Conditional ? : operator with class constructor
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 '18 at 14:55
Deduplicator
34.5k64888
asked Nov 23 '18 at 14:30
geniculatageniculata
1668
add a comment |
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 '18 at 14:55
Deduplicator
34.5k64888
asked Nov 23 '18 at 14:30
geniculatageniculata
1668
add a comment |
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 '18 at 14:55
Deduplicator
34.5k64888
asked Nov 23 '18 at 14:30
geniculatageniculata
1668
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 '18 at 14:55
Deduplicator
34.5k64888
asked Nov 23 '18 at 14:30
geniculatageniculata
1668
edited Nov 23 '18 at 14:55
Deduplicator
34.5k64888
asked Nov 23 '18 at 14:30
geniculatageniculata
1668
edited Nov 23 '18 at 14:55
Deduplicator
34.5k64888
edited Nov 23 '18 at 14:55
Deduplicator
34.5k64888
edited Nov 23 '18 at 14:55
Deduplicator
34.5k64888
34.5k64888
asked Nov 23 '18 at 14:30
geniculatageniculata
1668
asked Nov 23 '18 at 14:30
geniculatageniculata
1668
asked Nov 23 '18 at 14:30
geniculatageniculata
1668
1668
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
The type of a conditional expression is the common type of the two branches, and its value category also depends on them.
For
true ? x2 : x1, the common type isstd::vector<int>and the value category is lvalue. This can be tested with:
static_assert(std::is_same_v<decltype((true ? x2 : x1)), std::vector<int>&>);
For
true ? x2 : f(), the common type isstd::vector<int>, and the value category is prvalue. This can be tested with:
static_assert(std::is_same_v<decltype((true ? x2 : f())), std::vector<int>>);
Therefore you are storing a dangling reference in c1. Any access to c1.var is undefined behavior.
live example on godbolt.org
answered Nov 23 '18 at 14:41
Vittorio RomeoVittorio Romeo
58k17158299
4
...resulting in UB, if it is accessed after the declaration. That crucial part was omitted from OP's example though.
– Deduplicator
Nov 23 '18 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 '18 at 14:59
Oh C++. Oh you.
– Lightness Races in Orbit
Nov 23 '18 at 15:23
Don't conflate an expression's type with its value category. A?:never has reference type. Not even the adjusted-away-prior-to-any-further-analysis one.
– T.C.
Nov 29 '18 at 12:36
@T.C.: updated the answer, please let me know if it's correct now.
– Vittorio Romeo
Nov 29 '18 at 14:57
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
The type of a conditional expression is the common type of the two branches, and its value category also depends on them.
For
true ? x2 : x1, the common type isstd::vector<int>and the value category is lvalue. This can be tested with:
static_assert(std::is_same_v<decltype((true ? x2 : x1)), std::vector<int>&>);
For
true ? x2 : f(), the common type isstd::vector<int>, and the value category is prvalue. This can be tested with:
static_assert(std::is_same_v<decltype((true ? x2 : f())), std::vector<int>>);
Therefore you are storing a dangling reference in c1. Any access to c1.var is undefined behavior.
live example on godbolt.org
answered Nov 23 '18 at 14:41
Vittorio RomeoVittorio Romeo
58k17158299
4
...resulting in UB, if it is accessed after the declaration. That crucial part was omitted from OP's example though.
– Deduplicator
Nov 23 '18 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 '18 at 14:59
Oh C++. Oh you.
– Lightness Races in Orbit
Nov 23 '18 at 15:23
Don't conflate an expression's type with its value category. A?:never has reference type. Not even the adjusted-away-prior-to-any-further-analysis one.
– T.C.
Nov 29 '18 at 12:36
@T.C.: updated the answer, please let me know if it's correct now.
– Vittorio Romeo
Nov 29 '18 at 14:57
add a comment |
The type of a conditional expression is the common type of the two branches, and its value category also depends on them.
For
true ? x2 : x1, the common type isstd::vector<int>and the value category is lvalue. This can be tested with:
static_assert(std::is_same_v<decltype((true ? x2 : x1)), std::vector<int>&>);
For
true ? x2 : f(), the common type isstd::vector<int>, and the value category is prvalue. This can be tested with:
static_assert(std::is_same_v<decltype((true ? x2 : f())), std::vector<int>>);
Therefore you are storing a dangling reference in c1. Any access to c1.var is undefined behavior.
live example on godbolt.org
answered Nov 23 '18 at 14:41
Vittorio RomeoVittorio Romeo
58k17158299
4
...resulting in UB, if it is accessed after the declaration. That crucial part was omitted from OP's example though.
