Make auto become a reference if reference is returned by function
I need the following code to work:
class Parent{
public:
virtual void fun(){throw 0};
};
class Child:public Parent{
public:
virtual void fun(){/*Success*/};
};
Parent& getChild(){
return *(new Child());
}
void main(){
auto child=getChild();
child.fun();
}
But for some reason auto creates a Parent instead of Parent & which would correctly use the derived method. Is there a way to force auto to create a reference instead of using a simple Instance?
c++ reference language-lawyer type-inference auto
edited Nov 28 '18 at 2:32
curiousguy
4,58623045
asked Nov 24 '18 at 17:09
user2741831user2741831
5881716
|
show 1 more comment
I need the following code to work:
class Parent{
public:
virtual void fun(){throw 0};
};
class Child:public Parent{
public:
virtual void fun(){/*Success*/};
};
Parent& getChild(){
return *(new Child());
}
void main(){
auto child=getChild();
child.fun();
}
But for some reason auto creates a Parent instead of Parent & which would correctly use the derived method. Is there a way to force auto to create a reference instead of using a simple Instance?
c++ reference language-lawyer type-inference auto
edited Nov 28 '18 at 2:32
curiousguy
4,58623045
asked Nov 24 '18 at 17:09
user2741831user2741831
5881716
1
Useautp&maybe? Also note that you return a pointer in that function not a reference.
– πάντα ῥεῖ
Nov 24 '18 at 17:10
2
auto&ordecltype(auto). "But for some reason auto creates a instead of which would correctly use the derived method" becauseautouses template type deduction.
– George
Nov 24 '18 at 17:11
@russian_symbols fixed
– user2741831
Nov 24 '18 at 17:16
@george that helps, thanks
– user2741831
Nov 24 '18 at 17:17
2
getChildis a very poor design as the caller is unaware they need to free it
– M.M
Nov 25 '18 at 20:58
|
show 1 more comment
I need the following code to work:
class Parent{
public:
virtual void fun(){throw 0};
};
class Child:public Parent{
public:
virtual void fun(){/*Success*/};
};
Parent& getChild(){
return *(new Child());
}
void main(){
auto child=getChild();
child.fun();
}
But for some reason auto creates a Parent instead of Parent & which would correctly use the derived method. Is there a way to force auto to create a reference instead of using a simple Instance?
c++ reference language-lawyer type-inference auto
edited Nov 28 '18 at 2:32
curiousguy
4,58623045
asked Nov 24 '18 at 17:09
user2741831user2741831
5881716
I need the following code to work:
class Parent{
public:
virtual void fun(){throw 0};
};
class Child:public Parent{
public:
virtual void fun(){/*Success*/};
};
Parent& getChild(){
return *(new Child());
}
void main(){
auto child=getChild();
child.fun();
}
But for some reason auto creates a Parent instead of Parent & which would correctly use the derived method. Is there a way to force auto to create a reference instead of using a simple Instance?
c++ reference language-lawyer type-inference auto
c++ reference language-lawyer type-inference auto
edited Nov 28 '18 at 2:32
curiousguy
4,58623045
asked Nov 24 '18 at 17:09
user2741831user2741831
5881716
edited Nov 28 '18 at 2:32
curiousguy
4,58623045
asked Nov 24 '18 at 17:09
user2741831user2741831
5881716
edited Nov 28 '18 at 2:32
curiousguy
4,58623045
edited Nov 28 '18 at 2:32
curiousguy
4,58623045
edited Nov 28 '18 at 2:32
curiousguy
4,58623045
4,58623045
asked Nov 24 '18 at 17:09
user2741831user2741831
5881716
asked Nov 24 '18 at 17:09
user2741831user2741831
5881716
asked Nov 24 '18 at 17:09
user2741831user2741831
5881716
5881716
1
Useautp&maybe? Also note that you return a pointer in that function not a reference.
– πάντα ῥεῖ
Nov 24 '18 at 17:10
2
auto&ordecltype(auto). "But for some reason auto creates a instead of which would correctly use the derived method" becauseautouses template type deduction.
– George
Nov 24 '18 at 17:11
@russian_symbols fixed
– user2741831
Nov 24 '18 at 17:16
@george that helps, thanks
– user2741831
Nov 24 '18 at 17:17
2
getChildis a very poor design as the caller is unaware they need to free it
– M.M
Nov 25 '18 at 20:58
|
show 1 more comment
1
Useautp&maybe? Also note that you return a pointer in that function not a reference.
– πάντα ῥεῖ
Nov 24 '18 at 17:10
2
auto&ordecltype(auto). "But for some reason auto creates a instead of which would correctly use the derived method" becauseautouses template type deduction.
