Conditional ? : operator with class constructor
up vote
10
down vote
favorite
could someone explain me why c
and c1
are constructed different way.
I understand that I have reference to copy created by '?' operator, which is destroyed after construction, but why in first case it behave other way.
I've tested if its optimization, but even with conditions read from console, I have same result. Thanks in advance
#include <vector>
class foo {
public:
foo(const std::vector<int>& var) :var{ var } {};
const std::vector<int> & var;
};
std::vector<int> f(){
std::vector<int> x{ 1,2,3,4,5 };
return x;
};
int main(){
std::vector<int> x1{ 1,2,3,4,5 ,7 };
std::vector<int> x2{ 1,2,3,4,5 ,6 };
foo c{ true ? x2 : x1 }; //c.var has expected values
foo c1{ true ? x2 : f() }; //c.var empty
foo c2{ false ? x2 : f() }; //c.var empty
foo c3{ x2 }; //c.var has expected values
}
c++ c++14 conditional-operator
add a comment |
up vote
10
down vote
favorite
could someone explain me why c
and c1
are constructed different way.
I understand that I have reference to copy created by '?' operator, which is destroyed after construction, but why in first case it behave other way.
I've tested if its optimization, but even with conditions read from console, I have same result. Thanks in advance
#include <vector>
class foo {
public:
foo(const std::vector<int>& var) :var{ var } {};
const std::vector<int> & var;
};
std::vector<int> f(){
std::vector<int> x{ 1,2,3,4,5 };
return x;
};
int main(){
std::vector<int> x1{ 1,2,3,4,5 ,7 };
std::vector<int> x2{ 1,2,3,4,5 ,6 };
foo c{ true ? x2 : x1 }; //c.var has expected values
foo c1{ true ? x2 : f() }; //c.var empty
foo c2{ false ? x2 : f() }; //c.var empty
foo c3{ x2 }; //c.var has expected values
}
c++ c++14 conditional-operator
add a comment |
up vote
10
down vote
favorite
up vote
10
down vote
favorite
could someone explain me why c
and c1
are constructed different way.
I understand that I have reference to copy created by '?' operator, which is destroyed after construction, but why in first case it behave other way.
I've tested if its optimization, but even with conditions read from console, I have same result. Thanks in advance
#include <vector>
class foo {
public:
foo(const std::vector<int>& var) :var{ var } {};
const std::vector<int> & var;
};
std::vector<int> f(){
std::vector<int> x{ 1,2,3,4,5 };
return x;
};
int main(){
std::vector<int> x1{ 1,2,3,4,5 ,7 };
std::vector<int> x2{ 1,2,3,4,5 ,6 };
foo c{ true ? x2 : x1 }; //c.var has expected values
foo c1{ true ? x2 : f() }; //c.var empty
foo c2{ false ? x2 : f() }; //c.var empty
foo c3{ x2 }; //c.var has expected values
}
c++ c++14 conditional-operator
could someone explain me why c
and c1
are constructed different way.
I understand that I have reference to copy created by '?' operator, which is destroyed after construction, but why in first case it behave other way.
I've tested if its optimization, but even with conditions read from console, I have same result. Thanks in advance
#include <vector>
class foo {
public:
foo(const std::vector<int>& var) :var{ var } {};
const std::vector<int> & var;
};
std::vector<int> f(){
std::vector<int> x{ 1,2,3,4,5 };
return x;
};
int main(){
std::vector<int> x1{ 1,2,3,4,5 ,7 };
std::vector<int> x2{ 1,2,3,4,5 ,6 };
foo c{ true ? x2 : x1 }; //c.var has expected values
foo c1{ true ? x2 : f() }; //c.var empty
foo c2{ false ? x2 : f() }; //c.var empty
foo c3{ x2 }; //c.var has expected values
}
c++ c++14 conditional-operator
c++ c++14 conditional-operator
edited yesterday
Deduplicator
33.8k64787
33.8k64787
asked yesterday
geniculata
1568
1568
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add a comment |
1 Answer
1
active
oldest
votes
up vote
18
down vote
The type of a conditional expression is the common type of the two branches. For true ? x2 : x1
, the common type is std::vector<int>&
. For true ? x2 : f()
, the common type is std::vector<int>
.
