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Generic lambda vs generic function give different behaviour



Unicorn Meta Zoo #1: Why another podcast?
Announcing the arrival of Valued Associate #679: Cesar Manara
Data science time! April 2019 and salary with experience
The Ask Question Wizard is Live!What is a lambda (function)?What is the difference between #include <filename> and #include “filename”?What are the differences between a pointer variable and a reference variable in C++?What is the difference between a 'closure' and a 'lambda'?Why are Python lambdas useful?Distinct() with lambda?list comprehension vs. lambda + filterWhat is a lambda expression in C++11?Calling `this` member function from generic lambda - clang vs gccConstructing std::function argument from lambda



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14















Take following code as an example



#include <algorithm>

namespace baz
template<class T>
void sort(T&&)


namespace boot
const auto sort = [](auto &&);


void foo ()
using namespace std;
using namespace baz;
sort(1);


void bar()
using namespace std;
using namespace boot;
sort(1);



I expected that since foo compiled, then bar shall compile as well. To my surprise, the foo compiles correctly and bar has problem with ambiguous call to sort function. Am I doing something illegal here or this is proper way compiler should behave? If so, why is it so different. I though generic lambda can be treated as syntactic sugar for generic function.



live example










share|improve this question



















  • 5





    Lambdas do not participate in ADL

    – Guillaume Racicot
    Mar 26 at 22:50






  • 10





    This isn't ADL. An int argument doesn't come from any namespace.

    – chris
    Mar 26 at 22:53






  • 4





    Should this really be ambiguous, though? std::sort() doesn't take 1 parameter as input, it takes at least 2, so why is the compiler even considering it as a candidate for a call that passes only 1 parameter value?

    – Remy Lebeau
    Mar 26 at 23:15












  • There must be something about the extra layer of indirection that the lambda introduces. With the first example, the call is made to ::baz::sort, but in the second example, it would have to find ::boot::mystery_lambda_type::operator(). That extra step might be what causes std::sort to be considered first. I don't have the standard in front of me so can't be sure about this.

    – alter igel
    Mar 26 at 23:22

















14















Take following code as an example



#include <algorithm>

namespace baz
template<class T>
void sort(T&&)


namespace boot
const auto sort = [](auto &&);


void foo ()
using namespace std;
using namespace baz;
sort(1);


void bar()
using namespace std;
using namespace boot;
sort(1);



I expected that since foo compiled, then bar shall compile as well. To my surprise, the foo compiles correctly and bar has problem with ambiguous call to sort function. Am I doing something illegal here or this is proper way compiler should behave? If so, why is it so different. I though generic lambda can be treated as syntactic sugar for generic function.



live example










share|improve this question



















  • 5





    Lambdas do not participate in ADL

    – Guillaume Racicot
    Mar 26 at 22:50






  • 10





    This isn't ADL. An int argument doesn't come from any namespace.

    – chris
    Mar 26 at 22:53






  • 4





    Should this really be ambiguous, though? std::sort() doesn't take 1 parameter as input, it takes at least 2, so why is the compiler even considering it as a candidate for a call that passes only 1 parameter value?

    – Remy Lebeau
    Mar 26 at 23:15












  • There must be something about the extra layer of indirection that the lambda introduces. With the first example, the call is made to ::baz::sort, but in the second example, it would have to find ::boot::mystery_lambda_type::operator(). That extra step might be what causes std::sort to be considered first. I don't have the standard in front of me so can't be sure about this.

    – alter igel
    Mar 26 at 23:22













14












14








14


1






Take following code as an example



#include <algorithm>

namespace baz
template<class T>
void sort(T&&)


namespace boot
const auto sort = [](auto &&);


void foo ()
using namespace std;
using namespace baz;
sort(1);


void bar()
using namespace std;
using namespace boot;
sort(1);



I expected that since foo compiled, then bar shall compile as well. To my surprise, the foo compiles correctly and bar has problem with ambiguous call to sort function. Am I doing something illegal here or this is proper way compiler should behave? If so, why is it so different. I though generic lambda can be treated as syntactic sugar for generic function.



