_func = func(Args::arguments...);
}
ApplicationPair() = delete;
void run();
boost::function<T(Args...)> _func;
};
#endif
And then, what I'd like to do is the following:
main.cpp
template<typename T, typename... Args>
void ApplicationPair<T,Args...>::run()
{
this->_func;
}
//TEST
int counter = 0;
void HelloWorld()
{
std::cout << "HelloWorld\n";
}
void printNumber(int i)
{
std::cout << "Print: " << i << std::endl;
}
void increaseCounter(int x)
{
counter+=x;
}
int main()
{
ApplicationPair<void> p1(HelloWorld);
ApplicationPair<void> p2(printNumber, 5);
ApplicationPair<void> p3(increaseCounter, 10);
p1.run();
p2.run();
p3.run();
return 0;
}
Basically, the methods I want to store shouldn't be modified or adapted in any way: I want to be able to create any kind of method without caring about the fact that struct ApplicationPair will store it for its own personal use.
All I get with this though is a long string of errors like:
error: in declaration ‘typename boost::enable_if_c<(! boost::is_integral::value), boost::function&>::type boost::function::operator=(Functor)’
In the below line:
ApplicationPair<void> p2(printNumber, 5);
you have to specify all types in template arguments list, not only void as return type, int as argument of constructor should also be added. Now args... is empty. What is wrong. The same with p3.
Make constructor as templated method taking paramters pack as argument for your callable:
template<class F, class ... Args>
ApplicationPair(F&& func, Args... arguments )
{
_func = boost::bind(std::forward<F>(func),arguments...);
}
then args... can be deduced when invoking constructor. Your class template takes only a type for return value.
template<class Ret>
struct ApplicationPair {
template<class F, class ... Args>
ApplicationPair(F&& func, Args... arguments )
{
_func = boost::bind(std::forward<F>(func),arguments...);
}
ApplicationPair() = delete;
void run() {
this->_func();
}
boost::function<Ret()> _func;
};
In constructor boost::bind is used to bind passed parameters to callable. You don't store parameters anywhere, therefore they must be bound in functor created by boost::bind.
Uses:
ApplicationPair<void> p1(HelloWorld);
ApplicationPair<void> p2(printNumber, 5);
ApplicationPair<void> p3(increaseCounter, 10);
Demo
Don't use boost::bind, it is limited to handle only max 9 arguments.
You've already gotten an answer but here's a C++17 alternative capable of deducing the return value type as well as the argument types of the function using a deduction guide, making both the return type and argument types part of the ApplicationPair<> type. I've chosen to store the arguments separately in a std::tuple<Args...>.
boost::function can be replaced with std::function in this example in case you later decide to go with the standard:
#include <boost/function.hpp>
#include <iostream>
#include <type_traits>
#include <tuple>
template<typename T, typename... Args>
struct ApplicationPair {
ApplicationPair() = delete;
ApplicationPair(Func func, Args... args) :
_func(func),
// store the arguments for later use
arguments(std::make_tuple(std::forward<Args>(args)...))
{}
decltype(auto) run() { // I'd rename this: decltype(auto) operator()()
return std::apply(_func, arguments);
}
boost::function<T(Args...)> _func;
std::tuple<Args...> arguments;
};
// deduction guide
template<typename Func, typename... Args>
ApplicationPair(Func, Args...) ->
ApplicationPair<std::invoke_result_t<Func, Args...>, Args...>;
int counter = 0;
void HelloWorld()
{
std::cout << "HelloWorld\n";
}
void printNumber(int i)
{
std::cout << "Print: " << i << std::endl;
}
int increaseCounter(int x) // changed return type for demo
{
counter+=x;
return counter;
}
int main()
{
// full deduction using the deduction guide
ApplicationPair p1(HelloWorld);
ApplicationPair p2(printNumber, 5);
ApplicationPair p3(increaseCounter, 10);
p1.run();
p2.run();
std::cout << p3.run() << '\n';
std::cout << p3.run() << '\n';
}
I'm currently reading a few books to get caught up on c++14 features. I am trying to use a variadic template to bind arguments to a function. I know how to do this using std::bind, but I would also like to implement this function with a c++14 lambda expression, just for common knowledge and understanding, and for any possible performance benefits. I've read that lambdas can be inlined while std::bind cannot inline because it takes place through a call to a function pointer.
