What is practical difference between `inline` and `template`?

up vote
15
down vote

favorite

We have 2 methods to declare function in header-only library. They are inline and template<class = void>. In boost source code I can see both variants. Example follows:

inline void my_header_only_function(void)

{

   // Do something...

   return;

}



template<class = void> void my_header_only_function(void)

{

   // Do something...

   return;

}

I know what is difference according to C++ standard. However, any C++ compiler is much more than just standard, and also standard is unclear often.

In situation where template argument is never used and where it is not related to recursive variadic template, is there (and what is) practical difference between 2 variants for mainstream compilers?

edited Nov 15 at 9:32

VTT

23k32345

asked Nov 15 at 9:07

Vitaliy

1,85032344

2

I fail to see where the template solution is better than the inline one.
– YSC
Nov 15 at 9:10

3

Then my_header_only_function<int>, my_header_only_function<char>, and my_header_only_function<> are three separate inline functions that implement the same logic but operate on different static data.
– StoryTeller
Nov 15 at 9:13

3

"In boost source code I can see both variants" - I think it would be better to post those variants, I feel like something is missing here
– VTT
Nov 15 at 9:13

1

@Vitaliy: it's worse because it allows a fake non-used parameter for no reason.
– geza
Nov 15 at 9:16

1

@chris, as far as I understand mainstream compilers ignore inline keyword often and in practice do inline on theirs own discretion. Real inlining depend on optimization level much more that on keyword; keyword is just hint, so I am not sure whether compiler will inline the function and will not inline template in practice.
– Vitaliy
Nov 15 at 9:20

|
show 11 more comments

up vote
15
down vote

favorite

We have 2 methods to declare function in header-only library. They are inline and template<class = void>. In boost source code I can see both variants. Example follows:

inline void my_header_only_function(void)

{

   // Do something...

   return;

}



template<class = void> void my_header_only_function(void)

{

   // Do something...

   return;

}

I know what is difference according to C++ standard. However, any C++ compiler is much more than just standard, and also standard is unclear often.

edited Nov 15 at 9:32

VTT

23k32345

asked Nov 15 at 9:07

Vitaliy

1,85032344

2

I fail to see where the template solution is better than the inline one.
– YSC
Nov 15 at 9:10

3

Then my_header_only_function<int>, my_header_only_function<char>, and my_header_only_function<> are three separate inline functions that implement the same logic but operate on different static data.
– StoryTeller
Nov 15 at 9:13

3

"In boost source code I can see both variants" - I think it would be better to post those variants, I feel like something is missing here
– VTT
Nov 15 at 9:13

1

@Vitaliy: it's worse because it allows a fake non-used parameter for no reason.
– geza
Nov 15 at 9:16

1

@chris, as far as I understand mainstream compilers ignore inline keyword often and in practice do inline on theirs own discretion. Real inlining depend on optimization level much more that on keyword; keyword is just hint, so I am not sure whether compiler will inline the function and will not inline template in practice.
– Vitaliy
Nov 15 at 9:20

|
show 11 more comments

up vote
15
down vote

favorite

We have 2 methods to declare function in header-only library. They are inline and template<class = void>. In boost source code I can see both variants. Example follows:

inline void my_header_only_function(void)

{

   // Do something...

   return;

}



template<class = void> void my_header_only_function(void)

{

   // Do something...

   return;

}

I know what is difference according to C++ standard. However, any C++ compiler is much more than just standard, and also standard is unclear often.

edited Nov 15 at 9:32

VTT

23k32345

asked Nov 15 at 9:07

Vitaliy

1,85032344

We have 2 methods to declare function in header-only library. They are inline and template<class = void>. In boost source code I can see both variants. Example follows:

inline void my_header_only_function(void)

{

   // Do something...

   return;

}



template<class = void> void my_header_only_function(void)

{

   // Do something...

   return;

}

I know what is difference according to C++ standard. However, any C++ compiler is much more than just standard, and also standard is unclear often.

c++

edited Nov 15 at 9:32

VTT

23k32345

asked Nov 15 at 9:07

Vitaliy

1,85032344

edited Nov 15 at 9:32

VTT

23k32345

asked Nov 15 at 9:07

Vitaliy

1,85032344

edited Nov 15 at 9:32

VTT

23k32345

edited Nov 15 at 9:32

VTT

23k32345

edited Nov 15 at 9:32

VTT

23k32345

asked Nov 15 at 9:07

Vitaliy

1,85032344

asked Nov 15 at 9:07

Vitaliy

1,85032344

asked Nov 15 at 9:07

Vitaliy

1,85032344

2

I fail to see where the template solution is better than the inline one.
– YSC
Nov 15 at 9:10