live example










share|improve this question
















Take following code as an example



#include <algorithm>

namespace baz
template<class T>
void sort(T&&)


namespace boot
const auto sort = [](auto &&);


void foo ()
using namespace std;
using namespace baz;
sort(1);


void bar()
using namespace std;
using namespace boot;
sort(1);



I expected that since foo compiled, then bar shall compile as well. To my surprise, the foo compiles correctly and bar has problem with ambiguous call to sort function. Am I doing something illegal here or this is proper way compiler should behave? If so, why is it so different. I though generic lambda can be treated as syntactic sugar for generic function.



live example







c++ lambda c++14






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Mar 26 at 22:57







bartop

















asked Mar 26 at 22:47









bartopbartop

3,5031132




3,5031132







  • 5





    Lambdas do not participate in ADL

    – Guillaume Racicot
    Mar 26 at 22:50






  • 10





    This isn't ADL. An int argument doesn't come from any namespace.

    – chris
    Mar 26 at 22:53






  • 4





    Should this really be ambiguous, though? std::sort() doesn't take 1 parameter as input, it takes at least 2, so why is the compiler even considering it as a candidate for a call that passes only 1 parameter value?

    – Remy Lebeau
    Mar 26 at 23:15












  • There must be something about the extra layer of indirection that the lambda introduces. With the first example, the call is made to ::baz::sort, but in the second example, it would have to find ::boot::mystery_lambda_type::operator(). That extra step might be what causes std::sort to be considered first. I don't have the standard in front of me so can't be sure about this.

    – alter igel
    Mar 26 at 23:22












  • 5





    Lambdas do not participate in ADL

    – Guillaume Racicot
    Mar 26 at 22:50






  • 10





    This isn't ADL. An int argument doesn't come from any namespace.

    – chris
    Mar 26 at 22:53






  • 4





    Should this really be ambiguous, though? std::sort() doesn't take 1 parameter as input, it takes at least 2, so why is the compiler even considering it as a candidate for a call that passes only 1 parameter value?

    – Remy Lebeau
    Mar 26 at 23:15












  • There must be something about the extra layer of indirection that the lambda introduces. With the first example, the call is made to ::baz::sort, but in the second example, it would have to find ::boot::mystery_lambda_type::operator(). That extra step might be what causes std::sort to be considered first. I don't have the standard in front of me so can't be sure about this.

    – alter igel
    Mar 26 at 23:22







5




5





Lambdas do not participate in ADL

– Guillaume Racicot
Mar 26 at 22:50





Lambdas do not participate in ADL

– Guillaume Racicot
Mar 26 at 22:50




10




10





This isn't ADL. An int argument doesn't come from any namespace.

– chris
Mar 26 at 22:53





This isn't ADL. An int argument doesn't come from any namespace.

– chris
Mar 26 at 22:53




4




4





Should this really be ambiguous, though? std::sort() doesn't take 1 parameter as input, it takes at least 2, so why is the compiler even considering it as a candidate for a call that passes only 1 parameter value?

– Remy Lebeau
Mar 26 at 23:15






Should this really be ambiguous, though? std::sort() doesn't take 1 parameter as input, it takes at least 2, so why is the compiler even considering it as a candidate for a call that passes only 1 parameter value?

– Remy Lebeau
Mar 26 at 23:15














There must be something about the extra layer of indirection that the lambda introduces. With the first example, the call is made to ::baz::sort, but in the second example, it would have to find ::boot::mystery_lambda_type::operator(). That extra step might be what causes std::sort to be considered first. I don't have the standard in front of me so can't be sure about this.

– alter igel
Mar 26 at 23:22





There must be something about the extra layer of indirection that the lambda introduces. With the first example, the call is made to ::baz::sort, but in the second example, it would have to find ::boot::mystery_lambda_type::operator(). That extra step might be what causes std::sort to be considered first. I don't have the standard in front of me so can't be sure about this.

– alter igel
Mar 26 at 23:22












1 Answer
1






active

oldest

votes


















16














The problem here is not that the call to sort is ambiguous, but that the name sort is ambiguous. Name lookup happens before overload resolution.



