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When moving a unique_ptr into a lambda, why is it not possible to call reset?

Why are Python lambdas useful?What is std::move(), and when should it be used?Why can't I create a vector of lambdas (of the same type) in C++11?How to capture std::unique_ptr “by move” for a lambda in std::for_eachunique_ptr<T> lambda custom deleter for array specializationTrouble move-capturing std::unique_ptr in a lambda using std::bindWhy can I not move unique_ptr from a set to a function argument using an iterator?How to move a unique_ptr?Move objects from a unique_ptr array to a vectorWhy does [=] have a lambda capture?

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7

When moving std::unique_ptr into the lambda, it is not possible to call reset() on it, because it seems to be const then:

error C2662: void std::unique_ptr<int,std::default_delete<_Ty>>::reset(int *) noexcept': cannot convert 'this' pointer from 'const std::unique_ptr<int,std::default_delete<_Ty>>' to 'std::unique_ptr<int,std::default_delete<_Ty>> &

#include <memory>
int main()

 auto u = std::unique_ptr<int>();
 auto l = [v = std::move(u)]
 v.reset(); // this doesn't compile
 ;

1. Why does this happen?


2. Is it possible to capture the std::unique_ptr in another way which allows calling reset() within the lambda (with C++17 or later)?

c++ c++11 lambda smart-pointers

share|improve this question

edited 4 hours ago

Boann

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asked 9 hours ago

Roi DantonRoi Danton

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add a comment | 

7

When moving std::unique_ptr into the lambda, it is not possible to call reset() on it, because it seems to be const then:

error C2662: void std::unique_ptr<int,std::default_delete<_Ty>>::reset(int *) noexcept': cannot convert 'this' pointer from 'const std::unique_ptr<int,std::default_delete<_Ty>>' to 'std::unique_ptr<int,std::default_delete<_Ty>> &

#include <memory>
int main()

 auto u = std::unique_ptr<int>();
 auto l = [v = std::move(u)]
 v.reset(); // this doesn't compile
 ;

1. Why does this happen?


2. Is it possible to capture the std::unique_ptr in another way which allows calling reset() within the lambda (with C++17 or later)?

c++ c++11 lambda smart-pointers

share|improve this question

edited 4 hours ago

Boann

38.3k1313 gold badges9292 silver badges123123 bronze badges

asked 9 hours ago

Roi DantonRoi Danton

2,52422 gold badges2525 silver badges3838 bronze badges

add a comment | 

When moving std::unique_ptr into the lambda, it is not possible to call reset() on it, because it seems to be const then:

error C2662: void std::unique_ptr<int,std::default_delete<_Ty>>::reset(int *) noexcept': cannot convert 'this' pointer from 'const std::unique_ptr<int,std::default_delete<_Ty>>' to 'std::unique_ptr<int,std::default_delete<_Ty>> &

#include <memory>
int main()

 auto u = std::unique_ptr<int>();
 auto l = [v = std::move(u)]
 v.reset(); // this doesn't compile
 ;

1. Why does this happen?


2. Is it possible to capture the std::unique_ptr in another way which allows calling reset() within the lambda (with C++17 or later)?

c++ c++11 lambda smart-pointers

share|improve this question

edited 4 hours ago

Boann

38.3k1313 gold badges9292 silver badges123123 bronze badges

asked 9 hours ago

Roi DantonRoi Danton

2,52422 gold badges2525 silver badges3838 bronze badges

error C2662: void std::unique_ptr<int,std::default_delete<_Ty>>::reset(int *) noexcept': cannot convert 'this' pointer from 'const std::unique_ptr<int,std::default_delete<_Ty>>' to 'std::unique_ptr<int,std::default_delete<_Ty>> &

