my question today is pretty simple: why can't the compiler infer template parameters from class constructors, much as it can do from function parameters? For example, why couldn't the following code be valid:
template <typename obj>
class Variable {
obj data;
public:
Variable(obj d) { data = d; }
};
int main() {
int num = 2;
Variable var(num); // would be equivalent to Variable<int> var(num),
return 0; // but actually a compile error
}
As I say, I understand that this isn't valid, so my question is why isn't it? Would allowing this create any major syntactic holes? Is there an instance where one wouldn't want this functionality (where inferring a type would cause issues)? I'm just trying to understand the logic behind allowing template inference for functions, yet not for suitably-constructed classes.
I think it is not valid because the constructor isn't always the only point of entry of the class (I am talking about copy constructor and operator=). So suppose you are using your class like this :
MyClass m(string s);
MyClass *pm;
*pm = m;
I am not sure if it would be so obvious for the parser to know what template type is the MyClass pm;
Not sure if what I said make sense but feel free to add some comment, that's an interesting question.
C++ 17
It is accepted that C++17 will have type deduction from constructor arguments.
Examples:
std::pair p(2, 4.5);
std::tuple t(4, 3, 2.5);
Accepted paper.
You can't do what you ask for reasons other people have addressed, but you can do this:
template<typename T>
class Variable {
public: Variable(T d) {}
};
template<typename T>
Variable<T> make_variable(T instance) {
return Variable<T>(instance);
}
which for all intent and purposes is the same thing you ask for.
If you love encapsulation you can make make_variable a static member function. That's what people call named constructor. So not only does it do what you want, but it's almost called what you want: the compiler is infering the template parameter from the (named) constructor.
NB: any reasonable compiler will optimize away the temporary object when you write something like
auto v = make_variable(instance);
In the enlightened age of 2016, with two new standards under our belt since this question was asked and a new one just around the corner, the crucial thing to know is that compilers supporting the C++17 standard will compile your code as-is.
Template-argument deduction for class templates in C++17
Here (courtesy of an edit by Olzhas Zhumabek of the accepted answer) is the paper detailing the relevant changes to the standard.
Addressing concerns from other answers
The current top-rated answer
This answer points out that "copy constructor and operator=" wouldn't know the correct template specializations.
This is nonsense, because the standard copy-constructor and operator= only exist for a known template type:
template <typename T>
class MyClass {
MyClass(const MyClass&) =default;
... etc...
};
// usage example modified from the answer
MyClass m(string("blah blah blah"));
MyClass *pm; // WHAT IS THIS?
*pm = m;
Here, as I noted in the comments, there is no reason for MyClass *pm to be a legal declaration with or without the new form of inference: MyClass is not a type (it's a template), so it doesn't make sense to declare a pointer of type MyClass. Here's one possible way to fix the example:
MyClass m(string("blah blah blah"));
decltype(m) *pm; // uses type inference!
*pm = m;
Here, pm is already of the correct type, and so the inference is trivial. Moreover, it's impossible to accidentally mix types when calling the copy-constructor:
MyClass m(string("blah blah blah"));
auto pm = &(MyClass(m));
Here, pm will be a pointer to a copy of m. Here, MyClass is being copy-constructed from m—which is of type MyClass<string> (and not of the nonexistent type MyClass). Thus, at the point where pm's type is inferred, there is sufficient information to know that the template-type of m, and therefore the template-type of pm, is string.
Moreover, the following will always raise a compile error:
MyClass s(string("blah blah blah"));
MyClass i(3);
i = s;
This is because the declaration of the copy constructor is not templated:
MyClass(const MyClass&);
Here, the copy-constructor argument's template-type matches the template-type of the class overall; i.e., when MyClass<string> is instantiated, MyClass<string>::MyClass(const MyClass<string>&); is instantiated with it, and when MyClass<int> is instantiated, MyClass<int>::MyClass(const MyClass<int>&); is instantiated. Unless it is explicitly specified or a templatized constructor is declared, there is no reason for the compiler to instantiate MyClass<int>::MyClass(const MyClass<string>&);, which would obviously be inappropriate.
The answer by Cătălin Pitiș
Pitiș gives an example deducing Variable<int> and Variable<double>, then states:
I have the same type name (Variable) in the code for two different types (Variable and Variable). From my subjective point of view, it affects the readability of the code pretty much.
As noted in the previous example, Variable itself is not a type name, even though the new feature makes it look like one syntactically.
