I'm trying to implement Functor and various other category-theoretic concepts using C++ concepts, but am getting compile errors:
http://coliru.stacked-crooked.com/a/e8b6eb387229bddf
Here's my full code (I know that requiring fmap<int, int> does not verify fmap for any two types, and I plan to change it to fmap<int, std::string> or something to achieve a slightly stronger test -- or instead, possibly alter the Functor concept so that it takes in addition to F, two types T and U and verifies the existence of fmap<T, U>, but that's all after I figure out how to fix the error that I'm getting):
#include <functional>
#include <iostream>
#include <vector>
// empty Functor_Impl struct - specialize for each functor
template<template<class> class F> struct Functor_Impl {};
// std::vector Functor implementation
template<>
struct Functor_Impl<std::vector> {
template<class T, class U>
static std::vector<U> fmap(std::vector<T> x, std::function<U(T)> f) {
std::vector<U> out;
out.reserve(x.size());
for (int i = 0; i < x.size(); i++) {
out.push_back(f(x[i]));
}
return out;
}
};
// Functor concept requires Functor_Impl<F> to have fmap
template<template<class> class F>
concept bool Functor = requires(F<int> x) {
{Functor_Impl<F>::template fmap<int, int>(x)} -> F<int>;
};
// Test function using constraint.
template<template<class> class F, class T>
requires Functor<F>
F<T> mult_by_2(F<T> a) {
return Functor_Impl<F>::template fmap<T, T>(a, [](T x) {
return x * 2;
});
}
int main() {
std::vector<int> x = {1, 2, 3};
std::vector<int> x2 = mult_by_2(x);
for (int i = 0; i < x2.size(); i++) {
std::cout << x2[i] << std::endl;
}
}
And the compile error:
lol#foldingmachinebox:~/p/website-editor$ g++ foo.cpp -std=c++17 -fconcepts -o foo
foo.cpp: In function ‘int main()’:
foo.cpp:39:38: error: cannot call function ‘F<T> mult_by_2(F<T>) [with F = std::vector; T = int]’
std::vector<int> x2 = mult_by_2(x);
^
foo.cpp:31:6: note: constraints not satisfied
F<T> mult_by_2(F<T> a) {
^~~~~~~~~
foo.cpp:24:14: note: within ‘template<template<class> class F> concept const bool Functor<F> [with F = std::vector]’
concept bool Functor = requires(F<int> x) {
^~~~~~~
foo.cpp:24:14: note: with ‘std::vector<int> x’
foo.cpp:24:14: note: the required expression ‘Functor_Impl<F>::fmap<int, int>(x)’ would be ill-formed
I'm guessing that my syntax for the concept itself is wrong - that it's treating a variable as a function, or vice versa, since I'm not very familiar with the concept syntax, and in addition some of the example code on cppreference.com does not compile under GCC's implementation (e.g. concept EqualityComparable does not compile, it must be changed to concept bool EqualityComparable).
If I remove requires Functor<F> from the mult_by_2 function declaration, then the code compiles and runs.
The problem is exactly what the error message says: Functor_Impl<F>::template fmap<int, int>(x) is not a valid expression. Functor_Impl<std::vector>::fmap has two parameters, not one.
Related
when reading the document of std::span, I see there is no method to remove the first element from the std::span<T>.
Can you suggest a way to solve my issue?
The large picture of my problem(I asked in another question: How to instantiatiate a std::basic_string_view with custom class T, I got is_trivial_v<_CharT> assert error) is that I would like to have a std::basic_string_view<Token>, while the Token is not a trivial class, so I can't use std::basic_string_view, and someone suggested me to use std::span<Token> instead.
Since the basic_string_view has a method named remove_prefix which remove the first element, while I also need such kinds of function because I would like to use std::span<Token> as a parser input, so the Tokens will be matched, and consumed one by one.
Thanks.
