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6

2

I would like to have a default lambda for a functor argument in my function.

I am aware it is possible using a struct and operator() like this:

struct AddOne 
 int operator()(int a) 
 return a+1;
 
;

template <typename Functor = AddOne>
int run_old(int x, Functor func = AddOne()) 

 return func(x);

But I was wondering if there was a modern way, given the changes in the standard in either c++14/17/20, to make this work?

template <typename Functor>
int run_new(int x, Functor func = [](int a) return a+1; ) 

 return func(x);

I'm not sure what one would use as the default type to Functor, or if there is syntax i'm unaware of.

https://godbolt.org/z/Hs6vQs

c++ lambda c++14 c++17 template-deduction

share|improve this question

edited 9 hours ago

max66

45.4k77 gold badges4848 silver badges8080 bronze badges

asked 9 hours ago

SalgarSalgar

6,37111 gold badge2222 silver badges3636 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  "modern way" What's wrong with the old way? The struct method is much easier to read at the function site.
  
  – Nicol Bolas
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  In the old way, you likely don't want to provide both default template and function arguments this way: wandbox.org/permlink/PtSwiQwsNDyDrshu. Use Functor func = Functor() instead.
  
  – Daniel Langr
  9 hours ago
  
  

  
  
  
  

add a comment | 

6

2

I would like to have a default lambda for a functor argument in my function.

I am aware it is possible using a struct and operator() like this:

struct AddOne 
 int operator()(int a) 
 return a+1;
 
;

template <typename Functor = AddOne>
int run_old(int x, Functor func = AddOne()) 

 return func(x);

But I was wondering if there was a modern way, given the changes in the standard in either c++14/17/20, to make this work?

template <typename Functor>
int run_new(int x, Functor func = [](int a) return a+1; ) 

 return func(x);

I'm not sure what one would use as the default type to Functor, or if there is syntax i'm unaware of.

https://godbolt.org/z/Hs6vQs

c++ lambda c++14 c++17 template-deduction

share|improve this question

edited 9 hours ago

max66

45.4k77 gold badges4848 silver badges8080 bronze badges

asked 9 hours ago

SalgarSalgar

6,37111 gold badge2222 silver badges3636 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  "modern way" What's wrong with the old way? The struct method is much easier to read at the function site.
  
  – Nicol Bolas
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  In the old way, you likely don't want to provide both default template and function arguments this way: wandbox.org/permlink/PtSwiQwsNDyDrshu. Use Functor func = Functor() instead.
  
  – Daniel Langr
  9 hours ago
  
  

  
  
  
  

add a comment | 

I would like to have a default lambda for a functor argument in my function.

I am aware it is possible using a struct and operator() like this:

struct AddOne 
 int operator()(int a) 
 return a+1;
 
;

template <typename Functor = AddOne>
int run_old(int x, Functor func = AddOne()) 

 return func(x);

But I was wondering if there was a modern way, given the changes in the standard in either c++14/17/20, to make this work?

template <typename Functor>
int run_new(int x, Functor func = [](int a) return a+1; ) 

 return func(x);

I'm not sure what one would use as the default type to Functor, or if there is syntax i'm unaware of.

https://godbolt.org/z/Hs6vQs

c++ lambda c++14 c++17 template-deduction

share|improve this question

edited 9 hours ago

max66

45.4k77 gold badges4848 silver badges8080 bronze badges

asked 9 hours ago

SalgarSalgar

6,37111 gold badge2222 silver badges3636 bronze badges

struct AddOne 
 int operator()(int a) 
 return a+1;
 
;

template <typename Functor = AddOne>
int run_old(int x, Functor func = AddOne()) 

 return func(x);

template <typename Functor>
int run_new(int x, Functor func = [](int a) return a+1; ) 

 return func(x);

share|improve this question

edited 9 hours ago

max66

45.4k77 gold badges4848 silver badges8080 bronze badges

asked 9 hours ago

SalgarSalgar

6,37111 gold badge2222 silver badges3636 bronze badges

share|improve this question

edited 9 hours ago

max66

45.4k77 gold badges4848 silver badges8080 bronze badges

asked 9 hours ago

SalgarSalgar

6,37111 gold badge2222 silver badges3636 bronze badges

edited 9 hours ago

max66

45.4k77 gold badges4848 silver badges8080 bronze badges

edited 9 hours ago

max66

45.4k77 gold badges4848 silver badges8080 bronze badges

asked 9 hours ago

SalgarSalgar

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asked 9 hours ago

SalgarSalgar

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6 Answers
6

active

oldest

votes

8

From C++11 you can already do that:

template <typename Functor = int(int)>
int run_new(int x, Functor func = [](int a) return a+1; ) 

 return func(x);

share|improve this answer

answered 9 hours ago

Biagio FestaBiagio Festa

6,46722 gold badges1414 silver badges4242 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  works only with non-capturing lambda, no? (not relevant for the question as asked)
  
  – formerlyknownas_463035818
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  @formerlyknownas_463035818: If you pass a capturing lambda, then the template function will deduce the type of this lambda, thus overriding the default type of int(int).
  
  – Nicol Bolas
  9 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  
  @formerlyknownas_463035818 I don't think you can capture anything in this context since the lambda is used as a default argument.
  
  – Biagio Festa
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  @formerlyknownas_463035818 You couldn't capture a global anyway. Capturing requires automatic storage.
  
  – Biagio Festa
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  Note that GCC and Clang are smart enough to inline the lambda call even though it gets converted into a function pointer. godbolt.org/z/nl4qEj
  
  – Brian
  9 hours ago
  
  

  
  
  
  

 | 
show 1 more comment

3

Just add an overload for this.

template <typename Functor>
int run_new(int x, Functor func) 

 return func(x);


int run_new(int x) 

 return run_new(x, [](int a) return a+1; );

Allows you to get around not beening able to have a lambda expression as a default function argument.

share|improve this answer

edited 9 hours ago

answered 9 hours ago

NathanOliverNathanOliver

114k1919 gold badges181181 silver badges260260 bronze badges

add a comment | 

1

Not quite "modern" but you could use plain old overloading with a non-template method taking only a single parameter:

int run_new(int x) 

 return func(x,[](int a) return a+1;); // calls the template overload

share|improve this answer

answered 9 hours ago

formerlyknownas_463035818formerlyknownas_463035818

24.2k44 gold badges3232 silver badges8181 bronze badges

add a comment | 

1

Just for fun, in C++20 we have both (1) lambdas in unevaluated contexts and (2) lambdas without capture are default constructible. Combine those two and you get:

template <typename Functor = decltype([](int a) return a+1; )>
int run_new(int x, Functor func = )

 return func(x);

Obligatory godbolt.

share|improve this answer

answered 9 hours ago

BarryBarry

199k2222 gold badges369369 silver badges660660 bronze badges

add a comment | 

0

Alternatively to other questions, you can even avoid templates:

int run_new(int x, std::function<int(int)> func = [](int a) return a + 1; )

 return func(x);

share|improve this answer

edited 9 hours ago

Biagio Festa

6,46722 gold badges1414 silver badges4242 bronze badges

answered 9 hours ago

Daniel LangrDaniel Langr

8,8032828 silver badges5353 bronze badges

• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  
  Do note that you do incur a cost for this. std::function uses type erasure so you have dynamic allocation and more than likely lose the ability to inline the code.
  
  – NathanOliver
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  @NathanOliver Both GCC and Clang seem to be smart enough to generate the same inlined assembly code for f, which is minimal: godbolt.org/z/uuEeNE. Need to admit I was surprised.
  
  – Daniel Langr
  9 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  
  @DanielLangr But now the functor has to be copyable
  
  – Artyer
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @Artyer That's the drawback. The advantage is that you don't need to expose the function definition in a header file.
  
  – Daniel Langr
  9 hours ago
  
  

  
  
  
  

add a comment | 

0

The best I can imagine pass through a variable

static constexpr auto defFunc = [](int a) return a+1; ;

template <typename Functor = decltype(defFunc)>
int run_new(int x, Functor func = defFunc) 

 return func(x);

share|improve this answer

answered 9 hours ago

max66max66

45.4k77 gold badges4848 silver badges8080 bronze badges

add a comment | 

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8

From C++11 you can already do that:

template <typename Functor = int(int)>
int run_new(int x, Functor func = [](int a) return a+1; ) 

 return func(x);

share|improve this answer

answered 9 hours ago

Biagio FestaBiagio Festa

6,46722 gold badges1414 silver badges4242 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  works only with non-capturing lambda, no? (not relevant for the question as asked)
  
  – formerlyknownas_463035818
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  @formerlyknownas_463035818: If you pass a capturing lambda, then the template function will deduce the type of this lambda, thus overriding the default type of int(int).
  
