What's the lifetime of temporary objects in a range-for?
Consider this class:
class Foo
{
public:
~ Foo ()
{
std::cout << "~Foon";
}
typedef std::vector<std::string> Words;
const Words & words ()
{
return m_words;
}
private:
Words m_words = {"foo", "bar", "baz"};
};
Section 12.2 of the C++ standard specifies lifetimes of temporary objects. I thought this would be okay:
for (auto w : Foo () .words ())
std::cout << w << "n";
But it wasn't
~Foo
terminate called after throwing an instance of 'std::logic_error'
what(): basic_string::_M_construct null not valid
[1] 10290 abort (core dumped) ./a.out
The standard is confusing me. Why is ~Foo being called before the loop runs?
c++ c++11 for-loop temporary-objects
edited Jan 2 at 7:33
xeros
1068
asked Jan 1 at 19:28
spraffspraff
18k1486167
add a comment |
Consider this class:
class Foo
{
public:
~ Foo ()
{
std::cout << "~Foon";
}
typedef std::vector<std::string> Words;
const Words & words ()
{
return m_words;
}
private:
Words m_words = {"foo", "bar", "baz"};
};
Section 12.2 of the C++ standard specifies lifetimes of temporary objects. I thought this would be okay:
for (auto w : Foo () .words ())
std::cout << w << "n";
But it wasn't
~Foo
terminate called after throwing an instance of 'std::logic_error'
what(): basic_string::_M_construct null not valid
[1] 10290 abort (core dumped) ./a.out
The standard is confusing me. Why is ~Foo being called before the loop runs?
c++ c++11 for-loop temporary-objects
edited Jan 2 at 7:33
xeros
1068
asked Jan 1 at 19:28
spraffspraff
18k1486167
add a comment |
Consider this class:
class Foo
{
public:
~ Foo ()
{
std::cout << "~Foon";
}
typedef std::vector<std::string> Words;
const Words & words ()
{
return m_words;
}
private:
Words m_words = {"foo", "bar", "baz"};
};
Section 12.2 of the C++ standard specifies lifetimes of temporary objects. I thought this would be okay:
for (auto w : Foo () .words ())
std::cout << w << "n";
But it wasn't
~Foo
terminate called after throwing an instance of 'std::logic_error'
what(): basic_string::_M_construct null not valid
[1] 10290 abort (core dumped) ./a.out
The standard is confusing me. Why is ~Foo being called before the loop runs?
c++ c++11 for-loop temporary-objects
edited Jan 2 at 7:33
xeros
1068
asked Jan 1 at 19:28
spraffspraff
18k1486167
Consider this class:
class Foo
{
public:
~ Foo ()
{
std::cout << "~Foon";
}
typedef std::vector<std::string> Words;
const Words & words ()
{
return m_words;
}
private:
Words m_words = {"foo", "bar", "baz"};
};
Section 12.2 of the C++ standard specifies lifetimes of temporary objects. I thought this would be okay:
for (auto w : Foo () .words ())
std::cout << w << "n";
But it wasn't
~Foo
terminate called after throwing an instance of 'std::logic_error'
what(): basic_string::_M_construct null not valid
[1] 10290 abort (core dumped) ./a.out
The standard is confusing me. Why is ~Foo being called before the loop runs?
c++ c++11 for-loop temporary-objects
c++ c++11 for-loop temporary-objects
edited Jan 2 at 7:33
xeros
1068
asked Jan 1 at 19:28
spraffspraff
18k1486167
edited Jan 2 at 7:33
xeros
1068
asked Jan 1 at 19:28
spraffspraff
18k1486167
edited Jan 2 at 7:33
xeros
1068
edited Jan 2 at 7:33
xeros
1068
edited Jan 2 at 7:33
xeros
1068
1068
asked Jan 1 at 19:28
spraffspraff
18k1486167
asked Jan 1 at 19:28
spraffspraff
18k1486167
asked Jan 1 at 19:28
spraffspraff
18k1486167
18k1486167
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
The current standard says in The range-based for statement [stmt.ranged] that
The range-based for statement
for ( init-statementopt for-range-declaration : for-range-initializer )statement
is equivalent to
{
init-statementopt
auto &&__range = for-range-initializer ;
auto __begin = begin-expr ;
auto __end = end-expr ;
for ( ; __begin != __end; ++__begin ) {
for-range-declaration = *__begin;
statement
}
}
This means that your Foo().words() is only used in the assignment auto &&__range = Foo().words(); and that the temporary object not lives until the code reaches the for loop.
Please note that I copied from the latest C++20 draft. In C++11 the code is a bit different, but the relevant part is the same.
answered Jan 1 at 19:39
Werner HenzeWerner Henze
10.7k72854
add a comment |
I think that the answer is to be found in the way the range-for is defined and to what __range is bound to.
[class.temporary]/6 - There are three contexts in which temporaries are destroyed at a different point than the end of the full-expression.
- The third context is when a reference is bound to a temporary object. The temporary object to which the reference is bound or the temporary object that is the complete object of a subobject to which the reference is bound persists for the lifetime of the reference ...
If you change the for-loop expression to bind directly to the subobject m_words(assuming that it were public), then the lifetime of Foo() will be extended and the following will work
for (auto w : Foo ().m_words)
std::cout << w << "n";
answered Jan 1 at 19:45
JansJans
9,02422635
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 current standard says in The range-based for statement [stmt.ranged] that
The range-based for statement
for ( init-statementopt for-range-declaration : for-range-initializer )statement
is equivalent to
{
init-statementopt
auto &&__range = for-range-initializer ;
auto __begin = begin-expr ;
auto __end = end-expr ;
for ( ; __begin != __end; ++__begin ) {
for-range-declaration = *__begin;
statement
}
}
This means that your Foo().words() is only used in the assignment auto &&__range = Foo().words(); and that the temporary object not lives until the code reaches the for loop.