EDIT 2023-02-04
I try to derive a class named Span from std::span, and add the remove_prefix member function, but it looks like I still have build issues:
#include <string_view>
#include <vector>
#include <span>
// derived class, add remove_prefix function to std::span
template<typename T>
class Span : public std::span<T>
{
public:
// Inheriting constructors
using std::span<T>::span;
// add a public function which is similar to std::string_view::remove_prefix
constexpr void remove_prefix(std::size_t n) {
*this = subspan(n);
}
};
struct Token
{
Token(){};
Token(const Token& other)
{
lexeme = other.lexeme;
type = other.type;
}
std::string_view lexeme;
int type;
// equal operator
bool operator==(const Token& other)const {
return (this->lexeme == other.lexeme) ;
}
};
template <typename T>
struct Viewer;
template <>
struct Viewer<Token>
{
using type = Span<Token>; // std::span or derived class
};
template <>
struct Viewer<char>
{
using type = std::string_view;
};
template <typename T> using ViewerT = typename Viewer<T>::type;
template <typename T>
class Parser
{
using v = ViewerT<T>;
};
// a simple parser demo
template <typename Base, typename T>
struct parser_base {
using v = ViewerT<T>;
constexpr auto operator[](v& output) const noexcept;
};
template<typename T>
struct char_ final : public parser_base<char_<T>, T> {
using v = ViewerT<T>;
constexpr explicit char_(const T ch) noexcept
: ch(ch)
{}
constexpr inline bool visit(v& sv) const& noexcept {
if (!sv.empty() && sv.front() == ch) {
sv.remove_prefix(1);
return true;
}
return false;
}
private:
T ch;
};
template <typename Parser, typename T>
constexpr bool parse(Span<T> &input, Parser const& parser) noexcept {
return parser.visit(input);
}
int main()
{
Token kw_class;
kw_class.lexeme = "a";
std::vector<Token> token_stream;
token_stream.push_back(kw_class);
token_stream.push_back(kw_class);
token_stream.push_back(kw_class);
Span<Token> token_stream_view{&token_stream[0], 3};
auto p = char_(kw_class);
parse(token_stream_view, p);
return 0;
}
The build error looks like below:
[ 50.0%] g++.exe -Wall -std=c++20 -fexceptions -g -c F:\code\test_crtp_twoargs\main.cpp -o obj\Debug\main.o
F:\code\test_crtp_twoargs\main.cpp: In member function 'constexpr void Span<T>::remove_prefix(std::size_t)':
F:\code\test_crtp_twoargs\main.cpp:52:17: error: there are no arguments to 'subspan' that depend on a template parameter, so a declaration of 'subspan' must be available [-fpermissive]
52 | *this = subspan(n);
| ^~~~~~~
F:\code\test_crtp_twoargs\main.cpp:52:17: note: (if you use '-fpermissive', G++ will accept your code, but allowing the use of an undeclared name is deprecated)
F:\code\test_crtp_twoargs\main.cpp: In instantiation of 'constexpr void Span<T>::remove_prefix(std::size_t) [with T = Token; std::size_t = long long unsigned int]':
F:\code\test_crtp_twoargs\main.cpp:113:29: required from 'constexpr bool char_<T>::visit(v&) const & [with T = Token; v = Span<Token>]'
F:\code\test_crtp_twoargs\main.cpp:125:24: required from 'constexpr bool parse(Span<T>&, const Parser&) [with Parser = char_<Token>; T = Token]'
F:\code\test_crtp_twoargs\main.cpp:141:10: required from here
F:\code\test_crtp_twoargs\main.cpp:52:24: error: 'subspan' was not declared in this scope, and no declarations were found by argument-dependent lookup at the point of instantiation [-fpermissive]
52 | *this = subspan(n);
| ~~~~~~~^~~
F:\code\test_crtp_twoargs\main.cpp:52:24: note: declarations in dependent base 'std::span<Token, 18446744073709551615>' are not found by unqualified lookup
F:\code\test_crtp_twoargs\main.cpp:52:24: note: use 'this->subspan' instead
F:\code\test_crtp_twoargs\main.cpp:52:15: error: no match for 'operator=' (operand types are 'Span<Token>' and 'std::span<Token, 18446744073709551615>')
52 | *this = subspan(n);
| ~~~~~~^~~~~~~~~~~~
F:\code\test_crtp_twoargs\main.cpp:44:7: note: candidate: 'constexpr Span<Token>& Span<Token>::operator=(const Span<Token>&)'
44 | class Span : public std::span<T>
| ^~~~
F:\code\test_crtp_twoargs\main.cpp:44:7: note: no known conversion for argument 1 from 'std::span<Token, 18446744073709551615>' to 'const Span<Token>&'
F:\code\test_crtp_twoargs\main.cpp:44:7: note: candidate: 'constexpr Span<Token>& Span<Token>::operator=(Span<Token>&&)'
F:\code\test_crtp_twoargs\main.cpp:44:7: note: no known conversion for argument 1 from 'std::span<Token, 18446744073709551615>' to 'Span<Token>&&'
Any idea on how to fix this issue?