– George
Nov 24 '18 at 17:11
@russian_symbols fixed
– user2741831
Nov 24 '18 at 17:16
@george that helps, thanks
– user2741831
Nov 24 '18 at 17:17
2
getChildis a very poor design as the caller is unaware they need to free it
– M.M
Nov 25 '18 at 20:58
1
1
Use
autp& maybe? Also note that you return a pointer in that function not a reference.– πάντα ῥεῖ
Nov 24 '18 at 17:10
Use
autp& maybe? Also note that you return a pointer in that function not a reference.– πάντα ῥεῖ
Nov 24 '18 at 17:10
2
2
auto& or decltype(auto). "But for some reason auto creates a instead of which would correctly use the derived method" because auto uses template type deduction.– George
Nov 24 '18 at 17:11
auto& or decltype(auto). "But for some reason auto creates a instead of which would correctly use the derived method" because auto uses template type deduction.– George
Nov 24 '18 at 17:11
@russian_symbols fixed
– user2741831
Nov 24 '18 at 17:16
@russian_symbols fixed
– user2741831
Nov 24 '18 at 17:16
@george that helps, thanks
– user2741831
Nov 24 '18 at 17:17
@george that helps, thanks
– user2741831
Nov 24 '18 at 17:17
2
2
getChild is a very poor design as the caller is unaware they need to free it– M.M
Nov 25 '18 at 20:58
getChild is a very poor design as the caller is unaware they need to free it– M.M
Nov 25 '18 at 20:58
|
show 1 more comment
1 Answer
1
active
oldest
votes
The C++ standard explains that auto is deducted from the initilizer value:
Declarations
[dcl.spec.auto]/1:Theautoanddecltype(auto)
type-specifiers are used to designate a placeholder type that will be
replaced later by deduction from an initializer.
Initializers
[dcl.init]/1:A declarator can specify an initial
value for the identifier being declared.
So when you write the following declarators:
int t = f(10);
auto x = y;
auto child = getChild();
the right side is evaluated as an expression that yields a value. And this value is used to initialize the variable declared on the left side.
A value is not a reference. It's not like pointers and adresses. In an expression evaluation, whenever a reference is used, it's the referred value that is taken into consideration. You can see this with the following simple snippet:
int x=15;
int&y = x;
cout << x <<", " << y <<", " << &x<<endl;
cout << typeid(x).name() <<", " << typeid(y).name() <<", " << typeid(&x).name() <<endl;
In your case, the expression value is therefore a Parent and not Parent &. This is why your auto will be deduced to have the Parent type.
The C++ standard does however not consider all the values in the same manner. In [basic.lval] it makes an important distinction between different kind of values, which can be lvalues, xvalues and prvalues. In your case getChild() evaluates to an lvalue that is the object to which you referred and not some temporary copy of it.
When you have such an lvalue, you can create a reference to it:
[dcl.ref]/2: A reference type that is declared using&is called an
lvalue reference, (...)
But you need to do this explicitly by saying that your auto should be a reference:
auto &child = getChild();
Edit: More infos on the way auto works: Herb Sutter's GotW blog
answered Nov 24 '18 at 18:28
ChristopheChristophe
40.5k43576
add a comment |
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oldest
votes
The C++ standard explains that auto is deducted from the initilizer value:
Declarations
[dcl.spec.auto]/1:Theautoanddecltype(auto)
type-specifiers are used to designate a placeholder type that will be
replaced later by deduction from an initializer.
Initializers
[dcl.init]/1:A declarator can specify an initial
value for the identifier being declared.
So when you write the following declarators:
int t = f(10);
auto x = y;
auto child = getChild();
the right side is evaluated as an expression that yields a value. And this value is used to initialize the variable declared on the left side.
A value is not a reference. It's not like pointers and adresses. In an expression evaluation, whenever a reference is used, it's the referred value that is taken into consideration. You can see this with the following simple snippet:
int x=15;
int&y = x;
cout << x <<", " << y <<", " << &x<<endl;
cout << typeid(x).name() <<", " << typeid(y).name() <<", " << typeid(&x).name() <<endl;
In your case, the expression value is therefore a Parent and not Parent &. This is why your auto will be deduced to have the Parent type.
The C++ standard does however not consider all the values in the same manner. In [basic.lval] it makes an important distinction between different kind of values, which can be lvalues, xvalues and prvalues. In your case getChild() evaluates to an lvalue that is the object to which you referred and not some temporary copy of it.
When you have such an lvalue, you can create a reference to it:
[dcl.ref]/2: A reference type that is declared using&is called an
lvalue reference, (...)
But you need to do this explicitly by saying that your auto should be a reference:
auto &child = getChild();
Edit: More infos on the way auto works: Herb Sutter's GotW blog
answered Nov 24 '18 at 18:28
ChristopheChristophe
40.5k43576
add a comment |
The C++ standard explains that auto is deducted from the initilizer value:
Declarations
[dcl.spec.auto]/1:Theautoanddecltype(auto)
type-specifiers are used to designate a placeholder type that will be
replaced later by deduction from an initializer.
Initializers
[dcl.init]/1:A declarator can specify an initial
value for the identifier being declared.
So when you write the following declarators:
int t = f(10);
auto x = y;
auto child = getChild();
the right side is evaluated as an expression that yields a value. And this value is used to initialize the variable declared on the left side.