#include <memory>
int main()

 auto u = std::unique_ptr<int>();
 auto l = [v = std::move(u)]
 v.reset(); // this doesn't compile
 ;

share|improve this question

edited 4 hours ago

Boann

38.3k1313 gold badges9292 silver badges123123 bronze badges

asked 9 hours ago

Roi DantonRoi Danton

2,52422 gold badges2525 silver badges3838 bronze badges

share|improve this question

edited 4 hours ago

Boann

38.3k1313 gold badges9292 silver badges123123 bronze badges

asked 9 hours ago

Roi DantonRoi Danton

2,52422 gold badges2525 silver badges3838 bronze badges

edited 4 hours ago

Boann

38.3k1313 gold badges9292 silver badges123123 bronze badges

edited 4 hours ago

Boann

38.3k1313 gold badges9292 silver badges123123 bronze badges

asked 9 hours ago

Roi DantonRoi Danton

2,52422 gold badges2525 silver badges3838 bronze badges

asked 9 hours ago

Roi DantonRoi Danton

2,52422 gold badges2525 silver badges3838 bronze badges

4 Answers
4

active

oldest

votes

15

1. Why does this happen?

Because the function-call operator of lambda,

Unless the keyword mutable was used in the lambda-expression, the function-call operator is const-qualified and the objects that were captured by copy are non-modifiable from inside this operator().

and

2. Is it possible to capture the std::unique_ptr in another way which allows to call reset() within the lambda

You need to mark it mutable.

mutable: allows body to modify the parameters captured by copy, and to call their non-const member functions

e.g.

auto l = [v = std::move(u)] mutable 
 v.reset();
;

share|improve this answer

edited 9 hours ago

answered 9 hours ago

songyuanyaosongyuanyao

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add a comment | 

5

1. Why does this happen?

Because lambdas are by default non-mutable. Therefore all captured objects are const. reset is a non-const member function that modifies the unique pointer.

2. Is it possible to capture the std::unique_ptr in another way which allows to call reset() within the lambda (with C++17 or later)?

Yes. Declare the lambda mutable:

[captures](arguments) mutable body 
 ^^^^^^^

This is possible since C++11 where lambdas were introduced. All captured non-const objects of a mutable lambda are non-const copies.

share|improve this answer

answered 9 hours ago

eerorikaeerorika

97.9k66 gold badges7777 silver badges148148 bronze badges

add a comment | 

5

Within the lambda its data members are immutable by default. You need to append the mutable specifier to the lambda expression.

As an alternative, you could capture the unique_ptr by reference, as for example:

#include <memory>

int main()

 auto u = std::unique_ptr<int>();
 auto l = [&v = u]
 v.reset(); 
 ;

share|improve this answer

edited 6 hours ago

Remy Lebeau

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answered 8 hours ago

Vlad from MoscowVlad from Moscow

144k1313 gold badges8888 silver badges184184 bronze badges

add a comment | 

2

To mutate "member" of the lambda, you need mutable keyword:

auto l = [v = std::move(u)] () mutable 
 v.reset();
;

share|improve this answer

edited 9 hours ago

answered 9 hours ago

Jarod42Jarod42

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add a comment | 

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15

1. Why does this happen?

Because the function-call operator of lambda,

Unless the keyword mutable was used in the lambda-expression, the function-call operator is const-qualified and the objects that were captured by copy are non-modifiable from inside this operator().

and

2. Is it possible to capture the std::unique_ptr in another way which allows to call reset() within the lambda

You need to mark it mutable.

mutable: allows body to modify the parameters captured by copy, and to call their non-const member functions

e.g.

auto l = [v = std::move(u)] mutable 
 v.reset();
;

share|improve this answer

edited 9 hours ago

answered 9 hours ago

songyuanyaosongyuanyao

99.6k1111 gold badges195195 silver badges263263 bronze badges

add a comment | 

15

1. Why does this happen?

Because the function-call operator of lambda,

Unless the keyword mutable was used in the lambda-expression, the function-call operator is const-qualified and the objects that were captured by copy are non-modifiable from inside this operator().

and

2. Is it possible to capture the std::unique_ptr in another way which allows to call reset() within the lambda

You need to mark it mutable.