Please note that I copied from the latest C++20 draft. In C++11 the code is a bit different, but the relevant part is the same.
answered Jan 1 at 19:39
Werner HenzeWerner Henze
10.7k72854
add a comment |
The current standard says in The range-based for statement [stmt.ranged] that
The range-based for statement
for ( init-statementopt for-range-declaration : for-range-initializer )statement
is equivalent to
{
init-statementopt
auto &&__range = for-range-initializer ;
auto __begin = begin-expr ;
auto __end = end-expr ;
for ( ; __begin != __end; ++__begin ) {
for-range-declaration = *__begin;
statement
}
}
This means that your Foo().words() is only used in the assignment auto &&__range = Foo().words(); and that the temporary object not lives until the code reaches the for loop.
Please note that I copied from the latest C++20 draft. In C++11 the code is a bit different, but the relevant part is the same.
answered Jan 1 at 19:39
Werner HenzeWerner Henze
10.7k72854
add a comment |
The current standard says in The range-based for statement [stmt.ranged] that
The range-based for statement
for ( init-statementopt for-range-declaration : for-range-initializer )statement
is equivalent to
{
init-statementopt
auto &&__range = for-range-initializer ;
auto __begin = begin-expr ;
auto __end = end-expr ;
for ( ; __begin != __end; ++__begin ) {
for-range-declaration = *__begin;
statement
}
}
This means that your Foo().words() is only used in the assignment auto &&__range = Foo().words(); and that the temporary object not lives until the code reaches the for loop.
Please note that I copied from the latest C++20 draft. In C++11 the code is a bit different, but the relevant part is the same.
answered Jan 1 at 19:39
Werner HenzeWerner Henze
10.7k72854
The current standard says in The range-based for statement [stmt.ranged] that
The range-based for statement
for ( init-statementopt for-range-declaration : for-range-initializer )statement
is equivalent to
{
init-statementopt
auto &&__range = for-range-initializer ;
auto __begin = begin-expr ;
auto __end = end-expr ;
for ( ; __begin != __end; ++__begin ) {
for-range-declaration = *__begin;
statement
}
}
This means that your Foo().words() is only used in the assignment auto &&__range = Foo().words(); and that the temporary object not lives until the code reaches the for loop.
Please note that I copied from the latest C++20 draft. In C++11 the code is a bit different, but the relevant part is the same.
answered Jan 1 at 19:39
Werner HenzeWerner Henze
10.7k72854
edited Jan 1 at 19:45
answered Jan 1 at 19:39
Werner HenzeWerner Henze
10.7k72854
answered Jan 1 at 19:39
Werner HenzeWerner Henze
10.7k72854
answered Jan 1 at 19:39
Werner HenzeWerner Henze
10.7k72854
10.7k72854
add a comment |
add a comment |
I think that the answer is to be found in the way the range-for is defined and to what __range is bound to.
[class.temporary]/6 - There are three contexts in which temporaries are destroyed at a different point than the end of the full-expression.
- The third context is when a reference is bound to a temporary object. The temporary object to which the reference is bound or the temporary object that is the complete object of a subobject to which the reference is bound persists for the lifetime of the reference ...
If you change the for-loop expression to bind directly to the subobject m_words(assuming that it were public), then the lifetime of Foo() will be extended and the following will work
for (auto w : Foo ().m_words)
std::cout << w << "n";
answered Jan 1 at 19:45
JansJans
9,02422635
add a comment |
I think that the answer is to be found in the way the range-for is defined and to what __range is bound to.
[class.temporary]/6 - There are three contexts in which temporaries are destroyed at a different point than the end of the full-expression.
- The third context is when a reference is bound to a temporary object. The temporary object to which the reference is bound or the temporary object that is the complete object of a subobject to which the reference is bound persists for the lifetime of the reference ...
If you change the for-loop expression to bind directly to the subobject m_words(assuming that it were public), then the lifetime of Foo() will be extended and the following will work
for (auto w : Foo ().m_words)
std::cout << w << "n";
answered Jan 1 at 19:45
JansJans
9,02422635
add a comment |
I think that the answer is to be found in the way the range-for is defined and to what __range is bound to.
[class.temporary]/6 - There are three contexts in which temporaries are destroyed at a different point than the end of the full-expression.
- The third context is when a reference is bound to a temporary object. The temporary object to which the reference is bound or the temporary object that is the complete object of a subobject to which the reference is bound persists for the lifetime of the reference ...
If you change the for-loop expression to bind directly to the subobject m_words(assuming that it were public), then the lifetime of Foo() will be extended and the following will work
for (auto w : Foo ().m_words)
std::cout << w << "n";
answered Jan 1 at 19:45
JansJans
9,02422635
I think that the answer is to be found in the way the range-for is defined and to what __range is bound to.
[class.temporary]/6 - There are three contexts in which temporaries are destroyed at a different point than the end of the full-expression.
- The third context is when a reference is bound to a temporary object. The temporary object to which the reference is bound or the temporary object that is the complete object of a subobject to which the reference is bound persists for the lifetime of the reference ...
If you change the for-loop expression to bind directly to the subobject m_words(assuming that it were public), then the lifetime of Foo() will be extended and the following will work
for (auto w : Foo ().m_words)
std::cout << w << "n";
answered Jan 1 at 19:45
JansJans
9,02422635
edited Jan 2 at 1:31
answered Jan 1 at 19:45
JansJans
9,02422635
answered Jan 1 at 19:45
JansJans
9,02422635
answered Jan 1 at 19:45
JansJans
9,02422635
9,02422635
add a comment |
add a comment |
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