Also, I don't know how to make a general parse function:
template <typename Parser, typename T>
constexpr bool parse(Span<T> &input, Parser const& parser) noexcept {
return parser.visit(input);
}
Currently, the first argument of the parse should be a Viewer like type?
EDIT2023-02-05
Change the function as below, the above code can build correctly. This is from Benjamin Buch's answer.
constexpr void remove_prefix(std::size_t n) {
auto& self = static_cast<std::span<T>&>(*this);
self = self.subspan(n);
}
There is still one thing remains: How to generalize the parse function to accept both input types of std::string_view and Span<Token>?
If I change the parse function to this:
template <typename Parser, typename T>
constexpr bool parse(ViewerT<T> &input, Parser const& parser) noexcept {
return parser.visit(input);
}
I got such compile error:
[ 50.0%] g++.exe -Wall -std=c++20 -fexceptions -g -c F:\code\test_crtp_twoargs\main.cpp -o obj\Debug\main.o
F:\code\test_crtp_twoargs\main.cpp: In function 'int main()':
F:\code\test_crtp_twoargs\main.cpp:143:24: error: no matching function for call to 'parse(Span<Token>&, char_<Token>&)'
143 | bool result = parse(token_stream_view, p);
| ~~~~~^~~~~~~~~~~~~~~~~~~~~~
F:\code\test_crtp_twoargs\main.cpp:125:16: note: candidate: 'template<class Parser, class T> constexpr bool parse(ViewerT<T>&, const Parser&)'
125 | constexpr bool parse(ViewerT<T> &input, Parser const& parser) noexcept {
| ^~~~~
F:\code\test_crtp_twoargs\main.cpp:125:16: note: template argument deduction/substitution failed:
F:\code\test_crtp_twoargs\main.cpp:143:24: note: couldn't deduce template parameter 'T'
143 | bool result = parse(token_stream_view, p);
| ~~~~~^~~~~~~~~~~~~~~~~~~~~~
Any ideas?
Thanks.
BTW: I have to explicitly instantiation of the parse function call like:
bool result = parse<decltype(p), Token>(token_stream_view, p);
to workaround this issue.
Call subspan with 1 as only (template) argument to get a new span, which doesn't contain the first element.
If you use a span with a static extend, you need a new variable because the data type changes by subspan.
#include <string_view>
#include <iostream>
#include <span>
int main() {
std::span<char const, 12> text_a("a test-span");
std::cout << std::string_view(text_a) << '\n';
std::span<char const, 10> text_b = text_a.subspan<2>();
std::cout << std::string_view(text_b) << '\n';
}
If you have a dynamic extend, you can assign the result to the original variable.
#include <string_view>
#include <iostream>
#include <span>
int main() {
std::span<char const> text("a test-span");
std::cout << std::string_view(text) << '\n';
text = text.subspan(2);
std::cout << std::string_view(text) << '\n';
}
The implementation of a modifying inplace subspan version is only possible for spans with a dynamic extend. It can be implemented as a free function.
#include <string_view>
#include <iostream>
#include <span>
template <typename T>
constexpr void remove_front(std::span<T>& self, std::size_t const n) noexcept {
self = self.subspan(n);
}
int main() {
std::span<char const> text("a test-span");
std::cout << std::string_view(text) << '\n';
remove_front(text, 2);
std::cout << std::string_view(text) << '\n';
}
You can use your own spans derived from std::span if you prefer the dot-call.
#include <string_view>
#include <iostream>
#include <span>
template <typename T>
struct my_span: std::span<T> {
using std::span<T>::span;
constexpr void remove_front(std::size_t const n) noexcept {
auto& self = static_cast<std::span<T>&>(*this);
self = self.subspan(n);
}
};
int main() {
my_span<char const> my_text("a test-span");
std::cout << std::string_view(my_text) << '\n';
my_text.remove_front(2);
std::cout << std::string_view(my_text) << '\n';
}
You can also write a wrapper class to call via dot syntax. This way you can additionally implement cascadable modification calls by always returning the a reference modifier class.