mutable: allows body to modify the parameters captured by copy, and to call their non-const member functions

e.g.

auto l = [v = std::move(u)] mutable 
 v.reset();
;

share|improve this answer

edited 9 hours ago

answered 9 hours ago

songyuanyaosongyuanyao

99.6k1111 gold badges195195 silver badges263263 bronze badges

add a comment | 

1. Why does this happen?

Because the function-call operator of lambda,

Unless the keyword mutable was used in the lambda-expression, the function-call operator is const-qualified and the objects that were captured by copy are non-modifiable from inside this operator().

and

2. Is it possible to capture the std::unique_ptr in another way which allows to call reset() within the lambda

You need to mark it mutable.

mutable: allows body to modify the parameters captured by copy, and to call their non-const member functions

e.g.

auto l = [v = std::move(u)] mutable 
 v.reset();
;

share|improve this answer

edited 9 hours ago

answered 9 hours ago

songyuanyaosongyuanyao

99.6k1111 gold badges195195 silver badges263263 bronze badges

auto l = [v = std::move(u)] mutable 
 v.reset();
;

share|improve this answer

edited 9 hours ago

answered 9 hours ago

songyuanyaosongyuanyao

99.6k1111 gold badges195195 silver badges263263 bronze badges

answered 9 hours ago

songyuanyaosongyuanyao

99.6k1111 gold badges195195 silver badges263263 bronze badges

answered 9 hours ago

songyuanyaosongyuanyao

99.6k1111 gold badges195195 silver badges263263 bronze badges

5

1. Why does this happen?

Because lambdas are by default non-mutable. Therefore all captured objects are const. reset is a non-const member function that modifies the unique pointer.

2. Is it possible to capture the std::unique_ptr in another way which allows to call reset() within the lambda (with C++17 or later)?

Yes. Declare the lambda mutable:

[captures](arguments) mutable body 
 ^^^^^^^

This is possible since C++11 where lambdas were introduced. All captured non-const objects of a mutable lambda are non-const copies.

share|improve this answer

answered 9 hours ago

eerorikaeerorika

97.9k66 gold badges7777 silver badges148148 bronze badges

add a comment | 

5

1. Why does this happen?

Because lambdas are by default non-mutable. Therefore all captured objects are const. reset is a non-const member function that modifies the unique pointer.

2. Is it possible to capture the std::unique_ptr in another way which allows to call reset() within the lambda (with C++17 or later)?

Yes. Declare the lambda mutable:

[captures](arguments) mutable body 
 ^^^^^^^

This is possible since C++11 where lambdas were introduced. All captured non-const objects of a mutable lambda are non-const copies.

share|improve this answer

answered 9 hours ago

eerorikaeerorika

97.9k66 gold badges7777 silver badges148148 bronze badges

add a comment | 

1. Why does this happen?

Because lambdas are by default non-mutable. Therefore all captured objects are const. reset is a non-const member function that modifies the unique pointer.

2. Is it possible to capture the std::unique_ptr in another way which allows to call reset() within the lambda (with C++17 or later)?

Yes. Declare the lambda mutable:

[captures](arguments) mutable body 
 ^^^^^^^

This is possible since C++11 where lambdas were introduced. All captured non-const objects of a mutable lambda are non-const copies.

share|improve this answer

answered 9 hours ago

eerorikaeerorika

97.9k66 gold badges7777 silver badges148148 bronze badges

[captures](arguments) mutable body 
 ^^^^^^^

share|improve this answer

answered 9 hours ago

eerorikaeerorika

97.9k66 gold badges7777 silver badges148148 bronze badges

answered 9 hours ago

eerorikaeerorika

97.9k66 gold badges7777 silver badges148148 bronze badges

answered 9 hours ago

eerorikaeerorika

97.9k66 gold badges7777 silver badges148148 bronze badges

5

Within the lambda its data members are immutable by default. You need to append the mutable specifier to the lambda expression.