Here is the code from myFunctions.h:
#include <functional>
int simpleAdd(int x, int y) {
return x + y;
}
//function signatures
template<class Func, class... Args>
decltype(auto) funcBind(Func&& func, Args&&...args);
template<class Func, class... Args>
decltype(auto) funcLambda(Func&& func, Args&&...args);
/////////////////////////////////////////////////////////////////
//function definitions
template<class Func, class... Args>
inline decltype(auto) funcBind(Func&& func, Args&&... args)
{
return bind(forward<Func>(func), forward<Args>(args)...);
}
template<class Func, class ...Args>
inline decltype(auto) funcLambda(Func && func, Args && ...args)
{ //The error is caused by the lambda below:
return [func, args...]() {
forward<Func>(func)(forward<Args>(args)...);
};
}
Here is the main code I am running:
#include<iostream>
#include<functional>
#include "myFunctions.h"
using namespace std;
int main()
{
cout << "Application start" << endl;
cout << simpleAdd(5,7) << endl;
auto f1 = funcBind(simpleAdd,3, 4);
cout << f1() << endl;
//error is occurring below
auto f2 = funcLambda(simpleAdd, 10, -2);
cout << f2() << endl;
cout << "Application complete" << endl;
Error C2665 'std::forward': none of the 2 overloads could convert all the argument types
Error C2198 'int (__cdecl &)(int,int)': too few arguments for call
I think the error might be occurring when the variadic arguments are getting forwarded to the lambda, but I'm not really sure.
My question is how do I properly formulate this code so that I can use a lambda to capture the function and its arguments, and call it later.
I've read that lambdas can be inlined while std::bind cannot inline
because it takes place through a call to a function pointer.
If you pass simpleAdd to something that then binds the arguments, then whether you use bind or not doesn't matter. What do you think the lambda captures with func? It's a function pointer.
The lambda-vs-function-pointer case is about writing bind(simpleAdd, 2, 3) vs. [] { return simpleAdd(2, 3); }. Or binding a lambda like [](auto&&...args) -> decltype(auto) { return simpleAdd(decltype(args)(args)...); } vs. binding simpleAdd directly (which will use a function pointer).
In any event, implementing it is surprisingly tricky. You can't use by-reference capture because things can easily get dangling, you can't use a simple by-value capture because that would always copy the arguments even for rvalues, and you can't do a pack expansion in an init-capture.
This follows std::bind's semantics (invoking the function object and passing all bound arguments as lvalues) except that 1) it doesn't handle placeholders or nested binds, and 2) the function call operator is always const:
template<class Func, class ...Args>
inline decltype(auto) funcLambda(Func && func, Args && ...args)
{
return [func = std::forward<Func>(func),
args = std::make_tuple(std::forward<Args>(args)...)] {
return std::experimental::apply(func, args);
};
}
cppreference has an implementation of std::experimental::apply.
Note that this does unwrap reference_wrappers, like bind, because make_tuple does it.
Your original code breaks down because args are const in the lambda's function call operator (which is const by default), and the forward ends up attempting to cast away constness.
You use a tuple:
template<class Func, class ...Args>
inline decltype(auto) funcLambda(Func && func, Args && ...args)
{ //The error is caused by the lambda below:
auto tpl = make_tuple(std::forward<Args>(args)...);
//Use move just in case Args has move-only types.
return [func, tpl = move(tpl)]() {
apply(func, tpl);
};
}
Where apply is defined something like this:
namespace detail {
template <class F, class Tuple, std::size_t... I>
constexpr decltype(auto) apply_impl( F&& f, Tuple&& t, std::index_sequence<I...> )
{
return f(std::get<I>(std::forward<Tuple>(t))...);
}
} // namespace detail
template <class F, class Tuple>
constexpr decltype(auto) apply(F&& f, Tuple&& t)
{
return detail::apply_impl(std::forward<F>(f), std::forward<Tuple>(t),
std::make_index_sequence<std::tuple_size<std::decay_t<Tuple>>>::value);
}
apply is a feature of one of the library TS versions. With C++17, apply_impl could call invoke, which would work for any callable.
I have the following problem. Say you want to write a generic function that can take a lambda expression. I understand that if the parameter is of type std::function, then I could not only use lambdas, but also functions and even pointers to functions. So at a first step, I did the following:
void print(std::function<void(int, int)> fn) {
fn(1,2);
}
int main() {
print([](int i, int j) { std::cout << j <<','<<i<<'\n'; });
return 0;
}
Now the problem is that I want to make this function generic, meaning that I don't want the lambda expression to have only two parameters.