#include <string_view>
#include <iostream>
#include <span>
template <typename T>
class span_modifier {
public:
constexpr span_modifier(std::span<T>& span) noexcept: span_(span) {}
constexpr span_modifier& remove_front(std::size_t const n) noexcept {
span_ = span_.subspan(n);
return *this;
}
private:
std::span<T>& span_;
};
template <typename T>
constexpr span_modifier<T> modify(std::span<T>& span) noexcept {
return span;
}
int main() {
std::span<char const> text("a test-span");
std::cout << std::string_view(text) << '\n';
modify(text).remove_front(2).remove_front(5);
std::cout << std::string_view(text) << '\n';
}
Note I use the template function modify to create an object of the wrapper class, because the names of classes cannot be overloaded. Therefore class names should always be a bit more specific. The function modify can also be overloaded for other data types, which then return a different wrapper class. This results in a simple intuitive and consistent interface for modification wrappers.
You can write remove_prefix of your version,
template <typename T>
constexpr void remove_prefix(std::span<T>& sp, std::size_t n) {
sp = sp.subspan(n);
}
Demo
I want to write a function that returns an instance of a type T but behaves differently depending on how T can be constructed. Say I have structs like these
#include <type_traits>
#include <iostream>
struct A {};
struct B {};
struct C {
C(A a) {
std::cout << "C" << std::endl;
}
};
I want to create Cs by giving them an A. I have a struct like so that uses enable_if to choose one of two functions:
struct E {
template< bool condition = std::is_constructible<C, A>::value,std::enable_if_t<condition,int> = 0>
C get() {
return C{A{}};
}
template< bool condition = std::is_constructible<C, B>::value,std::enable_if_t<condition,bool> = false>
C get() {
return C{B{}};
}
};
This compiles fine with g++82 (and I think also g++9), but clang9 gives me the error
$ clang++ --std=c++17 main.cpp
main.cpp:26:12: error: no matching constructor for initialization of 'C'
return C{B{}};
^~~~~~
main.cpp:6:8: note: candidate constructor (the implicit copy constructor) not viable: no known conversion from 'B' to 'const C' for 1st argument
struct C {
^
main.cpp:6:8: note: candidate constructor (the implicit move constructor) not viable: no known conversion from 'B' to 'C' for 1st argument
struct C {
^
main.cpp:7:3: note: candidate constructor not viable: no known conversion from 'B' to 'A' for 1st argument
C(A a) {
^
1 error generated.
even though the enable_if should hide that function. (I call E e; auto c = e.get();). If I don't hardcode C but instead use a template to pass in C it works in both compilers.
template<typename T>
struct F {
template< bool condition = std::is_constructible<T, A>::value,std::enable_if_t<condition,int> = 0>
T get() {
return T{A{}};
}
template< bool condition = std::is_constructible<T, B>::value,std::enable_if_t<condition,bool> = false>
T get() {
return T{B{}};
}
};
I don't understand why clang apparently typechecks the body of the function even though the function should be disabled by enable_if.
Both compiler are right,
http://eel.is/c++draft/temp.res#8.1
The validity of a template may be checked prior to any instantiation. [ Note: Knowing which names are type names allows the syntax of every template to be checked in this way. — end note ] The program is ill-formed, no diagnostic required, if:
(8.1)
- no valid specialization can be generated for a template or a substatement of a constexpr if statement within a template and the template is not instantiated, or
[..]
(8.4)
- a hypothetical instantiation of a template immediately following its definition would be ill-formed due to a construct that does not depend on a template parameter, or
return C{B{}}; is not template dependent, and wrong. clang is fine by diagnosing the issue.
Since you seem to have access to compilers supporting c++17 you could use if constexpr instead of enable_if to accomplish what you want.
#include <iostream>
#include <type_traits>
struct A {};
struct B {};
struct C {
explicit C(A a) {
std::cout << "C" << std::endl;
}
};
template<typename T>
struct False : std::false_type {};
struct E {
template<typename T = void>
C get() const {
if constexpr (std::is_constructible_v<C, A>) {
return C{A{}};
} else if constexpr (std::is_constructible_v<C, B>) {
return C{B{}};
} else {
static_assert(False<T>::value, "Error");
}
}
};
int main() {
const auto C{E{}.get()};
return 0;
}
The following code compiles perfectly if:
I don't include <iostream> or
I name operator== as alp::operator==.
I suppose there is a problem with <iostream> and operator==, but I don't know what.
