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Lambda functions with template parameters, not in function parameters

What is a lambda (function)?How do I test a private function or a class that has private methods, fields or inner classes?Why can templates only be implemented in the header file?Where and why do I have to put the “template” and “typename” keywords?Distinct() with lambda?Can lambda functions be templated?What is a lambda expression in C++11?Replacing a 32-bit loop counter with 64-bit introduces crazy performance deviations

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6

2

Why does the first call not compile?

auto get1 = []<int B>() return B; ;
auto get2 = []<typename B>(B b) return b; ;

int main()

 get1<5>(); // error: no match for operator<
 get2(5); // ok

The reason I use this, is an expression repeated many times in code.

Of course I can use a real function function, but just I am curious WHY.

c++ templates lambda c++20

share|improve this question

edited 9 hours ago

Barry

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6

2

Why does the first call not compile?

auto get1 = []<int B>() return B; ;
auto get2 = []<typename B>(B b) return b; ;

int main()

 get1<5>(); // error: no match for operator<
 get2(5); // ok

The reason I use this, is an expression repeated many times in code.

Of course I can use a real function function, but just I am curious WHY.

c++ templates lambda c++20

share|improve this question

edited 9 hours ago

Barry

200k2222 gold badges375375 silver badges669669 bronze badges

asked 9 hours ago

ChameleonChameleon

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add a comment
 | 

Why does the first call not compile?

auto get1 = []<int B>() return B; ;
auto get2 = []<typename B>(B b) return b; ;

int main()

 get1<5>(); // error: no match for operator<
 get2(5); // ok

The reason I use this, is an expression repeated many times in code.

Of course I can use a real function function, but just I am curious WHY.

c++ templates lambda c++20

share|improve this question

edited 9 hours ago

Barry

200k2222 gold badges375375 silver badges669669 bronze badges

asked 9 hours ago

ChameleonChameleon

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auto get1 = []<int B>() return B; ;
auto get2 = []<typename B>(B b) return b; ;

int main()

 get1<5>(); // error: no match for operator<
 get2(5); // ok

share|improve this question

edited 9 hours ago

Barry

200k2222 gold badges375375 silver badges669669 bronze badges

asked 9 hours ago

ChameleonChameleon

61911 gold badge55 silver badges1313 bronze badges

share|improve this question

edited 9 hours ago

Barry

200k2222 gold badges375375 silver badges669669 bronze badges

asked 9 hours ago

ChameleonChameleon

61911 gold badge55 silver badges1313 bronze badges

edited 9 hours ago

Barry

200k2222 gold badges375375 silver badges669669 bronze badges

edited 9 hours ago

Barry

200k2222 gold badges375375 silver badges669669 bronze badges

asked 9 hours ago

ChameleonChameleon

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asked 9 hours ago

ChameleonChameleon

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1 Answer
1

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10

This is easier to understand if you consider what the equivalent class type looks like to your get1:

struct get1_t 
 template <int B> operator()() const return B; 
;

get1_t get1;

get1<5>(); // error

You're trying to provide an explicit template parameter to the call operator, but syntactically you're doing what looks like providing template parameters for get1 itself (i.e. as if get1 were a variable template). In order to provide the template parameter for the call operator, you have to do that directly:

get1.operator()<5>(); // ok

Or restructure the call operator to take something deducible:

template <int B> struct constant ;
get1(constant<5>);

---

Or restructure the whole thing to actually be the variable template that it looks like it is:

template <int B>
auto get1 = [] return B; ;

Now, get1<5> is itself a lambda, that you're invoking. That is, rather than a lambda with a call operator template we have a variable template lambda that is itself not a template.

share|improve this answer

edited 8 hours ago

T.C.

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answered 9 hours ago

BarryBarry

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10

This is easier to understand if you consider what the equivalent class type looks like to your get1:

struct get1_t 
 template <int B> operator()() const return B; 
;

get1_t get1;

get1<5>(); // error

You're trying to provide an explicit template parameter to the call operator, but syntactically you're doing what looks like providing template parameters for get1 itself (i.e. as if get1 were a variable template). In order to provide the template parameter for the call operator, you have to do that directly:

get1.operator()<5>(); // ok

Or restructure the call operator to take something deducible:

template <int B> struct constant ;
get1(constant<5>);

---

Or restructure the whole thing to actually be the variable template that it looks like it is:

template <int B>
auto get1 = [] return B; ;

Now, get1<5> is itself a lambda, that you're invoking. That is, rather than a lambda with a call operator template we have a variable template lambda that is itself not a template.

share|improve this answer

edited 8 hours ago

T.C.

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answered 9 hours ago

BarryBarry

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add a comment
 | 

10

This is easier to understand if you consider what the equivalent class type looks like to your get1:

struct get1_t 
 template <int B> operator()() const return B; 
;

get1_t get1;

get1<5>(); // error

You're trying to provide an explicit template parameter to the call operator, but syntactically you're doing what looks like providing template parameters for get1 itself (i.e. as if get1 were a variable template). In order to provide the template parameter for the call operator, you have to do that directly:

get1.operator()<5>(); // ok

Or restructure the call operator to take something deducible:

template <int B> struct constant ;
get1(constant<5>);

---

Or restructure the whole thing to actually be the variable template that it looks like it is:

template <int B>
auto get1 = [] return B; ;

Now, get1<5> is itself a lambda, that you're invoking. That is, rather than a lambda with a call operator template we have a variable template lambda that is itself not a template.

share|improve this answer

edited 8 hours ago

T.C.

111k1414 gold badges229229 silver badges338338 bronze badges

answered 9 hours ago

BarryBarry

200k2222 gold badges375375 silver badges669669 bronze badges

add a comment
 | 

This is easier to understand if you consider what the equivalent class type looks like to your get1:

struct get1_t 
 template <int B> operator()() const return B; 
;

get1_t get1;

get1<5>(); // error

You're trying to provide an explicit template parameter to the call operator, but syntactically you're doing what looks like providing template parameters for get1 itself (i.e. as if get1 were a variable template). In order to provide the template parameter for the call operator, you have to do that directly:

get1.operator()<5>(); // ok

Or restructure the call operator to take something deducible:

template <int B> struct constant ;
get1(constant<5>);

---

Or restructure the whole thing to actually be the variable template that it looks like it is:

template <int B>
auto get1 = [] return B; ;

Now, get1<5> is itself a lambda, that you're invoking. That is, rather than a lambda with a call operator template we have a variable template lambda that is itself not a template.

share|improve this answer

edited 8 hours ago

T.C.

111k1414 gold badges229229 silver badges338338 bronze badges

answered 9 hours ago

BarryBarry

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struct get1_t 
 template <int B> operator()() const return B; 
;

get1_t get1;

get1<5>(); // error

get1.operator()<5>(); // ok

template <int B> struct constant ;
get1(constant<5>);

template <int B>
auto get1 = [] return B; ;

share|improve this answer

edited 8 hours ago

T.C.

111k1414 gold badges229229 silver badges338338 bronze badges

answered 9 hours ago

BarryBarry

200k2222 gold badges375375 silver badges669669 bronze badges

edited 8 hours ago

T.C.

111k1414 gold badges229229 silver badges338338 bronze badges

edited 8 hours ago

T.C.

111k1414 gold badges229229 silver badges338338 bronze badges

answered 9 hours ago

BarryBarry

200k2222 gold badges375375 silver badges669669 bronze badges

answered 9 hours ago

BarryBarry

200k2222 gold badges375375 silver badges669669 bronze badges