Therefore you are storing a dangling reference in c1
. Any access to c1.var
is undefined behavior.
4
...resulting in UB, if it is accessed after the declaration. That crucial part was omitted from OP's example though.
– Deduplicator
yesterday
5
It'd be good to explain why these are the common types (ie: because the second case involves an lvalue and prvalue, while the first case is two lvalues).
– Nicol Bolas
yesterday
Oh C++. Oh you.
– Lightness Races in Orbit
yesterday
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
18
down vote
The type of a conditional expression is the common type of the two branches. For true ? x2 : x1
, the common type is std::vector<int>&
. For true ? x2 : f()
, the common type is std::vector<int>
.
Therefore you are storing a dangling reference in c1
. Any access to c1.var
is undefined behavior.
4
...resulting in UB, if it is accessed after the declaration. That crucial part was omitted from OP's example though.
– Deduplicator
yesterday
5
It'd be good to explain why these are the common types (ie: because the second case involves an lvalue and prvalue, while the first case is two lvalues).
– Nicol Bolas
yesterday
Oh C++. Oh you.
– Lightness Races in Orbit
yesterday
add a comment |
up vote
18
down vote
The type of a conditional expression is the common type of the two branches. For true ? x2 : x1
, the common type is std::vector<int>&
. For true ? x2 : f()
, the common type is std::vector<int>
.
Therefore you are storing a dangling reference in c1
. Any access to c1.var
is undefined behavior.
4
...resulting in UB, if it is accessed after the declaration. That crucial part was omitted from OP's example though.
– Deduplicator
yesterday
5
It'd be good to explain why these are the common types (ie: because the second case involves an lvalue and prvalue, while the first case is two lvalues).
– Nicol Bolas
yesterday
Oh C++. Oh you.
– Lightness Races in Orbit
yesterday
add a comment |
up vote
18
down vote
up vote
18
down vote
The type of a conditional expression is the common type of the two branches. For true ? x2 : x1
, the common type is std::vector<int>&
. For true ? x2 : f()
, the common type is std::vector<int>
.
Therefore you are storing a dangling reference in c1
. Any access to c1.var
is undefined behavior.
The type of a conditional expression is the common type of the two branches. For true ? x2 : x1
, the common type is std::vector<int>&
. For true ? x2 : f()
, the common type is std::vector<int>
.
Therefore you are storing a dangling reference in c1
. Any access to c1.var
is undefined behavior.
edited yesterday
answered yesterday
Vittorio Romeo
55.2k17146286
55.2k17146286
4
...resulting in UB, if it is accessed after the declaration. That crucial part was omitted from OP's example though.
– Deduplicator
yesterday
5
It'd be good to explain why these are the common types (ie: because the second case involves an lvalue and prvalue, while the first case is two lvalues).
– Nicol Bolas
yesterday
Oh C++. Oh you.
– Lightness Races in Orbit
yesterday
add a comment |
4
...resulting in UB, if it is accessed after the declaration. That crucial part was omitted from OP's example though.
– Deduplicator
yesterday
5
It'd be good to explain why these are the common types (ie: because the second case involves an lvalue and prvalue, while the first case is two lvalues).
– Nicol Bolas
yesterday
Oh C++. Oh you.
– Lightness Races in Orbit
yesterday
4
4
...resulting in UB, if it is accessed after the declaration. That crucial part was omitted from OP's example though.
– Deduplicator
yesterday
...resulting in UB, if it is accessed after the declaration. That crucial part was omitted from OP's example though.
– Deduplicator
yesterday
5
5
It'd be good to explain why these are the common types (ie: because the second case involves an lvalue and prvalue, while the first case is two lvalues).
– Nicol Bolas
yesterday
It'd be good to explain why these are the common types (ie: because the second case involves an lvalue and prvalue, while the first case is two lvalues).
– Nicol Bolas
yesterday
Oh C++. Oh you.
– Lightness Races in Orbit
yesterday
Oh C++. Oh you.
– Lightness Races in Orbit
yesterday
add a comment |
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