Constant reference data member
Let's suppose we have a structure with a constant reference data member.
struct A {
A() : i{5} {}
const int& foo() const { return i; }
const int& i;
};
Do you have any idea why the output for an integer literal 5 is different?
A a{};
std::cout << a.i << std::endl;
std::cout << a.foo() << std::endl;
5
-858993460
c++ c++11
asked Nov 24 '18 at 8:22
dubravadubrava
92
add a comment |
Let's suppose we have a structure with a constant reference data member.
struct A {
A() : i{5} {}
const int& foo() const { return i; }
const int& i;
};
Do you have any idea why the output for an integer literal 5 is different?
A a{};
std::cout << a.i << std::endl;
std::cout << a.foo() << std::endl;
5
-858993460
c++ c++11
asked Nov 24 '18 at 8:22
dubravadubrava
92
5
I get a compiler warning for this. Clang actually gives an error. Are you asking why this isn't correct, or specifically why the two outputs are different knowing the program is incorrect?
– chris
Nov 24 '18 at 8:25
1
See here also. Looks like you have a case of undefined behavior.
– πάντα ῥεῖ
Nov 24 '18 at 8:29
BTW, VS 2017 doesn't output any warnings. I don't understand why an integer literal goes out of the scope only when accessing the referenced value in function foo.
– dubrava
Nov 24 '18 at 8:41
1
@dubrava UB means anything is possible; so you might get the result of5,-858993460, or42, or something else like segfault.
– songyuanyao
Nov 24 '18 at 8:47
add a comment |
Let's suppose we have a structure with a constant reference data member.
struct A {
A() : i{5} {}
const int& foo() const { return i; }
const int& i;
};
Do you have any idea why the output for an integer literal 5 is different?
A a{};
std::cout << a.i << std::endl;
std::cout << a.foo() << std::endl;
5
-858993460
c++ c++11
asked Nov 24 '18 at 8:22
dubravadubrava
92
Let's suppose we have a structure with a constant reference data member.
struct A {
A() : i{5} {}
const int& foo() const { return i; }
const int& i;
};
Do you have any idea why the output for an integer literal 5 is different?
A a{};
std::cout << a.i << std::endl;
std::cout << a.foo() << std::endl;
5
-858993460
c++ c++11
c++ c++11
asked Nov 24 '18 at 8:22
dubravadubrava
92
asked Nov 24 '18 at 8:22
dubravadubrava
92
asked Nov 24 '18 at 8:22
dubravadubrava
92
asked Nov 24 '18 at 8:22
dubravadubrava
92
asked Nov 24 '18 at 8:22
dubravadubrava
92
92
5
I get a compiler warning for this. Clang actually gives an error. Are you asking why this isn't correct, or specifically why the two outputs are different knowing the program is incorrect?
– chris
Nov 24 '18 at 8:25
1
See here also. Looks like you have a case of undefined behavior.
– πάντα ῥεῖ
Nov 24 '18 at 8:29
BTW, VS 2017 doesn't output any warnings. I don't understand why an integer literal goes out of the scope only when accessing the referenced value in function foo.
– dubrava
Nov 24 '18 at 8:41
1
@dubrava UB means anything is possible; so you might get the result of5,-858993460, or42, or something else like segfault.
– songyuanyao
Nov 24 '18 at 8:47
add a comment |
5
I get a compiler warning for this. Clang actually gives an error. Are you asking why this isn't correct, or specifically why the two outputs are different knowing the program is incorrect?
– chris
Nov 24 '18 at 8:25
1
See here also. Looks like you have a case of undefined behavior.
– πάντα ῥεῖ
Nov 24 '18 at 8:29
BTW, VS 2017 doesn't output any warnings. I don't understand why an integer literal goes out of the scope only when accessing the referenced value in function foo.
– dubrava
Nov 24 '18 at 8:41
1
@dubrava UB means anything is possible; so you might get the result of5,-858993460, or42, or something else like segfault.
– songyuanyao
Nov 24 '18 at 8:47
5
5
I get a compiler warning for this. Clang actually gives an error. Are you asking why this isn't correct, or specifically why the two outputs are different knowing the program is incorrect?
– chris
Nov 24 '18 at 8:25
I get a compiler warning for this. Clang actually gives an error. Are you asking why this isn't correct, or specifically why the two outputs are different knowing the program is incorrect?