Pitiș then asks what would happen if no constructor is given that would permit the appropriate inference. The answer is that no inference is permitted, because the inference is triggered by the constructor call. Without a constructor-call, there is no inference.
This is similar to asking what version of foo is deduced here:
template <typename T> foo();
foo();
The answer is that this code is illegal, for the reason stated.
MSalter's answer
This is, as far as I can tell, the only answer to bring up a legitimate concern about the proposed feature.
The example is:
Variable var(num); // If equivalent to Variable<int> var(num),
Variable var2(var); // Variable<int> or Variable<Variable<int>> ?
The key question is, does the compiler select the type-inferred constructor here or the copy constructor?
Trying the code out, we can see that the copy constructor is selected. To expand on the example:
Variable var(num); // infering ctor
Variable var2(var); // copy ctor
Variable var3(move(var)); // move ctor
// Variable var4(Variable(num)); // compiler error
I am not sure how the proposal and the new version of the standard specify this; it appears to be determined by "deduction guides," which are a new bit of standardese that I don't yet understand.
I am also not sure why the var4 deduction is illegal; the compiler error from g++ seems to indicate that the statement is being parsed as a function declaration.
Still missing: It makes the following code quite ambiguous:
int main()
{
int num = 2;
Variable var(num); // If equivalent to Variable<int> var(num),
Variable var2(var); //Variable<int> or Variable<Variable<int>> ?
}
Supposing that the compiler supports what you asked. Then this code is valid:
Variable v1( 10); // Variable<int>
// Some code here
Variable v2( 20.4); // Variable<double>
Now, I have the same type name (Variable) in the code for two different types (Variable and Variable). From my subjective point of view, it affects the readability of the code pretty much. Having same type name for two different types in the same namespace looks misleading to me.
Later update:
Another thing to consider: partial (or full) template specialization.
What if I specialize Variable and provide no constructor like you expect?
So I would have:
template<>
class Variable<int>
{
// Provide default constructor only.
};
Then I have the code:
Variable v( 10);
What should the compiler do? Use generic Variable class definition to deduce that it is Variable, then discover that Variable doesn't provide one parameter constructor?
The C++03 and the C++11 standard does not allow for template argument deduction from the parameters passed to the constuructor.
But there is a proposal for "Template parameter deduction for constructors" so you may get what you are asking for soon. Edit: indeed, this feature has been confirmed for C++17.
See: http://www.open-std.org/jtc1/sc22/wg21/docs/papers/2013/n3602.html and http://www.open-std.org/jtc1/sc22/wg21/docs/papers/2015/p0091r0.html
A lot of classes don't depend on constructor parameters. There are only a few classes that have only one constructor, and parameterize based on this constructor's type(s).
If you really need template inference, use a helper function:
template<typename obj>
class Variable
{
obj data;
public:
Variable(obj d)
: data(d)
{ }
};
template<typename obj>
inline Variable<obj> makeVariable(const obj& d)
{
return Variable<obj>(d);
}
Deduction of types is limited to template functions in current C++, but it's long been realised that type deduction in other contexts would be very useful. Hence C++0x's auto.
While exactly what you suggest won't be possible in C++0x, the following shows you can get pretty close:
template <class X>
Variable<typename std::remove_reference<X>::type> MakeVariable(X&& x)
{
// remove reference required for the case that x is an lvalue
return Variable<typename std::remove_reference<X>::type>(std::forward(x));
}
void test()
{
auto v = MakeVariable(2); // v is of type Variable<int>
}
You are right the compiler could easily guess, but it's not in the standard or C++0x as far as I know so you'll have to wait atleast 10 more years (ISO standards fixed turn around rate) before compiller providers add this feature
Let's look at the problem with reference to a class everyone should be familar with - std::vector.
Firstly, a very common use of vector is to use the constructor that takes no parameters:
vector <int> v;
In this case, obviously no inference can be performed.
A second common use is to create a pre-sized vector:
vector <string> v(100);
Here, if inference were used:
vector v(100);
we get a vector of ints, not strings, and presumably it isn't sized!
Lastly, consider constructors that take multiple parameters - with "inference":
vector v( 100, foobar() ); // foobar is some class
Which parameter should be used for inference? We would need some way of telling the compiler that it should be the second one.
With all these problems for a class as simple as vector, it's easy to see why inference is not used.