– Deduplicator
Nov 23 '18 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 '18 at 14:59
Oh C++. Oh you.
– Lightness Races in Orbit
Nov 23 '18 at 15:23
Don't conflate an expression's type with its value category. A?:never has reference type. Not even the adjusted-away-prior-to-any-further-analysis one.
– T.C.
Nov 29 '18 at 12:36
@T.C.: updated the answer, please let me know if it's correct now.
– Vittorio Romeo
Nov 29 '18 at 14:57
add a comment |
The type of a conditional expression is the common type of the two branches, and its value category also depends on them.
For
true ? x2 : x1, the common type isstd::vector<int>and the value category is lvalue. This can be tested with:
static_assert(std::is_same_v<decltype((true ? x2 : x1)), std::vector<int>&>);
For
true ? x2 : f(), the common type isstd::vector<int>, and the value category is prvalue. This can be tested with:
static_assert(std::is_same_v<decltype((true ? x2 : f())), std::vector<int>>);
Therefore you are storing a dangling reference in c1. Any access to c1.var is undefined behavior.
live example on godbolt.org
answered Nov 23 '18 at 14:41
Vittorio RomeoVittorio Romeo
58k17158299
The type of a conditional expression is the common type of the two branches, and its value category also depends on them.
For
true ? x2 : x1, the common type isstd::vector<int>and the value category is lvalue. This can be tested with:
static_assert(std::is_same_v<decltype((true ? x2 : x1)), std::vector<int>&>);
For
true ? x2 : f(), the common type isstd::vector<int>, and the value category is prvalue. This can be tested with:
static_assert(std::is_same_v<decltype((true ? x2 : f())), std::vector<int>>);
Therefore you are storing a dangling reference in c1. Any access to c1.var is undefined behavior.
live example on godbolt.org
answered Nov 23 '18 at 14:41
Vittorio RomeoVittorio Romeo
58k17158299
edited Nov 29 '18 at 14:57
answered Nov 23 '18 at 14:41
Vittorio RomeoVittorio Romeo
58k17158299
answered Nov 23 '18 at 14:41
Vittorio RomeoVittorio Romeo
58k17158299
answered Nov 23 '18 at 14:41
Vittorio RomeoVittorio Romeo
58k17158299
58k17158299
4
...resulting in UB, if it is accessed after the declaration. That crucial part was omitted from OP's example though.
– Deduplicator
Nov 23 '18 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 '18 at 14:59
Oh C++. Oh you.
– Lightness Races in Orbit
Nov 23 '18 at 15:23
Don't conflate an expression's type with its value category. A?:never has reference type. Not even the adjusted-away-prior-to-any-further-analysis one.
– T.C.
Nov 29 '18 at 12:36
@T.C.: updated the answer, please let me know if it's correct now.
– Vittorio Romeo
Nov 29 '18 at 14:57
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 '18 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 '18 at 14:59
Oh C++. Oh you.
– Lightness Races in Orbit
Nov 23 '18 at 15:23
Don't conflate an expression's type with its value category. A?:never has reference type. Not even the adjusted-away-prior-to-any-further-analysis one.
– T.C.
Nov 29 '18 at 12:36
@T.C.: updated the answer, please let me know if it's correct now.
– Vittorio Romeo
Nov 29 '18 at 14:57
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 '18 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 '18 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 '18 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 '18 at 14:59
Oh C++. Oh you.
– Lightness Races in Orbit
Nov 23 '18 at 15:23
Oh C++. Oh you.
– Lightness Races in Orbit
Nov 23 '18 at 15:23
Don't conflate an expression's type with its value category. A
?: never has reference type. Not even the adjusted-away-prior-to-any-further-analysis one.– T.C.
Nov 29 '18 at 12:36
Don't conflate an expression's type with its value category. A
?: never has reference type. Not even the adjusted-away-prior-to-any-further-analysis one.– T.C.
Nov 29 '18 at 12:36
@T.C.: updated the answer, please let me know if it's correct now.
– Vittorio Romeo
Nov 29 '18 at 14:57
@T.C.: updated the answer, please let me know if it's correct now.
– Vittorio Romeo
Nov 29 '18 at 14:57
add a comment |
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