When is a reference to function useful? - c++

A reference to array parameter is useful because it prevents decay, allowing you to deduce the size of an array.
template <typename T, std::size_t N>
constexpr std::size_t array_size(T (&) [N]) noexcept
{
return N;
}
But when is this useful?
template<typename T>
void foo(T& t)
{
// we now have a reference to bar
}
void bar() { }
int main()
{
foo(bar);
}
When do we care about preventing function to pointer decay? I'm asking about when this is useful, not why it isn't disallowed.

Like with objects, we us a pointer if “no object” (“no function”) is a sensible value and a reference if we wish to make sure that (unless somebody insists to shoot himself in the foot) there is always a valid object (function) referred to.
Consider the following poor man's function wrapper (wealthy man goes over there).
template<typename>
class Function; // never defined
template<typename ReturnT, typename... ArgT>
class Function<ReturnT(ArgT...)>
{
private:
ReturnT (*func_)(ArgT...);
public:
// Will also accept a 'nullptr'
Function(ReturnT (*func)(ArgT...)) noexcept : func_ {func}
{
}
ReturnT
operator()(ArgT... args)
{
return this->func_(args...);
}
};
Now we can write the following program, which works just fine.
#include <iostream>
int
add(int a, int b)
{
return a + b;
}
int
main()
{
Function<int(int, int)> f {add}; // ok
std::cout << f(5, 7) << std::endl; // ok, prints 12
}
However, we can also write the following program, which doesn't work so nice.
int
main()
{
Function<int(int, int)> f {nullptr}; // compiles fine
std::cout << f(5, 7) << std::endl; // compiles fine, crashes at run-time
}
Instead, if we had replaced the (*func) with (&func) in the template's definition,
// Won't accept a 'nullptr'
Function(ReturnT (&func)(ArgT...)) noexcept : func_ {func}
{
}
the line
Function<int(int, int)> f {nullptr}; // compile-time error
would have triggered a compile-time error.

To me the difference is like between normal variable reference and a pointer. References are nicer to deal with than pointers.
I had a use-case when I to customize a parts of an algorithm. It had some phases and I could switch between several options for each phase. This is of course achievable with pointers, but hey, you can use pointers instead of references pretty much everywhere too.
It looked similar to this:
#include <iostream>
using namespace std;
class Foo
{
public:
Foo(void(&t)()) : barish(t) { };
void fun() {barish();};
private:
void(&barish)();
};
void bar() { cout << "meow\n";}
void bark() { cout << "woof\n";}
int main()
{
Foo foo1(bar);
Foo foo2(bark);
foo1.fun();
foo2.fun();
}
You can customize fun and don't have to deal with dereferencing and asserts that it's not null.

Related

C++20 function overloading with references [duplicate]