  – Nicol Bolas
  9 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  
  @formerlyknownas_463035818 I don't think you can capture anything in this context since the lambda is used as a default argument.
  
  – Biagio Festa
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  @formerlyknownas_463035818 You couldn't capture a global anyway. Capturing requires automatic storage.
  
  – Biagio Festa
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  Note that GCC and Clang are smart enough to inline the lambda call even though it gets converted into a function pointer. godbolt.org/z/nl4qEj
  
  – Brian
  9 hours ago
  
  

  
  
  
  

 | 
show 1 more comment

8

From C++11 you can already do that:

template <typename Functor = int(int)>
int run_new(int x, Functor func = [](int a) return a+1; ) 

 return func(x);

share|improve this answer

answered 9 hours ago

Biagio FestaBiagio Festa

6,46722 gold badges1414 silver badges4242 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  works only with non-capturing lambda, no? (not relevant for the question as asked)
  
  – formerlyknownas_463035818
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  @formerlyknownas_463035818: If you pass a capturing lambda, then the template function will deduce the type of this lambda, thus overriding the default type of int(int).
  
  – Nicol Bolas
  9 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  
  @formerlyknownas_463035818 I don't think you can capture anything in this context since the lambda is used as a default argument.
  
  – Biagio Festa
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  @formerlyknownas_463035818 You couldn't capture a global anyway. Capturing requires automatic storage.
  
  – Biagio Festa
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  Note that GCC and Clang are smart enough to inline the lambda call even though it gets converted into a function pointer. godbolt.org/z/nl4qEj
  
  – Brian
  9 hours ago
  
  

  
  
  
  

 | 
show 1 more comment

From C++11 you can already do that:

template <typename Functor = int(int)>
int run_new(int x, Functor func = [](int a) return a+1; ) 

 return func(x);

share|improve this answer

answered 9 hours ago

Biagio FestaBiagio Festa

6,46722 gold badges1414 silver badges4242 bronze badges

template <typename Functor = int(int)>
int run_new(int x, Functor func = [](int a) return a+1; ) 

 return func(x);

share|improve this answer

answered 9 hours ago

Biagio FestaBiagio Festa

6,46722 gold badges1414 silver badges4242 bronze badges

answered 9 hours ago

Biagio FestaBiagio Festa

6,46722 gold badges1414 silver badges4242 bronze badges

answered 9 hours ago

Biagio FestaBiagio Festa

6,46722 gold badges1414 silver badges4242 bronze badges

3

Just add an overload for this.

template <typename Functor>
int run_new(int x, Functor func) 

 return func(x);


int run_new(int x) 

 return run_new(x, [](int a) return a+1; );

Allows you to get around not beening able to have a lambda expression as a default function argument.

share|improve this answer

edited 9 hours ago

answered 9 hours ago

NathanOliverNathanOliver

114k1919 gold badges181181 silver badges260260 bronze badges

add a comment | 

3

Just add an overload for this.

template <typename Functor>
int run_new(int x, Functor func) 

 return func(x);


int run_new(int x) 

 return run_new(x, [](int a) return a+1; );

Allows you to get around not beening able to have a lambda expression as a default function argument.

share|improve this answer

edited 9 hours ago

answered 9 hours ago

NathanOliverNathanOliver

114k1919 gold badges181181 silver badges260260 bronze badges

add a comment | 

Just add an overload for this.

template <typename Functor>
int run_new(int x, Functor func) 

 return func(x);


int run_new(int x) 

 return run_new(x, [](int a) return a+1; );

Allows you to get around not beening able to have a lambda expression as a default function argument.