Making the ctor a template the Variable can have only one form but various ctors:
class Variable {
obj data; // let the compiler guess
public:
template<typename obj>
Variable(obj d)
{
data = d;
}
};
int main()
{
int num = 2;
Variable var(num); // Variable::data int?
float num2 = 2.0f;
Variable var2(num2); // Variable::data float?
return 0;
}
See? We can not have multiple Variable::data members.
See The C++ Template Argument Deduction for more info on this.
Related
my question today is pretty simple: why can't the compiler infer template parameters from class constructors, much as it can do from function parameters? For example, why couldn't the following code be valid:
template <typename obj>
class Variable {
obj data;
public:
Variable(obj d) { data = d; }
};
int main() {
int num = 2;
Variable var(num); // would be equivalent to Variable<int> var(num),
return 0; // but actually a compile error
}
As I say, I understand that this isn't valid, so my question is why isn't it? Would allowing this create any major syntactic holes? Is there an instance where one wouldn't want this functionality (where inferring a type would cause issues)? I'm just trying to understand the logic behind allowing template inference for functions, yet not for suitably-constructed classes.
I think it is not valid because the constructor isn't always the only point of entry of the class (I am talking about copy constructor and operator=). So suppose you are using your class like this :
MyClass m(string s);
MyClass *pm;
*pm = m;
I am not sure if it would be so obvious for the parser to know what template type is the MyClass pm;
Not sure if what I said make sense but feel free to add some comment, that's an interesting question.
C++ 17
It is accepted that C++17 will have type deduction from constructor arguments.
Examples:
std::pair p(2, 4.5);
std::tuple t(4, 3, 2.5);
Accepted paper.
You can't do what you ask for reasons other people have addressed, but you can do this:
template<typename T>
class Variable {
public: Variable(T d) {}
};
template<typename T>
Variable<T> make_variable(T instance) {
return Variable<T>(instance);
}
which for all intent and purposes is the same thing you ask for.
If you love encapsulation you can make make_variable a static member function. That's what people call named constructor. So not only does it do what you want, but it's almost called what you want: the compiler is infering the template parameter from the (named) constructor.
NB: any reasonable compiler will optimize away the temporary object when you write something like
auto v = make_variable(instance);
In the enlightened age of 2016, with two new standards under our belt since this question was asked and a new one just around the corner, the crucial thing to know is that compilers supporting the C++17 standard will compile your code as-is.
Template-argument deduction for class templates in C++17
Here (courtesy of an edit by Olzhas Zhumabek of the accepted answer) is the paper detailing the relevant changes to the standard.
Addressing concerns from other answers
The current top-rated answer
This answer points out that "copy constructor and operator=" wouldn't know the correct template specializations.
This is nonsense, because the standard copy-constructor and operator= only exist for a known template type:
template <typename T>
class MyClass {
MyClass(const MyClass&) =default;
... etc...
};
// usage example modified from the answer
MyClass m(string("blah blah blah"));
MyClass *pm; // WHAT IS THIS?
*pm = m;
Here, as I noted in the comments, there is no reason for MyClass *pm to be a legal declaration with or without the new form of inference: MyClass is not a type (it's a template), so it doesn't make sense to declare a pointer of type MyClass. Here's one possible way to fix the example:
MyClass m(string("blah blah blah"));
decltype(m) *pm; // uses type inference!
*pm = m;
Here, pm is already of the correct type, and so the inference is trivial. Moreover, it's impossible to accidentally mix types when calling the copy-constructor:
MyClass m(string("blah blah blah"));
auto pm = &(MyClass(m));
Here, pm will be a pointer to a copy of m. Here, MyClass is being copy-constructed from m—which is of type MyClass<string> (and not of the nonexistent type MyClass). Thus, at the point where pm's type is inferred, there is sufficient information to know that the template-type of m, and therefore the template-type of pm, is string.
Moreover, the following will always raise a compile error:
MyClass s(string("blah blah blah"));
MyClass i(3);
i = s;
This is because the declaration of the copy constructor is not templated:
MyClass(const MyClass&);
Here, the copy-constructor argument's template-type matches the template-type of the class overall; i.e., when MyClass<string> is instantiated, MyClass<string>::MyClass(const MyClass<string>&); is instantiated with it, and when MyClass<int> is instantiated, MyClass<int>::MyClass(const MyClass<int>&); is instantiated. Unless it is explicitly specified or a templatized constructor is declared, there is no reason for the compiler to instantiate MyClass<int>::MyClass(const MyClass<string>&);, which would obviously be inappropriate.