If we have this example functions code in C++
void foo(int x) { std::cout << "foo(int)" << std::endl; }
void foo(int& x) { std::cout << "foo(int &)" << std::endl; }
Is it possible to difference what function to call doing any modification in the calling arguments?
If the function foo is called in some of these ways:
foo( 10);
i = 10;
foo( static_cast<const int>(i));
foo( static_cast<const int&>(i));
it's called the first foo overloaded function, because it can't pass by reference a const argument to a non-const parameter.
But, how would you do to call the second foo overload function?
If I call the next way:
int i = 10;
foo( i);
It happens an ambiguous error because both functions are valid for this argument.
In this link https://stackoverflow.com/a/5465379/6717386 it's explained one way to resolve it: using objects instead of built-in types and doing private the copy constructor, so it can't do a copy of object value and it has to be called the second foo overload function and passing the object by reference. But, is there any way with the built-in types? I have to change the name of function to avoid the overloading?
You may do a cast (of the function) to select the overload function:
static_cast<void (&)(int&)>(foo)(i);
Demo
In most instance, function overloading involves distinct parameter types and different input parameter lengths.
Your attempt is generally a bad practice and the resulting compiled code is compiler dependent and code optimization may even worsen things even more.
You may consider simply adding a second parameter to the second method, something like this:
void foo(int x) { std::cout << "foo(int)" << std::endl; }
void foo(int& x, ...) { std::cout << "foo(int &, ...)" << std::endl; }
where ... could be a boolean type, say: bool anotherFunction
So calling foo(param1, param2) would simply call the second code and everybody is fine.
Very strange design, but if you want... I'll offer a solution as strange as your design Use Xreference in function signature. Then in the function you can check what you need to do using std::is_lvalue_reference, std::is_rvalue_reference.
Something like this
template<class T>
void foo(T&& x)
{
static_assert(std::is_same<std::decay_t<T>, int>::value, "!");
if (std::is_rvalue_reference<T&&>::value)
std::cout << "do here what you want in foo(int x)";
else
std::cout << "do here what you want in foo(int & x)";
}
int main()
{
int x = 5;
foo(x); //"do here what you want in foo(int x)" - will be printed
foo(std::move(x)); //"do here what you want in foo(int & x)" - will be printed
}
Despite the good answer of #Jarod42, as an alternative solution you can rely on a templated entry point and the overloading of an internal function (if you don't want to deal with explicit casts, of course).
It follows a minimal, working example:
#include<type_traits>
#include<iostream>
#include<utility>
void foo_i(char, int x) { std::cout << "foo(int)" << std::endl; }
void foo_i(int, int &x) { std::cout << "foo(int &)" << std::endl; }
template<typename T>
void foo(T &&t) {
static_assert(std::is_same<std::decay_t<T>, int>::value, "!");
foo_i(0, std::forward<T>(t));
}
int main() {
foo( 10);
int i = 10;
foo( static_cast<const int>(i));
foo( static_cast<const int &>(i));
foo(i);
}
The static_assert serves the purpose of checking the parameter to be something that involves int (that is int, int &, const int &, int &&`, and so on).
As you can see from the code above, foo(i) will print:
foo(int &)
As expected.
Another one:
#include <iostream>
#include <functional>
void foo(int x)
{
std::cout << "foo(int)\n";
}
template<typename T>
void foo(T&& x)
{
std::cout << "foo(int&)\n";
}
int main()
{
int i = 10;
foo(i); // foo(int)
foo(std::ref(i)); // foo(int&)
}
I just happened to have stumbled upon this post and was surprised not to find the typical SFINAE solution. So, there you go:
#include <iostream>
#include <type_traits>
template<typename T,
typename std::enable_if<!std::is_lvalue_reference<T>{}, int>::type = 0>
void foo(T)
{ std::cout << "foo(int)" << std::endl; }
template<typename T,
typename std::enable_if<std::is_lvalue_reference<T>{}, int>::type = 0>
void foo(T&)
{ std::cout << "foo(int &)" << std::endl; }
int main() {
int i = 42;
int& r = i;
foo<decltype(i)>(i);
foo<decltype(r)>(r);
}
Live example

c++ syntax: error conversion from some_type to non-scalar_type requested

Line 1 causes error: conversion from ‘C<void()>’ to non-scalar type ‘C<void (*)()>’ requested. I know I can write it as line 2, but how can I use the make_class() and assign it to a variable?
#include <iostream>
using namespace std;
template<class T> class C {
T f;
public:
C(T ff) : f(ff) {}
};
template<class Ft> C<Ft> make_class(const Ft& f)
{
return C<Ft>(f);
}
void f()
{
cout << "f()" << endl;
}
int main()
{
// C<void(*)()> v = make_class(f); // line 1
C<void(*)()> v(f); // line 2
return 0;
}
Another question is from this link. The code is shown as follows. How to understand line 3?
template <typename F>
struct foo {
F f;
void call() {
f();
}
};
void function() {
std::cout << "function called" << std::endl;
}
int main() {
foo<void(*)()> a = { function }; // line 3: { } is an array?
a.call();
}
Thanks.
A function type and a pointer to function types are different types in the language. While in most contexts the former will decay to the latter, when used as template arguments they generate two unrelated types (different instantiations of a template yield unrelated types).
The deduced type is const reference to function, not pointer to function. A simple workaround is dropping the const & from the function signature which will force the decay to a pointer to function (you cannot pass functions by value).
Regarding the second question, that is called aggregate-initialization, which is in fact the same initialization performed for arrays (arrays being a subset of aggregates).