– chris
Nov 24 '18 at 8:25
1
1
See here also. Looks like you have a case of undefined behavior.
– πάντα ῥεῖ
Nov 24 '18 at 8:29
See here also. Looks like you have a case of undefined behavior.
– πάντα ῥεῖ
Nov 24 '18 at 8:29
BTW, VS 2017 doesn't output any warnings. I don't understand why an integer literal goes out of the scope only when accessing the referenced value in function foo.
– dubrava
Nov 24 '18 at 8:41
BTW, VS 2017 doesn't output any warnings. I don't understand why an integer literal goes out of the scope only when accessing the referenced value in function foo.
– dubrava
Nov 24 '18 at 8:41
1
1
@dubrava UB means anything is possible; so you might get the result of
5, -858993460, or 42, or something else like segfault.– songyuanyao
Nov 24 '18 at 8:47
@dubrava UB means anything is possible; so you might get the result of
5, -858993460, or 42, or something else like segfault.– songyuanyao
Nov 24 '18 at 8:47
add a comment |
2 Answers
2
active
oldest
votes
The code is ill-formed. You're initializing i from literal 5, which requires a temporary object to be contructed and then bound to i. The temporary will be destroyed when the constructor exits, then i becomes dangled, any dereference on it later leads to UB, means anything is possible.
From the standard, [class.base.init]/8
A temporary expression bound to a reference member in a
mem-initializer is ill-formed. [ Example:
struct A {
A() : v(42) { } // error
const int& v;
};
— end example ]
BTW: Since the standard states it's ill-formed, the compilers are required to issue a diagnostic for it. Both the behavior of gcc (gives a warning) and clang (gives an error) are conforming; if VS2017 doesn't issue any diagnostic then it's non-conforming to the standard.
answered Nov 24 '18 at 8:36
songyuanyaosongyuanyao
92.1k11175240
OK, thank you for the reference and detailed explanation!
– dubrava
Nov 24 '18 at 9:20
add a comment |
Use of a constant value not a literal.
This is Class::Class() : member{arg1, arg2, ...} {... direct-list-initialization since c++11.
Try this :
struct A {
const int t = 5;
A() : i{ t } { }
const int& foo() const { return i; }
const int& i;
};
int main()
{
A a{};
std::cout << a.i << std::endl;
std::cout << a.foo() << std::endl;
return 0;
}
answered Nov 24 '18 at 8:55
Mohammadreza PanahiMohammadreza Panahi
2,73021433
I didn't mean to store a temporary literal, just was wondering if its scope is a class or constructor.
– dubrava
Nov 24 '18 at 9:14
@dubrava I improved my answer, please check it. But about your question, honestly i didnt work with c++11 but I offered this answer to solving your problem . by the way this link might help you.
– Mohammadreza Panahi
Nov 24 '18 at 13:10
If I'd like to initialize the const reference to default value, I'd use in-class initializer const int& i {5};
– dubrava
Nov 24 '18 at 16:45
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
The code is ill-formed. You're initializing i from literal 5, which requires a temporary object to be contructed and then bound to i. The temporary will be destroyed when the constructor exits, then i becomes dangled, any dereference on it later leads to UB, means anything is possible.
From the standard, [class.base.init]/8
A temporary expression bound to a reference member in a
mem-initializer is ill-formed. [ Example:
struct A {
A() : v(42) { } // error
const int& v;
};
— end example ]
BTW: Since the standard states it's ill-formed, the compilers are required to issue a diagnostic for it. Both the behavior of gcc (gives a warning) and clang (gives an error) are conforming; if VS2017 doesn't issue any diagnostic then it's non-conforming to the standard.
answered Nov 24 '18 at 8:36
songyuanyaosongyuanyao
92.1k11175240
OK, thank you for the reference and detailed explanation!
– dubrava
Nov 24 '18 at 9:20
add a comment |
The code is ill-formed. You're initializing i from literal 5, which requires a temporary object to be contructed and then bound to i. The temporary will be destroyed when the constructor exits, then i becomes dangled, any dereference on it later leads to UB, means anything is possible.
From the standard, [class.base.init]/8
A temporary expression bound to a reference member in a
mem-initializer is ill-formed. [ Example:
struct A {
A() : v(42) { } // error
const int& v;
};
— end example ]
BTW: Since the standard states it's ill-formed, the compilers are required to issue a diagnostic for it. Both the behavior of gcc (gives a warning) and clang (gives an error) are conforming; if VS2017 doesn't issue any diagnostic then it's non-conforming to the standard.
answered Nov 24 '18 at 8:36
songyuanyaosongyuanyao
92.1k11175240
OK, thank you for the reference and detailed explanation!