The answer by Cătălin Pitiș
Pitiș gives an example deducing Variable<int> and Variable<double>, then states:
I have the same type name (Variable) in the code for two different types (Variable and Variable). From my subjective point of view, it affects the readability of the code pretty much.
As noted in the previous example, Variable itself is not a type name, even though the new feature makes it look like one syntactically.
Pitiș then asks what would happen if no constructor is given that would permit the appropriate inference. The answer is that no inference is permitted, because the inference is triggered by the constructor call. Without a constructor-call, there is no inference.
This is similar to asking what version of foo is deduced here:
template <typename T> foo();
foo();
The answer is that this code is illegal, for the reason stated.
MSalter's answer
This is, as far as I can tell, the only answer to bring up a legitimate concern about the proposed feature.
The example is:
Variable var(num); // If equivalent to Variable<int> var(num),
Variable var2(var); // Variable<int> or Variable<Variable<int>> ?
The key question is, does the compiler select the type-inferred constructor here or the copy constructor?
Trying the code out, we can see that the copy constructor is selected. To expand on the example:
Variable var(num); // infering ctor
Variable var2(var); // copy ctor
Variable var3(move(var)); // move ctor
// Variable var4(Variable(num)); // compiler error
I am not sure how the proposal and the new version of the standard specify this; it appears to be determined by "deduction guides," which are a new bit of standardese that I don't yet understand.
I am also not sure why the var4 deduction is illegal; the compiler error from g++ seems to indicate that the statement is being parsed as a function declaration.
Still missing: It makes the following code quite ambiguous:
int main()
{
int num = 2;
Variable var(num); // If equivalent to Variable<int> var(num),
Variable var2(var); //Variable<int> or Variable<Variable<int>> ?
}
Supposing that the compiler supports what you asked. Then this code is valid:
Variable v1( 10); // Variable<int>
// Some code here
Variable v2( 20.4); // Variable<double>
Now, I have the same type name (Variable) in the code for two different types (Variable and Variable). From my subjective point of view, it affects the readability of the code pretty much. Having same type name for two different types in the same namespace looks misleading to me.
Later update:
Another thing to consider: partial (or full) template specialization.
What if I specialize Variable and provide no constructor like you expect?
So I would have:
template<>
class Variable<int>
{
// Provide default constructor only.
};
Then I have the code:
Variable v( 10);
What should the compiler do? Use generic Variable class definition to deduce that it is Variable, then discover that Variable doesn't provide one parameter constructor?
The C++03 and the C++11 standard does not allow for template argument deduction from the parameters passed to the constuructor.
But there is a proposal for "Template parameter deduction for constructors" so you may get what you are asking for soon. Edit: indeed, this feature has been confirmed for C++17.
See: http://www.open-std.org/jtc1/sc22/wg21/docs/papers/2013/n3602.html and http://www.open-std.org/jtc1/sc22/wg21/docs/papers/2015/p0091r0.html
A lot of classes don't depend on constructor parameters. There are only a few classes that have only one constructor, and parameterize based on this constructor's type(s).
If you really need template inference, use a helper function:
template<typename obj>
class Variable
{
obj data;
public:
Variable(obj d)
: data(d)
{ }
};
template<typename obj>
inline Variable<obj> makeVariable(const obj& d)
{
return Variable<obj>(d);
}
Deduction of types is limited to template functions in current C++, but it's long been realised that type deduction in other contexts would be very useful. Hence C++0x's auto.
While exactly what you suggest won't be possible in C++0x, the following shows you can get pretty close:
template <class X>
Variable<typename std::remove_reference<X>::type> MakeVariable(X&& x)
{
// remove reference required for the case that x is an lvalue
return Variable<typename std::remove_reference<X>::type>(std::forward(x));
}
void test()
{
auto v = MakeVariable(2); // v is of type Variable<int>
}
You are right the compiler could easily guess, but it's not in the standard or C++0x as far as I know so you'll have to wait atleast 10 more years (ISO standards fixed turn around rate) before compiller providers add this feature
Let's look at the problem with reference to a class everyone should be familar with - std::vector.
Firstly, a very common use of vector is to use the constructor that takes no parameters:
vector <int> v;
In this case, obviously no inference can be performed.
A second common use is to create a pre-sized vector:
vector <string> v(100);
Here, if inference were used:
vector v(100);
we get a vector of ints, not strings, and presumably it isn't sized!
