Generic lambda vs generic function give different behaviourWhat is a lambda (function)?What is the difference...

Is a roofing delivery truck likely to crack my driveway slab?

Is there a good way to store credentials outside of a password manager?

Where in the Bible does the greeting ("Dominus Vobiscum") used at Mass come from?

Best way to store options for panels

What will be the benefits of Brexit?

Coordinate position not precise

How do I rename a LINUX host without needing to reboot for the rename to take effect?

Products and sum of cubes in Fibonacci

How can a jailer prevent the Forge Cleric's Artisan's Blessing from being used?

How could Frankenstein get the parts for his _second_ creature?

Have I saved too much for retirement so far?

Finding all intervals that match predicate in vector

Implement the Thanos sorting algorithm

Is there any reason not to eat food that's been dropped on the surface of the moon?

Using parameter substitution on a Bash array

How can I replace every global instance of "x[2]" with "x_2"

Opposite of a diet

Is there a problem with hiding "forgot password" until it's needed?

Hostile work environment after whistle-blowing on coworker and our boss. What do I do?

Can criminal fraud exist without damages?

Can somebody explain Brexit in a few child-proof sentences?

Bash method for viewing beginning and end of file

Should my PhD thesis be submitted under my legal name?

Why did Kant, Hegel, and Adorno leave some words and phrases in the Greek alphabet?



Generic lambda vs generic function give different behaviour


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













9















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




















  • 4





    Lambdas do not participate in ADL

    – Guillaume Racicot
    2 hours ago






  • 5





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

    – chris
    2 hours ago






  • 2





    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
    2 hours ago













  • 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
    1 hour ago
















9















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




















  • 4





    Lambdas do not participate in ADL

    – Guillaume Racicot
    2 hours ago






  • 5





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

    – chris
    2 hours ago






  • 2





    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
    2 hours ago













  • 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
    1 hour ago














9












9








9


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 2 hours ago







bartop

















asked 2 hours ago









bartopbartop

3,2331030




3,2331030








  • 4





    Lambdas do not participate in ADL

    – Guillaume Racicot
    2 hours ago






  • 5





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

    – chris
    2 hours ago






  • 2





    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
    2 hours ago













  • 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
    1 hour ago














  • 4





    Lambdas do not participate in ADL

    – Guillaume Racicot
    2 hours ago






  • 5





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

    – chris
    2 hours ago






  • 2





    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
    2 hours ago













  • 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
    1 hour ago








4




4





Lambdas do not participate in ADL

– Guillaume Racicot
2 hours ago





Lambdas do not participate in ADL

– Guillaume Racicot
2 hours ago




5




5





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

– chris
2 hours ago





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

– chris
2 hours ago




2




2





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
2 hours ago







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
2 hours ago















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
1 hour ago





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
1 hour ago












1 Answer
1






active

oldest

votes


















3














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
    8 mins ago











Your Answer






StackExchange.ifUsing("editor", function () {
StackExchange.using("externalEditor", function () {
StackExchange.using("snippets", function () {
StackExchange.snippets.init();
});
});
}, "code-snippets");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "1"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f55367269%2fgeneric-lambda-vs-generic-function-give-different-behaviour%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









3














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
    8 mins ago
















3














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
    8 mins ago














3












3








3







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 7 mins ago

























answered 22 mins ago









Michael KenzelMichael Kenzel

5,05811020




5,05811020








  • 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
    8 mins ago














  • 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
    8 mins ago








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
8 mins ago





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
8 mins ago




















draft saved

draft discarded




















































Thanks for contributing an answer to Stack Overflow!


  • Please be sure to answer the question. Provide details and share your research!

But avoid



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f55367269%2fgeneric-lambda-vs-generic-function-give-different-behaviour%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







Popular posts from this blog

El tren de la libertad Índice Antecedentes "Porque yo decido" Desarrollo de la...

Puerta de Hutt Referencias Enlaces externos Menú de navegación15°58′00″S 5°42′00″O /...

Castillo d'Acher Características Menú de navegación