3

Then my_header_only_function<int>, my_header_only_function<char>, and my_header_only_function<> are three separate inline functions that implement the same logic but operate on different static data.
– StoryTeller
Nov 15 at 9:13

3

"In boost source code I can see both variants" - I think it would be better to post those variants, I feel like something is missing here
– VTT
Nov 15 at 9:13

1

@Vitaliy: it's worse because it allows a fake non-used parameter for no reason.
– geza
Nov 15 at 9:16

1

@chris, as far as I understand mainstream compilers ignore inline keyword often and in practice do inline on theirs own discretion. Real inlining depend on optimization level much more that on keyword; keyword is just hint, so I am not sure whether compiler will inline the function and will not inline template in practice.
– Vitaliy
Nov 15 at 9:20

|
show 11 more comments

2

I fail to see where the template solution is better than the inline one.
– YSC
Nov 15 at 9:10

3

Then my_header_only_function<int>, my_header_only_function<char>, and my_header_only_function<> are three separate inline functions that implement the same logic but operate on different static data.
– StoryTeller
Nov 15 at 9:13

3

"In boost source code I can see both variants" - I think it would be better to post those variants, I feel like something is missing here
– VTT
Nov 15 at 9:13

1

@Vitaliy: it's worse because it allows a fake non-used parameter for no reason.
– geza
Nov 15 at 9:16

1

@chris, as far as I understand mainstream compilers ignore inline keyword often and in practice do inline on theirs own discretion. Real inlining depend on optimization level much more that on keyword; keyword is just hint, so I am not sure whether compiler will inline the function and will not inline template in practice.
– Vitaliy
Nov 15 at 9:20

I fail to see where the template solution is better than the inline one.
– YSC
Nov 15 at 9:10

Then my_header_only_function<int>, my_header_only_function<char>, and my_header_only_function<> are three separate inline functions that implement the same logic but operate on different static data.
– StoryTeller
Nov 15 at 9:13

"In boost source code I can see both variants" - I think it would be better to post those variants, I feel like something is missing here
– VTT
Nov 15 at 9:13

@Vitaliy: it's worse because it allows a fake non-used parameter for no reason.
– geza
Nov 15 at 9:16

@chris, as far as I understand mainstream compilers ignore inline keyword often and in practice do inline on theirs own discretion. Real inlining depend on optimization level much more that on keyword; keyword is just hint, so I am not sure whether compiler will inline the function and will not inline template in practice.
– Vitaliy
Nov 15 at 9:20

|
show 11 more comments

3 Answers
3

active

oldest

votes

up vote
7
down vote

accepted

I think this can be used as a weird way to allow library extension (or mocking) from outside library code by providing specialization for void or a non-template version of the function in the same namespace:

#include <iostream>



template<class = void>

int

foo(int data)

{

    ::std::cout << "template" << std::endl;

    return data;

}

// somewhere else

int

foo(int data)

{

    ::std::cout << "non-template" << std::endl;

    return data;

}



int main()

{

    foo(1); // non template overload is selected

    return 0;

}

online compiler

edited Nov 15 at 9:45

answered Nov 15 at 9:25

VTT

23k32345

2

It looks like you are right, I didn't think about that. For sure it is practical difference, we can override template with inline or non-inline function.
– Vitaliy
Nov 15 at 9:27

This is actually a nice trick.
– YSC
Nov 15 at 9:30

1

I mean that template <> int foo<void>(int data) {mock();} is an alternative to int foo(int data) {mock();}. but both ways are risky with ODR or is partial mocking Demo.
– Jarod42
Nov 15 at 9:40

1

@geza - Neither is std a "special" namespace. Those restrictions are imposed to prevent libraries and user code from stepping on each other's toes. So "works in most cases" isn't refuting much. And just because this answer offers a nice explanation doesn't mean it's the explanation for boost itself.
– StoryTeller
Nov 15 at 9:57

1

@StoryTeller Well, std is a "special" namespace because the standard explicitly makes most cases of putting stuff inside std Undefined Behaviour.
– Angew
Nov 15 at 10:59

|
show 14 more comments

up vote
6
down vote

One difference is that binary code for the function may become part of the generated object file even if the function is never used in that file, but there will never be any code for the template if it's not used.

answered Nov 15 at 9:23

Angew

130k11245336

Can you please write any reference that proofs that any compiler will not discard unused function from binary file on optimization stage?
– Vitaliy
Nov 15 at 9:25