I compile the code with gcc 7.3.0, clang++-6.0 and goldbolt. Always the same error.
The problem is that the compiler is trying to cast the parameters of operator== to const_iterator, but why? (I suppose the compiler doesn't see my version of operator==, and looks for other versions).
#include <vector>
#include <iostream> // comment and compile
namespace alp{
template <typename It_base>
struct Iterator {
using const_iterator = Iterator<typename It_base::const_iterator>;
operator const_iterator() { return const_iterator{}; }
};
template <typename It_base>
bool operator==(const Iterator<It_base>& x, const Iterator<It_base>& y)
{ return true;}
}// namespace
struct Func{
int& operator()(int& p) const {return p;}
};
template <typename It, typename View>
struct View_iterator_base{
using return_type = decltype(View{}(*It{}));
using const_iterator =
View_iterator_base<std::vector<int>::const_iterator, Func>;
};
using view_it =
alp::Iterator<View_iterator_base<std::vector<int>::iterator, Func>>;
int main()
{
view_it p{};
view_it z{};
bool x = operator==(z, p); // only compiles if you remove <iostream>
bool y = alp::operator==(z,p); // always compile
}
Error message:
yy.cpp: In instantiation of ‘struct View_iterator_base<__gnu_cxx::__normal_iterator<const int*, std::vector<int> >, Func>’:
yy.cpp:9:73: required from ‘struct alp::Iterator<View_iterator_base<__gnu_cxx::__normal_iterator<const int*, std::vector<int> >, Func> >’
yy.cpp:44:29: required from here
yy.cpp:28:42: error: no match for call to ‘(Func) (const int&)’
using return_type = decltype(View{}(*It{}));
~~~~~~^~~~~~~
yy.cpp:22:10: note: candidate: int& Func::operator()(int&) const <near match>
int& operator()(int& p) const {return p;}
^~~~~~~~
yy.cpp:22:10: note: conversion of argument 1 would be ill-formed:
yy.cpp:28:42: error: binding reference of type ‘int&’ to ‘const int’ discards qualifiers
using return_type = decltype(View{}(*It{}));
~~~~~~^~~~~~~
I've made a more minimal test case here: https://godbolt.org/z/QQonMG .
The relevant details are:
A using type alias does not instantiate a template. So for example:
template<bool b>
struct fail_if_true {
static_assert(!b, "template parameter must be false");
};
using fail_if_used = fail_if_true<true>;
will not cause a compile time error (if fail_if_used isn't used)
ADL also inspects template parameter classes. In this case, std::vector<int>::iterator is __gnu_cxx::__normal_iterator<const int*, std::vector<int> >, Func>, which has a std::vector<int> in it's template. So, operator== will check in the global namespace (always), alp (As alp::Iterator is in alp), __gnu_cxx and std.
Your View_iterator_base::const_iterator is invalid. View_iterator_base::const_interator::result_type is defined as decltype(Func{}(*std::vector<int>::const_iterator{})). std::vector<int>::const_iterator{} will be a vectors const iterator, so *std::vector<int>::const_iterator{} is a const int&. Func::operator() takes an int&, so this means that the expression is invalid. But it won't cause a compile time error if not used, for the reasons stated above. This means that your conversion operator is to an invalid type.
Since you don't define it as explicit, the conversion operator (To an invalid type) will be used to try and match it to the function parameters if they don't already match. Obviously this will finally instantiate the invalid type, so it will throw a compile time error.
My guess is that iostream includes string, which defines std::operator== for strings.
Here's an example without the std namespace: https://godbolt.org/z/-wlAmv
// Avoid including headers for testing without std::
template<class T> struct is_const { static constexpr const bool value = false; } template<class T> struct is_const<const T> { static constexpr const bool value = true; }
namespace with_another_equals {
struct T {};
bool operator==(const T&, const T&) {
return true;
}
}
namespace ns {
template<class T>
struct wrapper {
using invalid_wrapper = wrapper<typename T::invalid>;
operator invalid_wrapper() {}
};
template<class T>
bool operator==(const wrapper<T>&, const wrapper<T>&) {
return true;
}
}
template<class T>
struct with_invalid {
static_assert(!is_const<T>::value, "Invalid if const");
using invalid = with_invalid<const T>;
};
template<class T>
void test() {
using wrapped = ns::wrapper<with_invalid<T>>;
wrapped a;
wrapped b;
bool x = operator==(a, b);
bool y = ns::operator==(a, b);
}
template void test<int*>();
// Will compile if this line is commented out
template void test<with_another_equals::T>();
Note that just declaring operator const_iterator() should instantiate the type. But it doesn't because it is within templates. My guess is that it is optimised out (where it does compile because it's unused) before it can be checked to show that it can't compile (It doesn't even warn with -Wall -pedantic that it doesn't have a return statement in my example).