I believe the relevant section is [basic.lookup]/1, specifically




[…] The declarations found by name lookup shall either all denote the same entity or shall all denote functions or function templates; in the latter case, the declarations are said to form a set of overloaded functions ([over.load]). […]




In your case, the name sort denotes both, the object boot::sort as well as the set of overloaded functions std::sort. Therefore, name lookup fails.



Your code is really no different from if you had written, for example



namespace baz 
int a;


namespace boot
int a;


void foo()
using namespace baz;
using namespace boot;
a = 42; // error: reference to 'a' is ambiguous



Try it out here; compare this to a case that actually has an ambiguous function call; note how the error message is the same as in your case, specifically referring to the name itself being ambiguous rather than the function call.






share|improve this answer




















  • 1





    I think this is actually the right answer. And I would like to add that if both the template function sort and the lambda sort were declared in the same namespace, it would be an error. You cannot have a function and non-function with the same name in the same namespace. So there could never be an overload set that has both true functions and function-like objects.

    – Mike
    Mar 27 at 1:09











  • Seems like a right answet to me. If there is something that can be done to workaround my problem I would be thankful for comment/answer edit.

    – bartop
    Mar 27 at 7:05






  • 1





    @Scheff indeed, I must've mixed up the URLs somehow. Should be fixed now. Thanks for pointing that out!

    – Michael Kenzel
    Mar 27 at 10:08












  • @bartop can you not just remove the using directives and/or use fully-qualified names, e.g., baz::sort!?

    – Michael Kenzel
    Mar 27 at 11:31












  • @MichaelKenzel Sure can, and I do it currently. Even though I still wonder if there is different way to avoid this ambiguity

    – bartop
    Mar 27 at 11:49











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









16














The problem here is not that the call to sort is ambiguous, but that the name sort is ambiguous. Name lookup happens before overload resolution.



I believe the relevant section is [basic.lookup]/1, specifically




[…] The declarations found by name lookup shall either all denote the same entity or shall all denote functions or function templates; in the latter case, the declarations are said to form a set of overloaded functions ([over.load]). […]




In your case, the name sort denotes both, the object boot::sort as well as the set of overloaded functions std::sort. Therefore, name lookup fails.



Your code is really no different from if you had written, for example



namespace baz 
int a;


namespace boot
int a;


void foo()
using namespace baz;
using namespace boot;
a = 42; // error: reference to 'a' is ambiguous



Try it out here; compare this to a case that actually has an ambiguous function call; note how the error message is the same as in your case, specifically referring to the name itself being ambiguous rather than the function call.






share|improve this answer




















  • 1





    I think this is actually the right answer. And I would like to add that if both the template function sort and the lambda sort were declared in the same namespace, it would be an error. You cannot have a function and non-function with the same name in the same namespace. So there could never be an overload set that has both true functions and function-like objects.

    – Mike
    Mar 27 at 1:09











  • Seems like a right answet to me. If there is something that can be done to workaround my problem I would be thankful for comment/answer edit.

    – bartop
    Mar 27 at 7:05






  • 1





    @Scheff indeed, I must've mixed up the URLs somehow. Should be fixed now. Thanks for pointing that out!

    – Michael Kenzel
    Mar 27 at 10:08












  • @bartop can you not just remove the using directives and/or use fully-qualified names, e.g., baz::sort!?

    – Michael Kenzel
    Mar 27 at 11:31












  • @MichaelKenzel Sure can, and I do it currently. Even though I still wonder if there is different way to avoid this ambiguity

    – bartop
    Mar 27 at 11:49















16














The problem here is not that the call to sort is ambiguous, but that the name sort is ambiguous. Name lookup happens before overload resolution.



I believe the relevant section is [basic.lookup]/1, specifically




[…] The declarations found by name lookup shall either all denote the same entity or shall all denote functions or function templates; in the latter case, the declarations are said to form a set of overloaded functions ([over.load]). […]




In your case, the name sort denotes both, the object boot::sort as well as the set of overloaded functions std::sort. Therefore, name lookup fails.