A value is not a reference. It's not like pointers and adresses. In an expression evaluation, whenever a reference is used, it's the referred value that is taken into consideration. You can see this with the following simple snippet:
int x=15;
int&y = x;
cout << x <<", " << y <<", " << &x<<endl;
cout << typeid(x).name() <<", " << typeid(y).name() <<", " << typeid(&x).name() <<endl;
In your case, the expression value is therefore a Parent and not Parent &. This is why your auto will be deduced to have the Parent type.
The C++ standard does however not consider all the values in the same manner. In [basic.lval] it makes an important distinction between different kind of values, which can be lvalues, xvalues and prvalues. In your case getChild() evaluates to an lvalue that is the object to which you referred and not some temporary copy of it.
When you have such an lvalue, you can create a reference to it:
[dcl.ref]/2: A reference type that is declared using&is called an
lvalue reference, (...)
But you need to do this explicitly by saying that your auto should be a reference:
auto &child = getChild();
Edit: More infos on the way auto works: Herb Sutter's GotW blog
answered Nov 24 '18 at 18:28
ChristopheChristophe
40.5k43576
add a comment |
The C++ standard explains that auto is deducted from the initilizer value:
Declarations
[dcl.spec.auto]/1:Theautoanddecltype(auto)
type-specifiers are used to designate a placeholder type that will be
replaced later by deduction from an initializer.
Initializers
[dcl.init]/1:A declarator can specify an initial
value for the identifier being declared.
So when you write the following declarators:
int t = f(10);
auto x = y;
auto child = getChild();
the right side is evaluated as an expression that yields a value. And this value is used to initialize the variable declared on the left side.
A value is not a reference. It's not like pointers and adresses. In an expression evaluation, whenever a reference is used, it's the referred value that is taken into consideration. You can see this with the following simple snippet:
int x=15;
int&y = x;
cout << x <<", " << y <<", " << &x<<endl;
cout << typeid(x).name() <<", " << typeid(y).name() <<", " << typeid(&x).name() <<endl;
In your case, the expression value is therefore a Parent and not Parent &. This is why your auto will be deduced to have the Parent type.
The C++ standard does however not consider all the values in the same manner. In [basic.lval] it makes an important distinction between different kind of values, which can be lvalues, xvalues and prvalues. In your case getChild() evaluates to an lvalue that is the object to which you referred and not some temporary copy of it.
When you have such an lvalue, you can create a reference to it:
[dcl.ref]/2: A reference type that is declared using&is called an
lvalue reference, (...)
But you need to do this explicitly by saying that your auto should be a reference:
auto &child = getChild();
Edit: More infos on the way auto works: Herb Sutter's GotW blog
answered Nov 24 '18 at 18:28
ChristopheChristophe
40.5k43576
The C++ standard explains that auto is deducted from the initilizer value:
Declarations
[dcl.spec.auto]/1:Theautoanddecltype(auto)
type-specifiers are used to designate a placeholder type that will be
replaced later by deduction from an initializer.
Initializers
[dcl.init]/1:A declarator can specify an initial
value for the identifier being declared.
So when you write the following declarators:
int t = f(10);
auto x = y;
auto child = getChild();
the right side is evaluated as an expression that yields a value. And this value is used to initialize the variable declared on the left side.
A value is not a reference. It's not like pointers and adresses. In an expression evaluation, whenever a reference is used, it's the referred value that is taken into consideration. You can see this with the following simple snippet:
int x=15;
int&y = x;
cout << x <<", " << y <<", " << &x<<endl;
cout << typeid(x).name() <<", " << typeid(y).name() <<", " << typeid(&x).name() <<endl;
In your case, the expression value is therefore a Parent and not Parent &. This is why your auto will be deduced to have the Parent type.
The C++ standard does however not consider all the values in the same manner. In [basic.lval] it makes an important distinction between different kind of values, which can be lvalues, xvalues and prvalues. In your case getChild() evaluates to an lvalue that is the object to which you referred and not some temporary copy of it.
When you have such an lvalue, you can create a reference to it:
[dcl.ref]/2: A reference type that is declared using&is called an
lvalue reference, (...)
But you need to do this explicitly by saying that your auto should be a reference:
auto &child = getChild();
Edit: More infos on the way auto works: Herb Sutter's GotW blog
answered Nov 24 '18 at 18:28
ChristopheChristophe
40.5k43576
edited Nov 25 '18 at 19:58
answered Nov 24 '18 at 18:28
ChristopheChristophe
40.5k43576
answered Nov 24 '18 at 18:28
ChristopheChristophe
40.5k43576
answered Nov 24 '18 at 18:28
ChristopheChristophe
40.5k43576
40.5k43576
add a comment |
add a comment |
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1
Use
autp&maybe? Also note that you return a pointer in that function not a reference.– πάντα ῥεῖ
Nov 24 '18 at 17:10
2
auto&ordecltype(auto). "But for some reason auto creates a instead of which would correctly use the derived method" becauseautouses template type deduction.– George
Nov 24 '18 at 17:11
@russian_symbols fixed
– user2741831
Nov 24 '18 at 17:16
@george that helps, thanks
– user2741831
Nov 24 '18 at 17:17
2
getChildis a very poor design as the caller is unaware they need to free it– M.M
Nov 25 '18 at 20:58