Lastly, consider constructors that take multiple parameters - with "inference":
vector v( 100, foobar() ); // foobar is some class
Which parameter should be used for inference? We would need some way of telling the compiler that it should be the second one.
With all these problems for a class as simple as vector, it's easy to see why inference is not used.
Making the ctor a template the Variable can have only one form but various ctors:
class Variable {
obj data; // let the compiler guess
public:
template<typename obj>
Variable(obj d)
{
data = d;
}
};
int main()
{
int num = 2;
Variable var(num); // Variable::data int?
float num2 = 2.0f;
Variable var2(num2); // Variable::data float?
return 0;
}
See? We can not have multiple Variable::data members.
See The C++ Template Argument Deduction for more info on this.
Why is that automatic type deduction is possible only for functions and not for Classes?
In specific cases you could always do like std::make_pair:
template<class T>
make_foo(T val) {
return foo<T>(val);
}
EDIT: I just found the following in "The C++ Programming Language, Third Edition", page 335. Bjarne says:
Note that class template arguments are
never deduced. The reason is that the
flexibility provided by several
constructors for a class would make
such deduction impossible in many
cases and obscure in many more.
This is of course very subjective. There's been some discussion about this in comp.std.c++ and the consensus seems to be that there's no reason why it couldn't be supported. Whether it would be a good idea or not is another question...
At Kona meeting Template parameter deduction for constructors (P0091R0) has been approved, which means that in C++17 we'll be able to we can write:
pair p1{"foo"s, 12};
auto p2 = pair{"foo"s, 12};
f(pair{"foo"s, 12});
In case of a function call, the compiler deduces the template type from the argument type. For example the std::max-function. The compiler uses the type of the arguments to deduce the template parameter. This does not allways work, as not all calls are unambigous.
int a = 5;
float b = 10;
double result1 = std::min( a, b ); // error: template parameter ambigous
double result2 = std::min< double >( a, b ); // explicit parameter enforces use of conversion
In case of a template class, that may not allways be possible. Take for instance this class:
template< class T>
class Foo {
public:
Foo();
void Bar( int a );
private:
T m_Member;
};
The type T never appears in any function call, so the compiler has no hint at all, what type should be used.
I think implicit type conversion is only applicable to function arguments only so compiler can deduce it to make the function call success.
But how can it deduce what type you wanna have it class.
We have to wait 4 such a day when we have AI based compiler to do read our minds.
This question already has answers here:
Why not infer template parameter from constructor?
(12 answers)
Closed 7 years ago.
template< class T >
class Foo {
public:
Foo( T t ) { }
};
int main () {
int i = 0;
Foo f( i );
}
In the above code, the compiler complains that template arguments are missing before 'f'. I understand that deducing template arguments for a class from the arguments to the constructor is not part of the standard, but my question is why? Doesn't the compiler have all the information it needs to implicitly instantiate Foo<int> and call its constructor?
Edited to make it clear that I'm calling the constructor with an int (as opposed to a short, long, void*, etc.)
Because nobody has specified how exactly that works. There is a current proposal to the standard committee to make it work. It also lists some of the difficulties:
http://open-std.org/JTC1/SC22/WG21/docs/papers/2015/n4471.html
Update: Here's the newest version of the proposal:
http://open-std.org/JTC1/SC22/WG21/docs/papers/2015/p0091r0.html
TL;DR: Template specialization
They can, but only template arguments on the function itself, not on the type.
The declaration Foo f(0); is illegal, because there is no type named Foo. Perhaps you were thinking of
auto f = Foo(0);
but that is not allowed either, because the compiler doesn't know what scope to search in (there are infinite potential types with a constructor named Foo and, with specialization, possibly more than one with a constructor Foo(int))
The usual method to do this is with a factory helper function:
auto f = make_foo(0);
where the factory function's return type depends on type deduction of its parameters.
You can imagine that the factory functions could be automatically generated in namespace scope and then the usual function overloading rules applied, but this runs into significant difficulty because there can be template arguments on both the type and the constructor itself. These could simply be concatenated, with the limitation that this would exclude class templates with variadic argument lists, because there would be no way to distinguish where the type parameters end and the function parameters begin.
Foo is a class template, not a class. Its type always needs to be supplied in some way or another for a class to be generated with the correct types. You can't do Foo because Foo isn't a type, but Foo<int> is. It creates a class like this:
class Foo {
public:
Foo( int t ) { }
};
If you only supplied Foo, the compiler wouldn't know how to generate the class. Foo<int> f(0) works because Foo<int> generates the class, substituting T with int. By the type you call the constructor, the compiler already knows that the constructor is accepting an int.