share|improve this answer

edited 9 hours ago

answered 9 hours ago

NathanOliverNathanOliver

114k1919 gold badges181181 silver badges260260 bronze badges

template <typename Functor>
int run_new(int x, Functor func) 

 return func(x);


int run_new(int x) 

 return run_new(x, [](int a) return a+1; );

share|improve this answer

edited 9 hours ago

answered 9 hours ago

NathanOliverNathanOliver

114k1919 gold badges181181 silver badges260260 bronze badges

answered 9 hours ago

NathanOliverNathanOliver

114k1919 gold badges181181 silver badges260260 bronze badges

answered 9 hours ago

NathanOliverNathanOliver

114k1919 gold badges181181 silver badges260260 bronze badges

1

Not quite "modern" but you could use plain old overloading with a non-template method taking only a single parameter:

int run_new(int x) 

 return func(x,[](int a) return a+1;); // calls the template overload

share|improve this answer

answered 9 hours ago

formerlyknownas_463035818formerlyknownas_463035818

24.2k44 gold badges3232 silver badges8181 bronze badges

add a comment | 

1

Not quite "modern" but you could use plain old overloading with a non-template method taking only a single parameter:

int run_new(int x) 

 return func(x,[](int a) return a+1;); // calls the template overload

share|improve this answer

answered 9 hours ago

formerlyknownas_463035818formerlyknownas_463035818

24.2k44 gold badges3232 silver badges8181 bronze badges

add a comment | 

Not quite "modern" but you could use plain old overloading with a non-template method taking only a single parameter:

int run_new(int x) 

 return func(x,[](int a) return a+1;); // calls the template overload

share|improve this answer

answered 9 hours ago

formerlyknownas_463035818formerlyknownas_463035818

24.2k44 gold badges3232 silver badges8181 bronze badges

int run_new(int x) 

 return func(x,[](int a) return a+1;); // calls the template overload

share|improve this answer

answered 9 hours ago

formerlyknownas_463035818formerlyknownas_463035818

24.2k44 gold badges3232 silver badges8181 bronze badges

answered 9 hours ago

formerlyknownas_463035818formerlyknownas_463035818

24.2k44 gold badges3232 silver badges8181 bronze badges

answered 9 hours ago

formerlyknownas_463035818formerlyknownas_463035818

24.2k44 gold badges3232 silver badges8181 bronze badges

1

Just for fun, in C++20 we have both (1) lambdas in unevaluated contexts and (2) lambdas without capture are default constructible. Combine those two and you get:

template <typename Functor = decltype([](int a) return a+1; )>
int run_new(int x, Functor func = )

 return func(x);

Obligatory godbolt.

share|improve this answer

answered 9 hours ago

BarryBarry

199k2222 gold badges369369 silver badges660660 bronze badges

add a comment | 

1

Just for fun, in C++20 we have both (1) lambdas in unevaluated contexts and (2) lambdas without capture are default constructible. Combine those two and you get:

template <typename Functor = decltype([](int a) return a+1; )>
int run_new(int x, Functor func = )

 return func(x);

Obligatory godbolt.

share|improve this answer

answered 9 hours ago

BarryBarry

199k2222 gold badges369369 silver badges660660 bronze badges

add a comment | 

Just for fun, in C++20 we have both (1) lambdas in unevaluated contexts and (2) lambdas without capture are default constructible. Combine those two and you get:

template <typename Functor = decltype([](int a) return a+1; )>
int run_new(int x, Functor func = )

 return func(x);

Obligatory godbolt.

share|improve this answer

answered 9 hours ago

BarryBarry

199k2222 gold badges369369 silver badges660660 bronze badges

template <typename Functor = decltype([](int a) return a+1; )>
int run_new(int x, Functor func = )

 return func(x);

share|improve this answer

answered 9 hours ago

BarryBarry

199k2222 gold badges369369 silver badges660660 bronze badges

answered 9 hours ago

BarryBarry

199k2222 gold badges369369 silver badges660660 bronze badges

answered 9 hours ago

BarryBarry

199k2222 gold badges369369 silver badges660660 bronze badges

0

Alternatively to other questions, you can even avoid templates:

int run_new(int x, std::function<int(int)> func = [](int a) return a + 1; )

 return func(x);

share|improve this answer

edited 9 hours ago

Biagio Festa

6,46722 gold badges1414 silver badges4242 bronze badges

answered 9 hours ago

Daniel LangrDaniel Langr

8,8032828 silver badges5353 bronze badges

• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  
  Do note that you do incur a cost for this. std::function uses type erasure so you have dynamic allocation and more than likely lose the ability to inline the code.
  