So I tried changing the signature of the print function to something more generic like:
template <class function_type>
void print(function_type fn);
But now the problem is that the function takes ANY object and I'm not ok with that.
But the main problem is that, I have no idea how many parameters the object fn can accept.
So in a way I'm looking for a compile time way to determine how many arguments fn has, and if possible to change the type of fn to std::function. And then, given that I know the number of parameters that fn accepts, is there a generic way to pack an arbitrary number of parameters to be passed to fn? I don't even know if this is possible within C++11. What I mean is that given the number of arguments, is there a way to pack parameters to pass to fn? So that if there are two arguments, then I would call
fn(arg1, arg2);
if there are three:
fn(arg1, arg2, arg3);
and so on.
Thank you all for your insight.
aa
The following snippets might be useful.
This gives the number of arguments that a std::function takes
template <typename Signature>
struct count_args;
template <typename Ret, typename... Args>
struct count_args<std::function<Ret(Args...)>> {
static constexpr size_t value = sizeof...(Args);
};
For example the following code compiles (clang 3.2, gcc 4.7.2 and icc 13.1.0)
static_assert(count_args<std::function<void() >>::value == 0, "Ops!");
static_assert(count_args<std::function<void(int) >>::value == 1, "Ops!");
static_assert(count_args<std::function<void(int, int)>>::value == 2, "Ops!");
As far as I understand, you want to call the function object passing the correct number of arguments, right? Then for each argument we need to provide a value which is convertible to its type. A solution with this generality is very hard (or even impossible). Hence, I'll present two alternatives.
1 Each argument is a value initialized object of its type. (This is what ecatmur suggested.)
template <typename Ret, typename... Args>
Ret call(const std::function<Ret(Args...)>& f) {
return f(Args{}...); // for the intel compiler replace {} with ()
}
2 A fixed value is given and all the arguments are implicitly initialized from this value:
template <typename Ret, typename... Args, typename Val, typename... Vals>
typename std::enable_if<sizeof...(Args) == sizeof...(Vals), Ret>::type
call(const std::function<Ret(Args...)>& f, const Val&, const Vals&... vals) {
return f(vals...);
}
template <typename Ret, typename... Args, typename Val, typename... Vals>
typename std::enable_if<(sizeof...(Args) > sizeof...(Vals)), Ret>::type
call(const std::function<Ret(Args...)>& f, const Val& val, const Vals&... vals) {
return call(f, val, val, vals...);
}
The three overloads are unambiguous and can be used as the following examples show:
{
std::function<char()> f = []() -> char {
std::cout << "f() ";
return 'A';
};
std::cout << call(f) << std::endl; // calls f()
std::cout << call(f, 0) << std::endl; // calls f()
}
{
std::function<char(int)> f = [](int i) -> char {
std::cout << "f(" << i << ") ";
return 'B';
};
std::cout << call(f) << std::endl; // calls f(0)
std::cout << call(f, 1) << std::endl; // calls f(1)
}
{
std::function<char(int, int)> f = [](int i, int j) -> char {
std::cout << "f(" << i << "," << j << ") ";
return 'C';
};
std::cout << call(f) << std::endl; // calls f(0, 0)
std::cout << call(f, 2) << std::endl; // calls f(2, 2)
}
Yes you can pack as many parameters to fn as you wish using variadic templates.
template <class function_type, class... Args>
void print(function_type fn, Args... args)
{
//Call fn with args
fn(std::forward<Args>(args...));
}
To find out how many args there are in the parameter pack, you can use sizeof...(args).
To determine the signature of a callable, you can use the solution from Inferring the call signature of a lambda or arbitrary callable for "make_function". You can then package the callable into a std::function, or create a tag and use parameter inference:
template<typename T> struct tag {};
template<typename F, typename... Args>
void print_impl(F &&fn, tag<void(Args...)>) {
fn(Args{}...);
}
template<typename F>
void print(F &&fn) {
print_impl(std::forward<F>(fn), tag<get_signature<F>>{});
}
Note this uses value-initialised arguments; if you want anything more complex you can build a std::tuple<Args...> and pass that along, invoking it per "unpacking" a tuple to call a matching function pointer.