– dubrava
Nov 24 '18 at 9:20
add a comment |
The code is ill-formed. You're initializing i from literal 5, which requires a temporary object to be contructed and then bound to i. The temporary will be destroyed when the constructor exits, then i becomes dangled, any dereference on it later leads to UB, means anything is possible.
From the standard, [class.base.init]/8
A temporary expression bound to a reference member in a
mem-initializer is ill-formed. [ Example:
struct A {
A() : v(42) { } // error
const int& v;
};
— end example ]
BTW: Since the standard states it's ill-formed, the compilers are required to issue a diagnostic for it. Both the behavior of gcc (gives a warning) and clang (gives an error) are conforming; if VS2017 doesn't issue any diagnostic then it's non-conforming to the standard.
answered Nov 24 '18 at 8:36
songyuanyaosongyuanyao
92.1k11175240
The code is ill-formed. You're initializing i from literal 5, which requires a temporary object to be contructed and then bound to i. The temporary will be destroyed when the constructor exits, then i becomes dangled, any dereference on it later leads to UB, means anything is possible.
From the standard, [class.base.init]/8
A temporary expression bound to a reference member in a
mem-initializer is ill-formed. [ Example:
struct A {
A() : v(42) { } // error
const int& v;
};
— end example ]
BTW: Since the standard states it's ill-formed, the compilers are required to issue a diagnostic for it. Both the behavior of gcc (gives a warning) and clang (gives an error) are conforming; if VS2017 doesn't issue any diagnostic then it's non-conforming to the standard.
answered Nov 24 '18 at 8:36
songyuanyaosongyuanyao
92.1k11175240
edited Nov 24 '18 at 8:55
answered Nov 24 '18 at 8:36
songyuanyaosongyuanyao
92.1k11175240
answered Nov 24 '18 at 8:36
songyuanyaosongyuanyao
92.1k11175240
answered Nov 24 '18 at 8:36
songyuanyaosongyuanyao
92.1k11175240
92.1k11175240
OK, thank you for the reference and detailed explanation!
– dubrava
Nov 24 '18 at 9:20
add a comment |
OK, thank you for the reference and detailed explanation!
– dubrava
Nov 24 '18 at 9:20
OK, thank you for the reference and detailed explanation!
– dubrava
Nov 24 '18 at 9:20
OK, thank you for the reference and detailed explanation!
– dubrava
Nov 24 '18 at 9:20
add a comment |
Use of a constant value not a literal.
This is Class::Class() : member{arg1, arg2, ...} {... direct-list-initialization since c++11.
Try this :
struct A {
const int t = 5;
A() : i{ t } { }
const int& foo() const { return i; }
const int& i;
};
int main()
{
A a{};
std::cout << a.i << std::endl;
std::cout << a.foo() << std::endl;
return 0;
}
answered Nov 24 '18 at 8:55
Mohammadreza PanahiMohammadreza Panahi
2,73021433
I didn't mean to store a temporary literal, just was wondering if its scope is a class or constructor.
– dubrava
Nov 24 '18 at 9:14
@dubrava I improved my answer, please check it. But about your question, honestly i didnt work with c++11 but I offered this answer to solving your problem . by the way this link might help you.
– Mohammadreza Panahi
Nov 24 '18 at 13:10
If I'd like to initialize the const reference to default value, I'd use in-class initializer const int& i {5};
– dubrava
Nov 24 '18 at 16:45
add a comment |
Use of a constant value not a literal.
This is Class::Class() : member{arg1, arg2, ...} {... direct-list-initialization since c++11.
Try this :
struct A {
const int t = 5;
A() : i{ t } { }
const int& foo() const { return i; }
const int& i;
};
int main()
{
A a{};
std::cout << a.i << std::endl;
std::cout << a.foo() << std::endl;
return 0;
}
answered Nov 24 '18 at 8:55
Mohammadreza PanahiMohammadreza Panahi
2,73021433
I didn't mean to store a temporary literal, just was wondering if its scope is a class or constructor.
– dubrava
Nov 24 '18 at 9:14
@dubrava I improved my answer, please check it. But about your question, honestly i didnt work with c++11 but I offered this answer to solving your problem . by the way this link might help you.