The name of the class you want to instantiate is
Foo<int> (as you indicate).
It is also possible to write Foo<short> f(0), or Foo<unsigned long> f(0),
or Foo<Foo<double>*> f(0). In those cases, however, you don't expect the compiler to be able to guess the type if you write only Foo f(0).
One could imagine that a C++ could have been specified with rules to make
some such guesses in certain ways (such as, the literal 0 implies type parameter int and no other), but then the language would be even more
complicated than it is now and there would be additional ways for
people to make programming errors.
Actually writing what you mean in a declaration like this
seems like not too much to ask.
Edit: After posting this, I noticed in another answer that there is
a proposal to make such a feature of C++, indeed as one could imagine.
I've been playing around with C++0x's auto keyword and tried the following.
std::unique_ptr<auto> ptr(new int(0));
I tried compiling it with g++ 4.4.5 and got
error: invalid use of auto
Judging by eye, auto can easily be inferred to int.
My guess is the type inference and the template engine don't talk to each other. Otherwise, the template engine would know to instantiate the template class with int as the type parameter.
Another guess is from the standard, I see this.
A member shall not be declared with auto, extern or register storage class.
But I thought that was the auto as in local variables, not as in auto used to deduce types.
And my last guess is that the compiler thinks this is an auto storage class, not auto for type deduction.
Is there a reason behind this stated in the standard?
That's because it has to determine the class on which to call a constructor before determining what to do with its arguments. If you make the constructor a template, it'll just work like any other template function - auto-deducing arguments.
#dascandy has correctly identified what's wrong with your code. I'll try to provide some rationale:
You're expecting the compiler to infer unique_ptr<int> because the argument is an int*, and unique_ptr<int> has a constructor which accepts int*. For a moment let's ignore the fact that we're using std::unique_ptr, and just talk about a template class we wrote (and can specialize).
Why should the compiler infer unique_ptr<int>? The argument isn't int, it's int*. Why shouldn't it guess unique_ptr<int*>? Of course that would result in a compiler error, since unique_ptr<int*>'s constructor won't accept an int*. Unless I add a specialization:
template<>
class unique_ptr<int*>
{
public:
unique_ptr(int*) {}
};
Now unique_ptr<int*> would compile. How should the compiler know which to choose, unique_ptr<int> or unique_ptr<int*>? What if I add another specialization?
template<>
class unique_ptr<double>
{
public:
unique_ptr(int*) {}
};
The compiler now has three options to choose from, and it has to instantiate the template with every possible argument in order to find them. Clearly this is not feasible, especially with multiple template arguments and template recursion.
What you can do, is make a factory function which connects the inferred type to exactly one template instance:
template<typename T>
std::unique_ptr<T> make_unique(T* arg) { return arg; }
(of course, this won't work because unique_ptr cannot be copied. But the idea is valid, and used in e.g.make_shared and make_pair.)
Some examples of extreme ugliness:
One could argue that unique_ptr<shared_ptr<int>> is a valid match for this code.
Or how about:
template<typename T>
class unique_ptr
{
public:
explicit unique_ptr(T* arg);
unique_ptr(int*, enable_if<(sizeof(T) > 16)>::type* = 0);
};
Just want to add that a solution already exists for most cases:
template <typename T>
std::unique_ptr<T> unique_ptr_auto(T* ptr)
{
// fails to handle std::unique_ptr<T[]>, not deducible from pointer
return std::unique_ptr<T>(ptr);
}
auto ptr = unique_ptr_auto(new int(0));
A bit more verbose, obviously, but you get the idea. These "generator functions" are quite common.
This (or similar) was proposed for the Standard. The proposed functionality looked something like:
std::vector<int> GetMahVector();
std::vector<auto> var = GetMahVector();
However, it was rejected. Why it was rejected, well, you'd have to dig up the relevant Standard process documents, if possible.
my question today is pretty simple: why can't the compiler infer template parameters from class constructors, much as it can do from function parameters? For example, why couldn't the following code be valid:
template <typename obj>
class Variable {
obj data;
public:
Variable(obj d) { data = d; }
};
int main() {
int num = 2;
Variable var(num); // would be equivalent to Variable<int> var(num),
return 0; // but actually a compile error
}
As I say, I understand that this isn't valid, so my question is why isn't it? Would allowing this create any major syntactic holes? Is there an instance where one wouldn't want this functionality (where inferring a type would cause issues)? I'm just trying to understand the logic behind allowing template inference for functions, yet not for suitably-constructed classes.