@Vitaliy Not really, that's why I wrote "may become", not "will become."
– Angew
Nov 15 at 9:26

It is reasonably common that compilers don't remove unused functions - it sort of breaks the separate compilation model, which means a function defined in one compilation unit needs to be called from another. Very few linkers (the program that combine objects into an executable) even check that any functions which are defined are never used, let alone remove unused functions. Whereas the compiler does not emit code for a template UNLESS the template is either used (resulting in implicit instantiation) or explicitly instantiated.
– Peter
Nov 15 at 9:35

@Peter: compilers don't emit code for non-used inline functions either. At least, I've never see one.
– geza
Nov 15 at 9:37

In that particular boost header all those template functions are always immediately instantiated to be used in non-template functions.
– VTT
Nov 15 at 9:47

add a comment |

up vote
2
down vote

I'm the author of Beast. Hopefully I will be able to shed some light on why you see one versus the other. It really is very simple, the template seems less likely to be inlined into calling functions, bloating the code needlessly. I know that "inline" is really only supposed to mean "remove duplicate definitions" but sometimes compiler implementors get overzealous. The template thing is a little bit harder on the compile (Travis craps out sometimes at only 2GB RAM). So I decided to try writing some new stuff using the "inline" keyword. I still don't know how I feel about it.

The short answer is that I was doing it one way for a long time and then I briefly did it the other way for no particularly strong reason. Sorry if that is not as exciting as the other theories! (which were very interesting in fact)

answered 2 days ago

Vinnie Falco

2,4101432

add a comment |

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',
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

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f53315816%2fwhat-is-practical-difference-between-inline-and-templateclass-void%23new-answer', 'question_page');
}
);

Post as a guest

Name

Required, but never shown

3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

up vote
7
down vote

accepted

#include <iostream>



template<class = void>

int

foo(int data)

{

    ::std::cout << "template" << std::endl;

    return data;

}

// somewhere else

int

foo(int data)

{

    ::std::cout << "non-template" << std::endl;

    return data;

}



int main()

{

    foo(1); // non template overload is selected

    return 0;

}

online compiler

edited Nov 15 at 9:45

answered Nov 15 at 9:25

VTT

23k32345

2

It looks like you are right, I didn't think about that. For sure it is practical difference, we can override template with inline or non-inline function.
– Vitaliy
Nov 15 at 9:27

This is actually a nice trick.
– YSC
Nov 15 at 9:30

1

I mean that template <> int foo<void>(int data) {mock();} is an alternative to int foo(int data) {mock();}. but both ways are risky with ODR or is partial mocking Demo.
– Jarod42
Nov 15 at 9:40

1

@geza - Neither is std a "special" namespace. Those restrictions are imposed to prevent libraries and user code from stepping on each other's toes. So "works in most cases" isn't refuting much. And just because this answer offers a nice explanation doesn't mean it's the explanation for boost itself.
– StoryTeller
Nov 15 at 9:57

1

@StoryTeller Well, std is a "special" namespace because the standard explicitly makes most cases of putting stuff inside std Undefined Behaviour.
– Angew
Nov 15 at 10:59

|
show 14 more comments

up vote
7
down vote

accepted

#include <iostream>



template<class = void>

int

foo(int data)

{

    ::std::cout << "template" << std::endl;

    return data;

}

// somewhere else

int

foo(int data)

{

    ::std::cout << "non-template" << std::endl;

    return data;

}



int main()

{

    foo(1); // non template overload is selected

    return 0;

}

online compiler

edited Nov 15 at 9:45

answered Nov 15 at 9:25

VTT

23k32345

2

It looks like you are right, I didn't think about that. For sure it is practical difference, we can override template with inline or non-inline function.
– Vitaliy
Nov 15 at 9:27

This is actually a nice trick.
– YSC
Nov 15 at 9:30

1

I mean that template <> int foo<void>(int data) {mock();} is an alternative to int foo(int data) {mock();}. but both ways are risky with ODR or is partial mocking Demo.
– Jarod42
Nov 15 at 9:40

1

@geza - Neither is std a "special" namespace. Those restrictions are imposed to prevent libraries and user code from stepping on each other's toes. So "works in most cases" isn't refuting much. And just because this answer offers a nice explanation doesn't mean it's the explanation for boost itself.
– StoryTeller
Nov 15 at 9:57

