Are std::tuple and std::tuple considered same type by std::vector?
I have a variable defined like this
auto drum = std::make_tuple
( std::make_tuple
( 0.3f
, ExampleClass
, (ExampleClass& instance) {return instance.eGetter ();}
)
);
I expect drum to be a tuple of tuple. (i.e. ((a, b, c))).
And I have another variable defined like this
auto base = std::make_tuple
( 0.48f
, ExampleClass
, (ExampleClass& instance) {return instance.eGetter ();}
);
which I expect to be just a tuple of three elements (i.e (a, b, c))
I also have a vector defined as follows
std::vector<std::tuple<std::tuple< float
, ExampleClass
, std::function<float (ExampleClass&)>
>>> listOfInstruments;
Now if I add drum to listOfInstruments I expect no errors.
Which was indeed the case with listOfInstruments.push_back(drum);
Where I expected an error was here listOfInstuments.push_back(base); but the code compiles just fine.
Since listOfInstruments has type 'tuple of tuples', should't adding just 'tuple' cause some error? Unless both () and (()) are considered same types by std::vector. Or am I completely wrong and there's something else at work here?
Can't seem to figure it out.
c++ tuples
asked 3 hours ago
atisatis
34619
add a comment |
I have a variable defined like this
auto drum = std::make_tuple
( std::make_tuple
( 0.3f
, ExampleClass
, (ExampleClass& instance) {return instance.eGetter ();}
)
);
I expect drum to be a tuple of tuple. (i.e. ((a, b, c))).
And I have another variable defined like this
auto base = std::make_tuple
( 0.48f
, ExampleClass
, (ExampleClass& instance) {return instance.eGetter ();}
);
which I expect to be just a tuple of three elements (i.e (a, b, c))
I also have a vector defined as follows
std::vector<std::tuple<std::tuple< float
, ExampleClass
, std::function<float (ExampleClass&)>
>>> listOfInstruments;
Now if I add drum to listOfInstruments I expect no errors.
Which was indeed the case with listOfInstruments.push_back(drum);
Where I expected an error was here listOfInstuments.push_back(base); but the code compiles just fine.
Since listOfInstruments has type 'tuple of tuples', should't adding just 'tuple' cause some error? Unless both () and (()) are considered same types by std::vector. Or am I completely wrong and there's something else at work here?
Can't seem to figure it out.
c++ tuples
asked 3 hours ago
atisatis
34619
As an aside, I'd strongly discourage you from declaring astd::tupleof one element, and instead use the element type.
– Caleth
1 hour ago
add a comment |
I have a variable defined like this
auto drum = std::make_tuple
( std::make_tuple
( 0.3f
, ExampleClass
, (ExampleClass& instance) {return instance.eGetter ();}
)
);
I expect drum to be a tuple of tuple. (i.e. ((a, b, c))).
And I have another variable defined like this
auto base = std::make_tuple
( 0.48f
, ExampleClass
, (ExampleClass& instance) {return instance.eGetter ();}
);
which I expect to be just a tuple of three elements (i.e (a, b, c))
I also have a vector defined as follows
std::vector<std::tuple<std::tuple< float
, ExampleClass
, std::function<float (ExampleClass&)>
>>> listOfInstruments;
Now if I add drum to listOfInstruments I expect no errors.
Which was indeed the case with listOfInstruments.push_back(drum);
Where I expected an error was here listOfInstuments.push_back(base); but the code compiles just fine.
Since listOfInstruments has type 'tuple of tuples', should't adding just 'tuple' cause some error? Unless both () and (()) are considered same types by std::vector. Or am I completely wrong and there's something else at work here?
Can't seem to figure it out.
c++ tuples
asked 3 hours ago
atisatis
34619
I have a variable defined like this
auto drum = std::make_tuple
( std::make_tuple
( 0.3f
, ExampleClass
, (ExampleClass& instance) {return instance.eGetter ();}
)
);
I expect drum to be a tuple of tuple. (i.e. ((a, b, c))).
And I have another variable defined like this
auto base = std::make_tuple
( 0.48f
, ExampleClass
, (ExampleClass& instance) {return instance.eGetter ();}
);
which I expect to be just a tuple of three elements (i.e (a, b, c))
I also have a vector defined as follows
std::vector<std::tuple<std::tuple< float
, ExampleClass
, std::function<float (ExampleClass&)>
>>> listOfInstruments;
Now if I add drum to listOfInstruments I expect no errors.
Which was indeed the case with listOfInstruments.push_back(drum);
Where I expected an error was here listOfInstuments.push_back(base); but the code compiles just fine.