Your code is really no different from if you had written, for example



namespace baz 
int a;


namespace boot
int a;


void foo()
using namespace baz;
using namespace boot;
a = 42; // error: reference to 'a' is ambiguous



Try it out here; compare this to a case that actually has an ambiguous function call; note how the error message is the same as in your case, specifically referring to the name itself being ambiguous rather than the function call.






share|improve this answer




















  • 1





    I think this is actually the right answer. And I would like to add that if both the template function sort and the lambda sort were declared in the same namespace, it would be an error. You cannot have a function and non-function with the same name in the same namespace. So there could never be an overload set that has both true functions and function-like objects.

    – Mike
    Mar 27 at 1:09











  • Seems like a right answet to me. If there is something that can be done to workaround my problem I would be thankful for comment/answer edit.

    – bartop
    Mar 27 at 7:05






  • 1





    @Scheff indeed, I must've mixed up the URLs somehow. Should be fixed now. Thanks for pointing that out!

    – Michael Kenzel
    Mar 27 at 10:08












  • @bartop can you not just remove the using directives and/or use fully-qualified names, e.g., baz::sort!?

    – Michael Kenzel
    Mar 27 at 11:31












  • @MichaelKenzel Sure can, and I do it currently. Even though I still wonder if there is different way to avoid this ambiguity

    – bartop
    Mar 27 at 11:49













16












16








16







The problem here is not that the call to sort is ambiguous, but that the name sort is ambiguous. Name lookup happens before overload resolution.



I believe the relevant section is [basic.lookup]/1, specifically




[…] The declarations found by name lookup shall either all denote the same entity or shall all denote functions or function templates; in the latter case, the declarations are said to form a set of overloaded functions ([over.load]). […]




In your case, the name sort denotes both, the object boot::sort as well as the set of overloaded functions std::sort. Therefore, name lookup fails.



Your code is really no different from if you had written, for example



namespace baz 
int a;


namespace boot
int a;


void foo()
using namespace baz;
using namespace boot;
a = 42; // error: reference to 'a' is ambiguous



Try it out here; compare this to a case that actually has an ambiguous function call; note how the error message is the same as in your case, specifically referring to the name itself being ambiguous rather than the function call.






share|improve this answer















The problem here is not that the call to sort is ambiguous, but that the name sort is ambiguous. Name lookup happens before overload resolution.



I believe the relevant section is [basic.lookup]/1, specifically




[…] The declarations found by name lookup shall either all denote the same entity or shall all denote functions or function templates; in the latter case, the declarations are said to form a set of overloaded functions ([over.load]). […]




In your case, the name sort denotes both, the object boot::sort as well as the set of overloaded functions std::sort. Therefore, name lookup fails.



Your code is really no different from if you had written, for example



namespace baz 
int a;


namespace boot
int a;


void foo()
using namespace baz;
using namespace boot;
a = 42; // error: reference to 'a' is ambiguous



Try it out here; compare this to a case that actually has an ambiguous function call; note how the error message is the same as in your case, specifically referring to the name itself being ambiguous rather than the function call.







share|improve this answer














share|improve this answer



share|improve this answer








edited Mar 27 at 10:06

























answered Mar 27 at 0:55









Michael KenzelMichael Kenzel

8,39311424




8,39311424







  • 1





    I think this is actually the right answer. And I would like to add that if both the template function sort and the lambda sort were declared in the same namespace, it would be an error. You cannot have a function and non-function with the same name in the same namespace. So there could never be an overload set that has both true functions and function-like objects.

    – Mike
    Mar 27 at 1:09











  • Seems like a right answet to me. If there is something that can be done to workaround my problem I would be thankful for comment/answer edit.

    – bartop
    Mar 27 at 7:05






  • 1





    @Scheff indeed, I must've mixed up the URLs somehow. Should be fixed now. Thanks for pointing that out!

    – Michael Kenzel
    Mar 27 at 10:08












  • @bartop can you not just remove the using directives and/or use fully-qualified names, e.g., baz::sort!?

    – Michael Kenzel
    Mar 27 at 11:31












  • @MichaelKenzel Sure can, and I do it currently. Even though I still wonder if there is different way to avoid this ambiguity

    – bartop
    Mar 27 at 11:49












  • 1





    I think this is actually the right answer. And I would like to add that if both the template function sort and the lambda sort were declared in the same namespace, it would be an error. You cannot have a function and non-function with the same name in the same namespace. So there could never be an overload set that has both true functions and function-like objects.