– Mohammadreza Panahi
Nov 24 '18 at 13:10
If I'd like to initialize the const reference to default value, I'd use in-class initializer const int& i {5};
– dubrava
Nov 24 '18 at 16:45
add a comment |
Use of a constant value not a literal.
This is Class::Class() : member{arg1, arg2, ...} {... direct-list-initialization since c++11.
Try this :
struct A {
const int t = 5;
A() : i{ t } { }
const int& foo() const { return i; }
const int& i;
};
int main()
{
A a{};
std::cout << a.i << std::endl;
std::cout << a.foo() << std::endl;
return 0;
}
answered Nov 24 '18 at 8:55
Mohammadreza PanahiMohammadreza Panahi
2,73021433
Use of a constant value not a literal.
This is Class::Class() : member{arg1, arg2, ...} {... direct-list-initialization since c++11.
Try this :
struct A {
const int t = 5;
A() : i{ t } { }
const int& foo() const { return i; }
const int& i;
};
int main()
{
A a{};
std::cout << a.i << std::endl;
std::cout << a.foo() << std::endl;
return 0;
}
answered Nov 24 '18 at 8:55
Mohammadreza PanahiMohammadreza Panahi
2,73021433
edited Nov 24 '18 at 13:04
answered Nov 24 '18 at 8:55
Mohammadreza PanahiMohammadreza Panahi
2,73021433
answered Nov 24 '18 at 8:55
Mohammadreza PanahiMohammadreza Panahi
2,73021433
answered Nov 24 '18 at 8:55
Mohammadreza PanahiMohammadreza Panahi
2,73021433
2,73021433
I didn't mean to store a temporary literal, just was wondering if its scope is a class or constructor.
– dubrava
Nov 24 '18 at 9:14
@dubrava I improved my answer, please check it. But about your question, honestly i didnt work with c++11 but I offered this answer to solving your problem . by the way this link might help you.
– Mohammadreza Panahi
Nov 24 '18 at 13:10
If I'd like to initialize the const reference to default value, I'd use in-class initializer const int& i {5};
– dubrava
Nov 24 '18 at 16:45
add a comment |
I didn't mean to store a temporary literal, just was wondering if its scope is a class or constructor.
– dubrava
Nov 24 '18 at 9:14
@dubrava I improved my answer, please check it. But about your question, honestly i didnt work with c++11 but I offered this answer to solving your problem . by the way this link might help you.
– Mohammadreza Panahi
Nov 24 '18 at 13:10
If I'd like to initialize the const reference to default value, I'd use in-class initializer const int& i {5};
– dubrava
Nov 24 '18 at 16:45
I didn't mean to store a temporary literal, just was wondering if its scope is a class or constructor.
– dubrava
Nov 24 '18 at 9:14
I didn't mean to store a temporary literal, just was wondering if its scope is a class or constructor.
– dubrava
Nov 24 '18 at 9:14
@dubrava I improved my answer, please check it. But about your question, honestly i didnt work with c++11 but I offered this answer to solving your problem . by the way this link might help you.
– Mohammadreza Panahi
Nov 24 '18 at 13:10
@dubrava I improved my answer, please check it. But about your question, honestly i didnt work with c++11 but I offered this answer to solving your problem . by the way this link might help you.
– Mohammadreza Panahi
Nov 24 '18 at 13:10
If I'd like to initialize the const reference to default value, I'd use in-class initializer const int& i {5};
– dubrava
Nov 24 '18 at 16:45
If I'd like to initialize the const reference to default value, I'd use in-class initializer const int& i {5};
– dubrava
Nov 24 '18 at 16:45
add a comment |
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5
I get a compiler warning for this. Clang actually gives an error. Are you asking why this isn't correct, or specifically why the two outputs are different knowing the program is incorrect?
– chris
Nov 24 '18 at 8:25
1
See here also. Looks like you have a case of undefined behavior.
– πάντα ῥεῖ
Nov 24 '18 at 8:29
BTW, VS 2017 doesn't output any warnings. I don't understand why an integer literal goes out of the scope only when accessing the referenced value in function foo.
– dubrava
Nov 24 '18 at 8:41
1
@dubrava UB means anything is possible; so you might get the result of
5,-858993460, or42, or something else like segfault.– songyuanyao
Nov 24 '18 at 8:47