I think it is not valid because the constructor isn't always the only point of entry of the class (I am talking about copy constructor and operator=). So suppose you are using your class like this :
MyClass m(string s);
MyClass *pm;
*pm = m;
I am not sure if it would be so obvious for the parser to know what template type is the MyClass pm;
Not sure if what I said make sense but feel free to add some comment, that's an interesting question.
C++ 17
It is accepted that C++17 will have type deduction from constructor arguments.
Examples:
std::pair p(2, 4.5);
std::tuple t(4, 3, 2.5);
Accepted paper.
You can't do what you ask for reasons other people have addressed, but you can do this:
template<typename T>
class Variable {
public: Variable(T d) {}
};
template<typename T>
Variable<T> make_variable(T instance) {
return Variable<T>(instance);
}
which for all intent and purposes is the same thing you ask for.
If you love encapsulation you can make make_variable a static member function. That's what people call named constructor. So not only does it do what you want, but it's almost called what you want: the compiler is infering the template parameter from the (named) constructor.
NB: any reasonable compiler will optimize away the temporary object when you write something like
auto v = make_variable(instance);
In the enlightened age of 2016, with two new standards under our belt since this question was asked and a new one just around the corner, the crucial thing to know is that compilers supporting the C++17 standard will compile your code as-is.
Template-argument deduction for class templates in C++17
Here (courtesy of an edit by Olzhas Zhumabek of the accepted answer) is the paper detailing the relevant changes to the standard.
Addressing concerns from other answers
The current top-rated answer
This answer points out that "copy constructor and operator=" wouldn't know the correct template specializations.
This is nonsense, because the standard copy-constructor and operator= only exist for a known template type:
template <typename T>
class MyClass {
MyClass(const MyClass&) =default;
... etc...
};
// usage example modified from the answer
MyClass m(string("blah blah blah"));
MyClass *pm; // WHAT IS THIS?
*pm = m;
Here, as I noted in the comments, there is no reason for MyClass *pm to be a legal declaration with or without the new form of inference: MyClass is not a type (it's a template), so it doesn't make sense to declare a pointer of type MyClass. Here's one possible way to fix the example:
MyClass m(string("blah blah blah"));
decltype(m) *pm; // uses type inference!
*pm = m;
Here, pm is already of the correct type, and so the inference is trivial. Moreover, it's impossible to accidentally mix types when calling the copy-constructor:
MyClass m(string("blah blah blah"));
auto pm = &(MyClass(m));
Here, pm will be a pointer to a copy of m. Here, MyClass is being copy-constructed from m—which is of type MyClass<string> (and not of the nonexistent type MyClass). Thus, at the point where pm's type is inferred, there is sufficient information to know that the template-type of m, and therefore the template-type of pm, is string.
Moreover, the following will always raise a compile error:
MyClass s(string("blah blah blah"));
MyClass i(3);
i = s;
This is because the declaration of the copy constructor is not templated:
MyClass(const MyClass&);
Here, the copy-constructor argument's template-type matches the template-type of the class overall; i.e., when MyClass<string> is instantiated, MyClass<string>::MyClass(const MyClass<string>&); is instantiated with it, and when MyClass<int> is instantiated, MyClass<int>::MyClass(const MyClass<int>&); is instantiated. Unless it is explicitly specified or a templatized constructor is declared, there is no reason for the compiler to instantiate MyClass<int>::MyClass(const MyClass<string>&);, which would obviously be inappropriate.
The answer by Cătălin Pitiș
Pitiș gives an example deducing Variable<int> and Variable<double>, then states:
I have the same type name (Variable) in the code for two different types (Variable and Variable). From my subjective point of view, it affects the readability of the code pretty much.
As noted in the previous example, Variable itself is not a type name, even though the new feature makes it look like one syntactically.
Pitiș then asks what would happen if no constructor is given that would permit the appropriate inference. The answer is that no inference is permitted, because the inference is triggered by the constructor call. Without a constructor-call, there is no inference.
This is similar to asking what version of foo is deduced here:
template <typename T> foo();
foo();
The answer is that this code is illegal, for the reason stated.