    – Mike
    Mar 27 at 1:09











  • Seems like a right answet to me. If there is something that can be done to workaround my problem I would be thankful for comment/answer edit.

    – bartop
    Mar 27 at 7:05






  • 1





    @Scheff indeed, I must've mixed up the URLs somehow. Should be fixed now. Thanks for pointing that out!

    – Michael Kenzel
    Mar 27 at 10:08












  • @bartop can you not just remove the using directives and/or use fully-qualified names, e.g., baz::sort!?

    – Michael Kenzel
    Mar 27 at 11:31












  • @MichaelKenzel Sure can, and I do it currently. Even though I still wonder if there is different way to avoid this ambiguity

    – bartop
    Mar 27 at 11:49







1




1





I think this is actually the right answer. And I would like to add that if both the template function sort and the lambda sort were declared in the same namespace, it would be an error. You cannot have a function and non-function with the same name in the same namespace. So there could never be an overload set that has both true functions and function-like objects.

– Mike
Mar 27 at 1:09





I think this is actually the right answer. And I would like to add that if both the template function sort and the lambda sort were declared in the same namespace, it would be an error. You cannot have a function and non-function with the same name in the same namespace. So there could never be an overload set that has both true functions and function-like objects.

– Mike
Mar 27 at 1:09













Seems like a right answet to me. If there is something that can be done to workaround my problem I would be thankful for comment/answer edit.

– bartop
Mar 27 at 7:05





Seems like a right answet to me. If there is something that can be done to workaround my problem I would be thankful for comment/answer edit.

– bartop
Mar 27 at 7:05




1




1





@Scheff indeed, I must've mixed up the URLs somehow. Should be fixed now. Thanks for pointing that out!

– Michael Kenzel
Mar 27 at 10:08






@Scheff indeed, I must've mixed up the URLs somehow. Should be fixed now. Thanks for pointing that out!

– Michael Kenzel
Mar 27 at 10:08














@bartop can you not just remove the using directives and/or use fully-qualified names, e.g., baz::sort!?

– Michael Kenzel
Mar 27 at 11:31






@bartop can you not just remove the using directives and/or use fully-qualified names, e.g., baz::sort!?

– Michael Kenzel
Mar 27 at 11:31














@MichaelKenzel Sure can, and I do it currently. Even though I still wonder if there is different way to avoid this ambiguity

– bartop
Mar 27 at 11:49





@MichaelKenzel Sure can, and I do it currently. Even though I still wonder if there is different way to avoid this ambiguity