Since listOfInstruments has type 'tuple of tuples', should't adding just 'tuple' cause some error? Unless both () and (()) are considered same types by std::vector. Or am I completely wrong and there's something else at work here?
Can't seem to figure it out.
c++ tuples
c++ tuples
asked 3 hours ago
atisatis
34619
asked 3 hours ago
atisatis
34619
asked 3 hours ago
atisatis
34619
asked 3 hours ago
atisatis
34619
asked 3 hours ago
atisatis
34619
34619
As an aside, I'd strongly discourage you from declaring astd::tupleof one element, and instead use the element type.
– Caleth
1 hour ago
add a comment |
As an aside, I'd strongly discourage you from declaring astd::tupleof one element, and instead use the element type.
– Caleth
1 hour ago
As an aside, I'd strongly discourage you from declaring a
std::tuple of one element, and instead use the element type.– Caleth
1 hour ago
As an aside, I'd strongly discourage you from declaring a
std::tuple of one element, and instead use the element type.– Caleth
1 hour ago
add a comment |
1 Answer
1
active
oldest
votes
Tuples are mostly a red herring here. The way that works is simply that push_back can perform implicit conversions on its argument, as demonstrated by the following working snippet:
#include <vector>
struct A { };
struct B {
B(A const &) { }
};
int main() {
std::vector<B> v;
v.push_back(A{});
}
Going back to tuple, we can see that it has (among others) an implicit constructor (#2 here) which takes references to the tuple's members-to-be:
tuple( const Types&... args );
This means that std::tuple<...> is implicitly convertible to std::tuple<std::tuple<...>>, which is what you're observing.
answered 3 hours ago
QuentinQuentin
45.6k587142
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Tuples are mostly a red herring here. The way that works is simply that push_back can perform implicit conversions on its argument, as demonstrated by the following working snippet:
#include <vector>
struct A { };
struct B {
B(A const &) { }
};
int main() {
std::vector<B> v;
v.push_back(A{});
}
Going back to tuple, we can see that it has (among others) an implicit constructor (#2 here) which takes references to the tuple's members-to-be:
tuple( const Types&... args );
This means that std::tuple<...> is implicitly convertible to std::tuple<std::tuple<...>>, which is what you're observing.
answered 3 hours ago
QuentinQuentin
45.6k587142
add a comment |
Tuples are mostly a red herring here. The way that works is simply that push_back can perform implicit conversions on its argument, as demonstrated by the following working snippet:
#include <vector>
struct A { };
struct B {
B(A const &) { }
};
int main() {
std::vector<B> v;
v.push_back(A{});
}
Going back to tuple, we can see that it has (among others) an implicit constructor (#2 here) which takes references to the tuple's members-to-be:
tuple( const Types&... args );
This means that std::tuple<...> is implicitly convertible to std::tuple<std::tuple<...>>, which is what you're observing.
answered 3 hours ago
QuentinQuentin
45.6k587142
add a comment |
Tuples are mostly a red herring here. The way that works is simply that push_back can perform implicit conversions on its argument, as demonstrated by the following working snippet:
#include <vector>
struct A { };
struct B {
B(A const &) { }
};
int main() {
std::vector<B> v;
v.push_back(A{});
}
Going back to tuple, we can see that it has (among others) an implicit constructor (#2 here) which takes references to the tuple's members-to-be:
tuple( const Types&... args );
This means that std::tuple<...> is implicitly convertible to std::tuple<std::tuple<...>>, which is what you're observing.
answered 3 hours ago
QuentinQuentin
45.6k587142
Tuples are mostly a red herring here. The way that works is simply that push_back can perform implicit conversions on its argument, as demonstrated by the following working snippet:
#include <vector>
struct A { };
struct B {
B(A const &) { }
};
int main() {
std::vector<B> v;
v.push_back(A{});
}
Going back to tuple, we can see that it has (among others) an implicit constructor (#2 here) which takes references to the tuple's members-to-be:
tuple( const Types&... args );
This means that std::tuple<...> is implicitly convertible to std::tuple<std::tuple<...>>, which is what you're observing.
answered 3 hours ago
QuentinQuentin
45.6k587142
answered 3 hours ago
QuentinQuentin
45.6k587142
answered 3 hours ago
QuentinQuentin
45.6k587142
answered 3 hours ago
QuentinQuentin
45.6k587142
45.6k587142
add a comment |
add a comment |
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As an aside, I'd strongly discourage you from declaring a
std::tupleof one element, and instead use the element type.– Caleth
1 hour ago