As an alternative, you could capture the unique_ptr by reference, as for example:

#include <memory>

int main()

 auto u = std::unique_ptr<int>();
 auto l = [&v = u]
 v.reset(); 
 ;

share|improve this answer

edited 6 hours ago

Remy Lebeau

354k1919 gold badges286286 silver badges480480 bronze badges

answered 8 hours ago

Vlad from MoscowVlad from Moscow

144k1313 gold badges8888 silver badges184184 bronze badges

add a comment | 

5

Within the lambda its data members are immutable by default. You need to append the mutable specifier to the lambda expression.

As an alternative, you could capture the unique_ptr by reference, as for example:

#include <memory>

int main()

 auto u = std::unique_ptr<int>();
 auto l = [&v = u]
 v.reset(); 
 ;

share|improve this answer

edited 6 hours ago

Remy Lebeau

354k1919 gold badges286286 silver badges480480 bronze badges

answered 8 hours ago

Vlad from MoscowVlad from Moscow

144k1313 gold badges8888 silver badges184184 bronze badges

add a comment | 

Within the lambda its data members are immutable by default. You need to append the mutable specifier to the lambda expression.

As an alternative, you could capture the unique_ptr by reference, as for example:

#include <memory>

int main()

 auto u = std::unique_ptr<int>();
 auto l = [&v = u]
 v.reset(); 
 ;

share|improve this answer

edited 6 hours ago

Remy Lebeau

354k1919 gold badges286286 silver badges480480 bronze badges

answered 8 hours ago

Vlad from MoscowVlad from Moscow

144k1313 gold badges8888 silver badges184184 bronze badges

#include <memory>

int main()

 auto u = std::unique_ptr<int>();
 auto l = [&v = u]
 v.reset(); 
 ;

share|improve this answer

edited 6 hours ago

Remy Lebeau

354k1919 gold badges286286 silver badges480480 bronze badges

answered 8 hours ago

Vlad from MoscowVlad from Moscow

144k1313 gold badges8888 silver badges184184 bronze badges

edited 6 hours ago

Remy Lebeau

354k1919 gold badges286286 silver badges480480 bronze badges

edited 6 hours ago

Remy Lebeau

354k1919 gold badges286286 silver badges480480 bronze badges

answered 8 hours ago

Vlad from MoscowVlad from Moscow

144k1313 gold badges8888 silver badges184184 bronze badges

answered 8 hours ago

Vlad from MoscowVlad from Moscow

144k1313 gold badges8888 silver badges184184 bronze badges

2

To mutate "member" of the lambda, you need mutable keyword:

auto l = [v = std::move(u)] () mutable 
 v.reset();
;

share|improve this answer

edited 9 hours ago

answered 9 hours ago

Jarod42Jarod42

125k1212 gold badges109109 silver badges196196 bronze badges

add a comment | 

2

To mutate "member" of the lambda, you need mutable keyword:

auto l = [v = std::move(u)] () mutable 
 v.reset();
;

share|improve this answer

edited 9 hours ago

answered 9 hours ago

Jarod42Jarod42

125k1212 gold badges109109 silver badges196196 bronze badges

add a comment | 

To mutate "member" of the lambda, you need mutable keyword:

auto l = [v = std::move(u)] () mutable 
 v.reset();
;

share|improve this answer

edited 9 hours ago

answered 9 hours ago

Jarod42Jarod42

125k1212 gold badges109109 silver badges196196 bronze badges

auto l = [v = std::move(u)] () mutable 
 v.reset();
;

share|improve this answer

edited 9 hours ago

answered 9 hours ago

Jarod42Jarod42

125k1212 gold badges109109 silver badges196196 bronze badges

answered 9 hours ago

Jarod42Jarod42

125k1212 gold badges109109 silver badges196196 bronze badges

answered 9 hours ago

Jarod42Jarod42

125k1212 gold badges109109 silver badges196196 bronze badges