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Generic lambda vs generic function give different behaviour
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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
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
add a comment |
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
4
Lambdas do not participate in ADL
– Guillaume Racicot
3 hours ago
5
This isn't ADL. Anint
argument doesn't come from any namespace.
– chris
3 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 causesstd::sort
to be considered first. I don't have the standard in front of me so can't be sure about this.
– alter igel
2 hours ago
add a comment |
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
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
c++ lambda c++14
edited 3 hours ago
bartop
asked 3 hours ago
bartopbartop
3,2381030
3,2381030
4
Lambdas do not participate in ADL
– Guillaume Racicot
3 hours ago
5
This isn't ADL. Anint
argument doesn't come from any namespace.
– chris
3 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 causesstd::sort
to be considered first. I don't have the standard in front of me so can't be sure about this.
– alter igel
2 hours ago
add a comment |
4
Lambdas do not participate in ADL
– Guillaume Racicot
3 hours ago
5
This isn't ADL. Anint
argument doesn't come from any namespace.
– chris
3 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 causesstd::sort
to be considered first. I don't have the standard in front of me so can't be sure about this.
– alter igel
2 hours ago
4
4
Lambdas do not participate in ADL
– Guillaume Racicot
3 hours ago
Lambdas do not participate in ADL
– Guillaume Racicot
3 hours ago
5
5
This isn't ADL. An
int
argument doesn't come from any namespace.– chris
3 hours ago
This isn't ADL. An
int
argument doesn't come from any namespace.– chris
3 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
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
2 hours ago
add a comment |
2 Answers
2
active
oldest
votes
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.
1
I think this is actually the right answer. And I would like to add that if both the template functionsort
and the lambdasort
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
53 mins ago
@Mike Kudos to you, Sir, for deleting your answer despite putting quite some effort into it and getting that many upvotes!
– Michael Kenzel
30 mins ago
add a comment |
I used the function typeid
to watch how compiler identify lambda
, function
, function pointer
, std::function
, and each in them is different from any other.
I have not clarified why compiler work in this way, but obviously, behavior compiled between lambda
and function
are different.lambda
must check it has the authority to use or change which value declared in this namespace when compiling, so the range compiled is limited, situation that can use all just like a function
is precious few. At least there is essential distinction between them.
Consider about when lambda is used anonymously, compiler may not have to remember the initial address of lambda, compare lambda is impossible.I thought that's one of reasons.
And there should be more practical reasons for compiler has no support for distinguish named lambda from function by parameter list or return type, if it appeared in future, it won't be named overload probably.
New contributor
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
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.
1
I think this is actually the right answer. And I would like to add that if both the template functionsort
and the lambdasort
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
53 mins ago
@Mike Kudos to you, Sir, for deleting your answer despite putting quite some effort into it and getting that many upvotes!
– Michael Kenzel
30 mins ago
add a comment |
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.
1
I think this is actually the right answer. And I would like to add that if both the template functionsort
and the lambdasort
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
53 mins ago
@Mike Kudos to you, Sir, for deleting your answer despite putting quite some effort into it and getting that many upvotes!
– Michael Kenzel
30 mins ago
add a comment |
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.
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.
edited 52 mins ago
answered 1 hour ago
Michael KenzelMichael Kenzel
5,09811020
5,09811020
1
I think this is actually the right answer. And I would like to add that if both the template functionsort
and the lambdasort
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
53 mins ago
@Mike Kudos to you, Sir, for deleting your answer despite putting quite some effort into it and getting that many upvotes!
– Michael Kenzel
30 mins ago
add a comment |
1
I think this is actually the right answer. And I would like to add that if both the template functionsort
and the lambdasort
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
53 mins ago
@Mike Kudos to you, Sir, for deleting your answer despite putting quite some effort into it and getting that many upvotes!
– Michael Kenzel
30 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
53 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
53 mins ago
@Mike Kudos to you, Sir, for deleting your answer despite putting quite some effort into it and getting that many upvotes!
– Michael Kenzel
30 mins ago
@Mike Kudos to you, Sir, for deleting your answer despite putting quite some effort into it and getting that many upvotes!
– Michael Kenzel
30 mins ago
add a comment |
I used the function typeid
to watch how compiler identify lambda
, function
, function pointer
, std::function
, and each in them is different from any other.
I have not clarified why compiler work in this way, but obviously, behavior compiled between lambda
and function
are different.lambda
must check it has the authority to use or change which value declared in this namespace when compiling, so the range compiled is limited, situation that can use all just like a function
is precious few. At least there is essential distinction between them.
Consider about when lambda is used anonymously, compiler may not have to remember the initial address of lambda, compare lambda is impossible.I thought that's one of reasons.
And there should be more practical reasons for compiler has no support for distinguish named lambda from function by parameter list or return type, if it appeared in future, it won't be named overload probably.
New contributor
add a comment |
I used the function typeid
to watch how compiler identify lambda
, function
, function pointer
, std::function
, and each in them is different from any other.
I have not clarified why compiler work in this way, but obviously, behavior compiled between lambda
and function
are different.lambda
must check it has the authority to use or change which value declared in this namespace when compiling, so the range compiled is limited, situation that can use all just like a function
is precious few. At least there is essential distinction between them.
Consider about when lambda is used anonymously, compiler may not have to remember the initial address of lambda, compare lambda is impossible.I thought that's one of reasons.
And there should be more practical reasons for compiler has no support for distinguish named lambda from function by parameter list or return type, if it appeared in future, it won't be named overload probably.
New contributor
add a comment |
I used the function typeid
to watch how compiler identify lambda
, function
, function pointer
, std::function
, and each in them is different from any other.
I have not clarified why compiler work in this way, but obviously, behavior compiled between lambda
and function
are different.lambda
must check it has the authority to use or change which value declared in this namespace when compiling, so the range compiled is limited, situation that can use all just like a function
is precious few. At least there is essential distinction between them.
Consider about when lambda is used anonymously, compiler may not have to remember the initial address of lambda, compare lambda is impossible.I thought that's one of reasons.
And there should be more practical reasons for compiler has no support for distinguish named lambda from function by parameter list or return type, if it appeared in future, it won't be named overload probably.
New contributor
I used the function typeid
to watch how compiler identify lambda
, function
, function pointer
, std::function
, and each in them is different from any other.
I have not clarified why compiler work in this way, but obviously, behavior compiled between lambda
and function
are different.lambda
must check it has the authority to use or change which value declared in this namespace when compiling, so the range compiled is limited, situation that can use all just like a function
is precious few. At least there is essential distinction between them.
Consider about when lambda is used anonymously, compiler may not have to remember the initial address of lambda, compare lambda is impossible.I thought that's one of reasons.
And there should be more practical reasons for compiler has no support for distinguish named lambda from function by parameter list or return type, if it appeared in future, it won't be named overload probably.
New contributor
edited 17 mins ago
New contributor
answered 25 mins ago
LuLiLuLi
233
233
New contributor
New contributor
add a comment |
add a comment |
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4
Lambdas do not participate in ADL
– Guillaume Racicot
3 hours ago
5
This isn't ADL. An
int
argument doesn't come from any namespace.– chris
3 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 causesstd::sort
to be considered first. I don't have the standard in front of me so can't be sure about this.– alter igel
2 hours ago