Conditional ? : operator with class constructor
up vote
11
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
edited Nov 23 at 14:55
Deduplicator
33.9k64787
asked Nov 23 at 14:30
geniculata
1618
add a comment |
up vote
11
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
edited Nov 23 at 14:55
Deduplicator
33.9k64787
asked Nov 23 at 14:30
geniculata
1618
add a comment |
up vote
11
down vote
favorite
up vote
11
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
edited Nov 23 at 14:55
Deduplicator
33.9k64787
asked Nov 23 at 14:30
geniculata
1618
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 Nov 23 at 14:55
Deduplicator
33.9k64787
asked Nov 23 at 14:30
geniculata
1618
edited Nov 23 at 14:55
Deduplicator
33.9k64787
asked Nov 23 at 14:30
geniculata
1618
edited Nov 23 at 14:55
Deduplicator
33.9k64787
edited Nov 23 at 14:55
Deduplicator
33.9k64787
edited Nov 23 at 14:55
Deduplicator
33.9k64787
33.9k64787
asked Nov 23 at 14:30
geniculata
1618
asked Nov 23 at 14:30
geniculata
1618
asked Nov 23 at 14:30
geniculata
1618
1618
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
up vote
19
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.
answered Nov 23 at 14:41
Vittorio Romeo
55.6k17146287
4
...resulting in UB, if it is accessed after the declaration. That crucial part was omitted from OP's example though.
– Deduplicator
Nov 23 at 14:57
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
Nov 23 at 14:59
Oh C++. Oh you.
– Lightness Races in Orbit
Nov 23 at 15:23
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
19
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.
answered Nov 23 at 14:41
Vittorio Romeo
55.6k17146287
4
...resulting in UB, if it is accessed after the declaration. That crucial part was omitted from OP's example though.
– Deduplicator
Nov 23 at 14:57
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
Nov 23 at 14:59
Oh C++. Oh you.
– Lightness Races in Orbit
Nov 23 at 15:23
add a comment |
up vote
19
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.
answered Nov 23 at 14:41
Vittorio Romeo
55.6k17146287
4
...resulting in UB, if it is accessed after the declaration. That crucial part was omitted from OP's example though.
– Deduplicator
Nov 23 at 14:57
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
Nov 23 at 14:59
Oh C++. Oh you.
– Lightness Races in Orbit
Nov 23 at 15:23
add a comment |
up vote
19
down vote
up vote
19
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.
answered Nov 23 at 14:41
Vittorio Romeo
55.6k17146287
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.
answered Nov 23 at 14:41
Vittorio Romeo
55.6k17146287
edited Nov 23 at 14:59
answered Nov 23 at 14:41
Vittorio Romeo
55.6k17146287
answered Nov 23 at 14:41
Vittorio Romeo
55.6k17146287
answered Nov 23 at 14:41
Vittorio Romeo
55.6k17146287
55.6k17146287
4
...resulting in UB, if it is accessed after the declaration. That crucial part was omitted from OP's example though.
– Deduplicator
Nov 23 at 14:57
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
Nov 23 at 14:59
Oh C++. Oh you.
– Lightness Races in Orbit
Nov 23 at 15:23
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
Nov 23 at 14:57
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
Nov 23 at 14:59
Oh C++. Oh you.
– Lightness Races in Orbit
Nov 23 at 15:23
4
4
...resulting in UB, if it is accessed after the declaration. That crucial part was omitted from OP's example though.
– Deduplicator
Nov 23 at 14:57
...resulting in UB, if it is accessed after the declaration. That crucial part was omitted from OP's example though.
– Deduplicator
Nov 23 at 14:57
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
Nov 23 at 14:59
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
Nov 23 at 14:59
Oh C++. Oh you.
– Lightness Races in Orbit
Nov 23 at 15:23
Oh C++. Oh you.
– Lightness Races in Orbit
Nov 23 at 15:23
add a comment |
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f53448536%2fconditional-operator-with-class-constructor%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