  – NathanOliver
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  @NathanOliver Both GCC and Clang seem to be smart enough to generate the same inlined assembly code for f, which is minimal: godbolt.org/z/uuEeNE. Need to admit I was surprised.
  
  – Daniel Langr
  9 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  
  @DanielLangr But now the functor has to be copyable
  
  – Artyer
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @Artyer That's the drawback. The advantage is that you don't need to expose the function definition in a header file.
  
  – Daniel Langr
  9 hours ago
  
  

  
  
  
  

add a comment | 

0

Alternatively to other questions, you can even avoid templates:

int run_new(int x, std::function<int(int)> func = [](int a) return a + 1; )

 return func(x);

share|improve this answer

edited 9 hours ago

Biagio Festa

6,46722 gold badges1414 silver badges4242 bronze badges

answered 9 hours ago

Daniel LangrDaniel Langr

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  Do note that you do incur a cost for this. std::function uses type erasure so you have dynamic allocation and more than likely lose the ability to inline the code.
  
  – NathanOliver
  9 hours ago
  

  
  
  
  


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  @NathanOliver Both GCC and Clang seem to be smart enough to generate the same inlined assembly code for f, which is minimal: godbolt.org/z/uuEeNE. Need to admit I was surprised.
  
  – Daniel Langr
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  @DanielLangr But now the functor has to be copyable
  
  – Artyer
  9 hours ago
  

  
  
  
  


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  @Artyer That's the drawback. The advantage is that you don't need to expose the function definition in a header file.
  
  – Daniel Langr
  9 hours ago
  
  

  
  
  
  

add a comment | 

Alternatively to other questions, you can even avoid templates:

int run_new(int x, std::function<int(int)> func = [](int a) return a + 1; )

 return func(x);

share|improve this answer

edited 9 hours ago

Biagio Festa

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answered 9 hours ago

Daniel LangrDaniel Langr

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int run_new(int x, std::function<int(int)> func = [](int a) return a + 1; )

 return func(x);

share|improve this answer

edited 9 hours ago

Biagio Festa

6,46722 gold badges1414 silver badges4242 bronze badges

answered 9 hours ago

Daniel LangrDaniel Langr

8,8032828 silver badges5353 bronze badges

edited 9 hours ago

Biagio Festa

6,46722 gold badges1414 silver badges4242 bronze badges

edited 9 hours ago

Biagio Festa

6,46722 gold badges1414 silver badges4242 bronze badges

answered 9 hours ago

Daniel LangrDaniel Langr

8,8032828 silver badges5353 bronze badges

answered 9 hours ago

Daniel LangrDaniel Langr

8,8032828 silver badges5353 bronze badges

0

The best I can imagine pass through a variable

static constexpr auto defFunc = [](int a) return a+1; ;

template <typename Functor = decltype(defFunc)>
int run_new(int x, Functor func = defFunc) 

 return func(x);

share|improve this answer

answered 9 hours ago

max66max66

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add a comment | 

0

The best I can imagine pass through a variable

static constexpr auto defFunc = [](int a) return a+1; ;

template <typename Functor = decltype(defFunc)>
int run_new(int x, Functor func = defFunc) 

 return func(x);

share|improve this answer

answered 9 hours ago

max66max66

45.4k77 gold badges4848 silver badges8080 bronze badges

add a comment | 

The best I can imagine pass through a variable

static constexpr auto defFunc = [](int a) return a+1; ;

template <typename Functor = decltype(defFunc)>
int run_new(int x, Functor func = defFunc) 

 return func(x);

share|improve this answer

answered 9 hours ago

max66max66

45.4k77 gold badges4848 silver badges8080 bronze badges

static constexpr auto defFunc = [](int a) return a+1; ;

template <typename Functor = decltype(defFunc)>
int run_new(int x, Functor func = defFunc) 

 return func(x);

share|improve this answer

answered 9 hours ago

max66max66

45.4k77 gold badges4848 silver badges8080 bronze badges

answered 9 hours ago

max66max66

45.4k77 gold badges4848 silver badges8080 bronze badges