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.
The following code is a minimum working (or perhaps non-working) example.
What it does is basically encapsulates a bunch of std::map structures as private members in a base class. To avoid writing a lot of setters and getters, they are implemented as template functions.
// test.cpp
#include <map>
#include <iostream>
enum class E0
{
F0, F1, F2,
};
The declaration of the base class.
using std::map;
class P_base
{
private:
map<E0, int> m_imap;
// ...
// ... Other std::map members with different key types and value types.
public:
map<E0, int> & imap;
// ...
// ... Other std::map references.
P_base() : imap(m_imap) {}
template<typename map_type, typename key_type, typename val_type>
void set(map_type & m, const key_type & k, const val_type & v)
{
m[k] = v;
}
template<typename map_type, typename key_type>
auto access_to_map(const map_type & m, const key_type & k) -> decltype(m.at(k))
{
return m.at(k);
}
};
class P : private P_base
{
public:
decltype(P_base::imap) & imap;
P() : P_base(), imap(P_base::imap) {}
template<typename map_type, typename key_type, typename val_type>
void set(map_type & m, const key_type & k, const val_type & v)
{
P_base::set(m, k, v);
}
template<typename map_type, typename key_type>
auto access_to_map(const map_type & m, const key_type & k) -> decltype(P_base::access_to_map(m, k))
{
return P_base::access_to_map(m, k);
}
};
main
int main(int argc, const char * argv[])
{
using std::cout;
using std::endl;
P op;
op.set(op.imap, E0::F0, 100);
op.set(op.imap, E0::F1, 101);
op.set(op.imap, E0::F2, 102);
cout << op.access_to_map(op.imap, E0::F1) << endl;
}
$ clang++ -std=c++11 test.cpp && ./a.out
101
But if I compile it with intel compiler (icpc version 15.0.3 (gcc version 5.1.0 compatibility)), the compiler gives me this error message (which I don't undertand at all, especially when clang will compile the code):
$ icpc -std=c++ test.cpp && ./a.out
test.cpp(67): error: no instance of function template "P::access_to_map" matches the argument list
argument types are: (std::__1::map<E0, int, std::__1::less<E0>, std::__1::allocator<std::__1::pair<const E0, int>>>, E0)
object type is: P
cout << op.access_to_map(op.imap, E0::F1) << endl;
And it also confuses me by not complaining about the set function.
Does anyone have any idea what is going on here?
Note: My answer applies to g++ - hopefully it's the same as icc.
Here is a smaller test case:
struct Base
{
int func(int t) { return t; }
};
struct Der : Base
{
template<typename T>
auto f(T t) -> decltype(Base::func(t))
{
return t;
}
};
int main(){ Der d; d.f(5); }
The error is:
mcv.cc: In function 'int main()':
mcv.cc:16:25: error: no matching function for call to 'Der::f(int)'
int main(){ Der d; d.f(5); }
^
mcv.cc:16:25: note: candidate is:
mcv.cc:9:7: note: template<class T> decltype (t->Base::func()) Der::f(T)
auto f(T t) -> decltype(Base::func(t))
^
mcv.cc:9:7: note: template argument deduction/substitution failed:
mcv.cc: In substitution of 'template<class T> decltype (t->Base::func()) Der::f(T) [with T = int]':
mcv.cc:16:25: required from here
mcv.cc:9:38: error: cannot call member function 'int Base::func(int)' without object
auto f(T t) -> decltype(Base::func(t))
This can be fixed by changing decltype(Base::func(t)) to decltype(this->Base::func(t)). A corresponding fix fixes your code sample, for me.
Apparently, the compiler doesn't consider that Base::func(t) should be called with *this as hidden argument. I don't know if this is a g++ bug, or if clang is going beyond the call of duty.
Note that in C++14, since the function has a single return statement, the trailing return type can be omitted entirely:
template<typename T>
auto f(T t)
{
return t;
}