MSalter's answer
This is, as far as I can tell, the only answer to bring up a legitimate concern about the proposed feature.
The example is:
Variable var(num); // If equivalent to Variable<int> var(num),
Variable var2(var); // Variable<int> or Variable<Variable<int>> ?
The key question is, does the compiler select the type-inferred constructor here or the copy constructor?
Trying the code out, we can see that the copy constructor is selected. To expand on the example:
Variable var(num); // infering ctor
Variable var2(var); // copy ctor
Variable var3(move(var)); // move ctor
// Variable var4(Variable(num)); // compiler error
I am not sure how the proposal and the new version of the standard specify this; it appears to be determined by "deduction guides," which are a new bit of standardese that I don't yet understand.
I am also not sure why the var4 deduction is illegal; the compiler error from g++ seems to indicate that the statement is being parsed as a function declaration.
Still missing: It makes the following code quite ambiguous:
int main()
{
int num = 2;
Variable var(num); // If equivalent to Variable<int> var(num),
Variable var2(var); //Variable<int> or Variable<Variable<int>> ?
}
Supposing that the compiler supports what you asked. Then this code is valid:
Variable v1( 10); // Variable<int>
// Some code here
Variable v2( 20.4); // Variable<double>
Now, I have the same type name (Variable) in the code for two different types (Variable and Variable). From my subjective point of view, it affects the readability of the code pretty much. Having same type name for two different types in the same namespace looks misleading to me.
Later update:
Another thing to consider: partial (or full) template specialization.
What if I specialize Variable and provide no constructor like you expect?
So I would have:
template<>
class Variable<int>
{
// Provide default constructor only.
};
Then I have the code:
Variable v( 10);
What should the compiler do? Use generic Variable class definition to deduce that it is Variable, then discover that Variable doesn't provide one parameter constructor?
The C++03 and the C++11 standard does not allow for template argument deduction from the parameters passed to the constuructor.
But there is a proposal for "Template parameter deduction for constructors" so you may get what you are asking for soon. Edit: indeed, this feature has been confirmed for C++17.
See: http://www.open-std.org/jtc1/sc22/wg21/docs/papers/2013/n3602.html and http://www.open-std.org/jtc1/sc22/wg21/docs/papers/2015/p0091r0.html
A lot of classes don't depend on constructor parameters. There are only a few classes that have only one constructor, and parameterize based on this constructor's type(s).
If you really need template inference, use a helper function:
template<typename obj>
class Variable
{
obj data;
public:
Variable(obj d)
: data(d)
{ }
};
template<typename obj>
inline Variable<obj> makeVariable(const obj& d)
{
return Variable<obj>(d);
}
Deduction of types is limited to template functions in current C++, but it's long been realised that type deduction in other contexts would be very useful. Hence C++0x's auto.
While exactly what you suggest won't be possible in C++0x, the following shows you can get pretty close:
template <class X>
Variable<typename std::remove_reference<X>::type> MakeVariable(X&& x)
{
// remove reference required for the case that x is an lvalue
return Variable<typename std::remove_reference<X>::type>(std::forward(x));
}
void test()
{
auto v = MakeVariable(2); // v is of type Variable<int>
}
You are right the compiler could easily guess, but it's not in the standard or C++0x as far as I know so you'll have to wait atleast 10 more years (ISO standards fixed turn around rate) before compiller providers add this feature
Let's look at the problem with reference to a class everyone should be familar with - std::vector.
Firstly, a very common use of vector is to use the constructor that takes no parameters:
vector <int> v;
In this case, obviously no inference can be performed.
A second common use is to create a pre-sized vector:
vector <string> v(100);
Here, if inference were used:
vector v(100);
we get a vector of ints, not strings, and presumably it isn't sized!
Lastly, consider constructors that take multiple parameters - with "inference":
vector v( 100, foobar() ); // foobar is some class
Which parameter should be used for inference? We would need some way of telling the compiler that it should be the second one.
With all these problems for a class as simple as vector, it's easy to see why inference is not used.
Making the ctor a template the Variable can have only one form but various ctors:
class Variable {
obj data; // let the compiler guess
public:
template<typename obj>
Variable(obj d)
{
data = d;
}
};
int main()
{
int num = 2;
Variable var(num); // Variable::data int?
float num2 = 2.0f;
Variable var2(num2); // Variable::data float?
return 0;
}
See? We can not have multiple Variable::data members.
See The C++ Template Argument Deduction for more info on this.