– bartop
Mar 27 at 11:49



















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Hall Of Fame””Slayer Wins 'Best Metal' Grammy Award””Slayer Guitarist Jeff Hanneman Dies””Bullet-For My Valentine booed at Metal Hammer Golden Gods Awards””Unholy Aliance””The End Of Slayer?””Slayer: We Could Thrash Out Two More Albums If We're Fast Enough...””'The Unholy Alliance: Chapter III' UK Dates Added”originalet”Megadeth And Slayer To Co-Headline 'Canadian Carnage' Trek”originalet”World Painted Blood””Release “World Painted Blood” by Slayer””Metallica Heading To Cinemas””Slayer, Megadeth To Join Forces For 'European Carnage' Tour - Dec. 18, 2010”originalet”Slayer's Hanneman Contracts Acute Infection; Band To Bring In Guest Guitarist””Cannibal Corpse's Pat O'Brien Will Step In As Slayer's Guest Guitarist”originalet”Slayer’s Jeff Hanneman Dead at 49””Dave Lombardo Says He Made Only $67,000 In 2011 While Touring With Slayer””Slayer: We Do Not Agree With Dave Lombardo's Substance Or Timeline Of Events””Slayer Welcomes Drummer Paul Bostaph Back To The Fold””Slayer Hope to Unveil Never-Before-Heard Jeff Hanneman Material on Next Album””Slayer Debut New Song 'Implode' During Surprise Golden Gods Appearance””Release group Repentless by Slayer””Repentless - Slayer - Credits””Slayer””Metal Storm Awards 2015””Slayer - to release comic book "Repentless #1"””Slayer To Release 'Repentless' 6.66" Vinyl Box Set””BREAKING NEWS: Slayer Announce Farewell Tour””Slayer Recruit Lamb of God, Anthrax, Behemoth + Testament for Final Tour””Slayer lägger ner efter 37 år””Slayer Announces Second North American Leg Of 'Final' Tour””Final World Tour””Slayer Announces Final European Tour With Lamb of God, Anthrax And Obituary””Slayer To Tour Europe With Lamb of God, Anthrax And Obituary””Slayer To Play 'Last French Show Ever' At Next Year's Hellfst””Slayer's Final World Tour Will Extend Into 2019””Death Angel's Rob Cavestany On Slayer's 'Farewell' Tour: 'Some Of Us Could See This Coming'””Testament Has No Plans To Retire Anytime Soon, Says Chuck Billy””Anthrax's Scott Ian On Slayer's 'Farewell' Tour Plans: 'I Was Surprised And I Wasn't Surprised'””Slayer””Slayer's Morbid Schlock””Review/Rock; For Slayer, the Mania Is the Message””Slayer - Biography””Slayer - Reign In Blood”originalet”Dave Lombardo””An exclusive oral history of Slayer”originalet”Exclusive! Interview With Slayer Guitarist Jeff Hanneman”originalet”Thinking Out Loud: Slayer's Kerry King on hair metal, Satan and being polite””Slayer Lyrics””Slayer - Biography””Most influential artists for extreme metal music””Slayer - Reign in Blood””Slayer guitarist Jeff Hanneman dies aged 49””Slatanic Slaughter: A Tribute to Slayer””Gateway to Hell: A Tribute to Slayer””Covered In Blood””Slayer: The Origins of Thrash in San Francisco, CA.””Why They Rule - #6 Slayer”originalet”Guitar World's 100 Greatest Heavy Metal Guitarists Of All Time”originalet”The fans have spoken: Slayer comes out on top in readers' polls”originalet”Tribute to Jeff Hanneman (1964-2013)””Lamb Of God Frontman: We Sound Like A Slayer Rip-Off””BEHEMOTH Frontman Pays Tribute To SLAYER's JEFF HANNEMAN””Slayer, Hatebreed Doing Double Duty On This Year's Ozzfest””System of a Down””Lacuna Coil’s Andrea Ferro Talks Influences, Skateboarding, Band Origins + More””Slayer - Reign in Blood””Into The Lungs of Hell””Slayer rules - en utställning om fans””Slayer and Their Fans Slashed Through a No-Holds-Barred Night at Gas Monkey””Home””Slayer””Gold & Platinum - The Big 4 Live from Sofia, Bulgaria””Exclusive! Interview With Slayer Guitarist Kerry King””2008-02-23: Wiltern, Los Angeles, CA, USA””Slayer's Kerry King To Perform With Megadeth Tonight! - Oct. 21, 2010”originalet”Dave Lombardo - Biography”Slayer Case DismissedArkiveradUltimate Classic Rock: Slayer guitarist Jeff Hanneman dead at 49.”Slayer: "We could never do any thing like Some Kind Of Monster..."””Cannibal Corpse'S Pat O'Brien Will Step In As Slayer'S Guest Guitarist | The Official Slayer Site”originalet”Slayer Wins 'Best Metal' Grammy Award””Slayer Guitarist Jeff Hanneman Dies””Kerrang! Awards 2006 Blog: Kerrang! Hall Of Fame””Kerrang! Awards 2013: Kerrang! Legend”originalet”Metallica, Slayer, Iron Maien Among Winners At Metal Hammer Awards””Metal Hammer Golden Gods Awards””Bullet For My Valentine Booed At Metal Hammer Golden Gods Awards””Metal Storm Awards 2006””Metal Storm Awards 2015””Slayer's Concert History””Slayer - Relationships””Slayer - Releases”Slayers officiella webbplatsSlayer på MusicBrainzOfficiell webbplatsSlayerSlayerr1373445760000 0001 1540 47353068615-5086262726cb13906545x(data)6033143kn20030215029