C++11 Dynamic Cast If Else Chain -> Switch

Consider the following:
struct B { };
template<typename T>
struct D : B
{
T t;
}
void g(int i) { ... }
void g(string s) { ... }
void g(char c) { ... }
void f(B* b)
{
if (dynamic_cast<D<int>*>(b))
{
g(dynamic_cast<D<int>*>(b)->t);
}
else if (dynamic_cast<D<string>*>(b))
{
g(dynamic_cast<D<string>*>(b)->t);
}
else if (dynamic_cast<D<char>*>(b))
{
g(dynamic_cast<D<char>*>(c)->t)
}
else
throw error;
};
Here there are only three possible types of T - int, string, char - but if the list of possible types were longer, say n, the if else chain would take O(n) to execute.
One way to deal with this would be to store an extra type code in D somehow and then switch on the type code.
The RTTI system must already have such a code. Is there someway to get access to it and switch on it?
Or is there a better way to do what I'm trying to do?
C++11 is almost there.
In C++03 it was impossible because the only way to get a compile-time constant (which case requires) was through the type system. Since typeid always returns the same type, it couldn't produce different alternatives for switch.
C++11 adds constexpr and type_info::hash_code as a unique identifier of types, but doesn't combine them. You can use typeid in a constant expression on a of type name or statically-typed expressions, but because hash_code is a non- constexpr function you cannot call it.
Of course there are various workarounds, one of which you describe, and the most general of which apply a visitor over a type vector using template metaprogramming.
Since only a few types are valid, you could solve this with virtual functions and template specialization instead:
struct B
{
virtual void g() = 0;
}
template<typename T>
struct D : public B
{
T t;
};
template<>
struct D<int> : public B
{
int t;
void g() { /* do something here */ }
};
template<>
struct D<std::string> : public B
{
std::string t;
void g() { /* do something here */ }
};
template<>
struct D<char> : public B
{
char t;
void g() { /* do something here */ }
};
void f(B* b)
{
b->g();
}
This will fail at compile-time if you provide the wrong types, instead or requiring runtime checks (which C++ is quite bad at).
The primary choice for run time switching on type in C++, is a virtual function.
It is dead simple:
#include <string>
#include <iostream>
using namespace std;
struct Base
{
virtual void g() const = 0;
};
template< class Type > void g( Type const& );
template<> void g( int const& ) { cout << "int" << endl; }
template<> void g( string const& ) { cout << "string" << endl; }
template<> void g( char const& ) { cout << "char" << endl; }
template< class Type >
struct Derived: Base
{
Type t;
virtual void g() const override { ::g<Type>( t ); }
};
void f( Base& b ) { b.g(); }
int main()
{
Derived<int>().g();
}
As you can it is also efficient, O(1) instead of the silly O(n). Plus, with static (compile time) type checking instead of dynamic (run time) type checking, saving a pretty annoying amount of testing. What more can I say? Really, forget about type code and enums and such. Remember that Bertrand Meyer chose to not support enums in Eiffel, for just this reason, that people tend to abuse them for type codes. Do use virtual functions.
Hey, virtual functions!
They're really useful when otherwise you'd want dynamic dispatch on type.
So, I recommend using virtual functions for that. :)
EDIT: templatized ::g in order to avoid possible ambiguities in the real code.

Static class member

What's wrong with the code below? Latest version of g++ and clang both give error. I am sure I am missing something basic here.
#include <iostream>
struct Z
{
static const int mysize = 10;
};
Z f2();
int main()
{
std::cout << f2()::mysize << std::endl;
}
The motivation here is to be able to find out the size of an array using templates using code such as below. I know there are many ways, but just stumbled upon this idea.
template<int N> struct S
{
enum { mysize = N };
};
template<class T, int N> S<N> f(T (&)[N]);
int main()
{
char buf[10];
std::cout << f(buf)::mysize << std::endl;
}
f2() returns a value, not a type. You'd need to use the . operator on the return value instead of ::
The :: operator requires a type to be named on the lefthand side, while . allows for a value to be named. Your expression f2() does not name a type so it cannot be used in conjunction with ::.
As a side note, with a little more detail in the question we might be able to solve your real problem.
Your program contains two mistakes:
You are using the :: operator to access the member of an object. Use operator . ("dot") instead;
You declare function f2() and invoke it without defining it (this will give you a linker error).
Also, since static member variables are shared among all instances of a class (Z in this case), you do not need an object to access it;
Here is how you could fix your program:
#include <iostream>
struct Z
{
static const int mysize = 10;
};
Z f2() { return Z(); }
int main()
{
// Don't need an object to access a static variable...
std::cout << Z::mysize << std::endl;
// ...but if you really want to, do it this way...
std::cout << f2().mysize << std::endl;
}
Why don't you use this way to find out the size of array by templates:
#include <iostream>
template<int N> struct S
{
enum { mysize = N };
};
template<class T, int N> int f1(T (&)[N])
{
return N;
}
int main()
{
char buf[10];
std::cout << f1(buf) << std::endl;
}
And this one is closer to your variant:
template<class T, int N> S<N> f(T (&)[N])
{
S<N> foo;
return foo;
}
int main()
{
char buf[10];
std::cout << f(buf).mysize << std::endl;
}
Anyway, you will need to return an object from f and access it's member by ., not by ::.
But it's more probable that second variant will be slower, because first variant is fully compile-time, but in the second variant compiler may miss the optimization and don't optimize out the run-time creation of foo.
I think you need to add const int Z::mysize; after class declaration.

c++ functor and function templates

consider this simple and pointless code.
#include <iostream>
struct A {
template<int N>
void test() {
std::cout << N << std::endl;
}
};
int main() {
A a;
a.test<1>();
}
It is a very simple example of a function template. What if however, I wanted to replace A::test with an overloaded operator() to make it a functor?
#include <iostream>
struct A {
template<int N>
void operator()() {
std::cout << N << std::endl;
}
};
int main() {
A a;
a<1>(); // <-- error, how do I do this?
}
Certainly if the operator() took parameters which were dependent on the template, the compiler could possibly deduce the template. But I just can't figure out the proper syntax to specify template parameters with a parameterless functor.
Is there a proper way to do this?
Obviously, this code would work since it bypasses the functor syntax:
a.operator()<1>();
but that kinda defeats the purpose of it being a functor :-P.
You can only call
a.operator()<1>();
but that would not be using a functor. Functors need a non template operator(), as they must be able to be called as varname() and that won't work with your code.
To make it a real functor change your code a template class (functors are classes):
#include <iostream>
template<int N>
struct A {
void operator()() {
std::cout << N << std::endl;
}
};
int main() {
A<1> a;
a();
}
There's not another "direct" way I know other than the:
a.operator()<1>();
syntax. If you're open to changing the code, moving the template parameter to the class would work, or using a (boost|tr1)::bind to make a (boost|tr1)::function object.
You are trying to pass a template parameter to an instance of an object, which as far as I know is not allowed. You can only pass templates parameters to template functions or template objects.
a.test<1>(); and a.operator()<1>(); work because they are serving as template functions.
Use boost::bind (check out boost libraries) to fix it though.
struct A {
void operator()(int n) {
std::cout << n << std::endl;
}
};
int main(int argc, char* argv[]) {
A a;
boost::function<void()> f = boost::bind<void>(a, 1);
f(); // prints 1
return 0;
}
And you don't even have to mess with templates!
You're stuck. Have you considered something like
struct A {
template<int N>
struct B
{
void operator()()
{ std::cout << N << std::endl; }
};
template<int N>
B<N> functor() {return B<N>();}
};
int main()
{
A a;
a.functor<1>()();
}
Nope, there's no way around it. Like you said, you have to either call the operator explicitly (which defeats the purpose), or the template arguments must be able to be deduced by the compiler.