1

@StoryTeller Well, std is a "special" namespace because the standard explicitly makes most cases of putting stuff inside std Undefined Behaviour.
– Angew
Nov 15 at 10:59

|
show 14 more comments

up vote
7
down vote

accepted

#include <iostream>



template<class = void>

int

foo(int data)

{

    ::std::cout << "template" << std::endl;

    return data;

}

// somewhere else

int

foo(int data)

{

    ::std::cout << "non-template" << std::endl;

    return data;

}



int main()

{

    foo(1); // non template overload is selected

    return 0;

}

online compiler

edited Nov 15 at 9:45

answered Nov 15 at 9:25

VTT

23k32345

#include <iostream>



template<class = void>

int

foo(int data)

{

    ::std::cout << "template" << std::endl;

    return data;

}

// somewhere else

int

foo(int data)

{

    ::std::cout << "non-template" << std::endl;

    return data;

}



int main()

{

    foo(1); // non template overload is selected

    return 0;

}

online compiler

edited Nov 15 at 9:45

answered Nov 15 at 9:25

VTT

23k32345

edited Nov 15 at 9:45

answered Nov 15 at 9:25

VTT

23k32345

answered Nov 15 at 9:25

VTT

23k32345

answered Nov 15 at 9:25

VTT

23k32345

2

It looks like you are right, I didn't think about that. For sure it is practical difference, we can override template with inline or non-inline function.
– Vitaliy
Nov 15 at 9:27

This is actually a nice trick.
– YSC
Nov 15 at 9:30

1

I mean that template <> int foo<void>(int data) {mock();} is an alternative to int foo(int data) {mock();}. but both ways are risky with ODR or is partial mocking Demo.
– Jarod42
Nov 15 at 9:40

1

@geza - Neither is std a "special" namespace. Those restrictions are imposed to prevent libraries and user code from stepping on each other's toes. So "works in most cases" isn't refuting much. And just because this answer offers a nice explanation doesn't mean it's the explanation for boost itself.
– StoryTeller
Nov 15 at 9:57

1

@StoryTeller Well, std is a "special" namespace because the standard explicitly makes most cases of putting stuff inside std Undefined Behaviour.
– Angew
Nov 15 at 10:59

|
show 14 more comments

2

It looks like you are right, I didn't think about that. For sure it is practical difference, we can override template with inline or non-inline function.
– Vitaliy
Nov 15 at 9:27

This is actually a nice trick.
– YSC
Nov 15 at 9:30

1

I mean that template <> int foo<void>(int data) {mock();} is an alternative to int foo(int data) {mock();}. but both ways are risky with ODR or is partial mocking Demo.
– Jarod42
Nov 15 at 9:40

1

@geza - Neither is std a "special" namespace. Those restrictions are imposed to prevent libraries and user code from stepping on each other's toes. So "works in most cases" isn't refuting much. And just because this answer offers a nice explanation doesn't mean it's the explanation for boost itself.
– StoryTeller
Nov 15 at 9:57

1

@StoryTeller Well, std is a "special" namespace because the standard explicitly makes most cases of putting stuff inside std Undefined Behaviour.
– Angew
Nov 15 at 10:59

It looks like you are right, I didn't think about that. For sure it is practical difference, we can override template with inline or non-inline function.
– Vitaliy
Nov 15 at 9:27

This is actually a nice trick.
– YSC
Nov 15 at 9:30

I mean that template <> int foo<void>(int data) {mock();} is an alternative to int foo(int data) {mock();}. but both ways are risky with ODR or is partial mocking Demo.
– Jarod42
Nov 15 at 9:40

@geza - Neither is std a "special" namespace. Those restrictions are imposed to prevent libraries and user code from stepping on each other's toes. So "works in most cases" isn't refuting much. And just because this answer offers a nice explanation doesn't mean it's the explanation for boost itself.
– StoryTeller
Nov 15 at 9:57

@StoryTeller Well, std is a "special" namespace because the standard explicitly makes most cases of putting stuff inside std Undefined Behaviour.
– Angew
Nov 15 at 10:59

|
show 14 more comments

up vote
6
down vote

answered Nov 15 at 9:23

Angew

130k11245336

Can you please write any reference that proofs that any compiler will not discard unused function from binary file on optimization stage?
– Vitaliy
Nov 15 at 9:25

@Vitaliy Not really, that's why I wrote "may become", not "will become."
– Angew
Nov 15 at 9:26

It is reasonably common that compilers don't remove unused functions - it sort of breaks the separate compilation model, which means a function defined in one compilation unit needs to be called from another. Very few linkers (the program that combine objects into an executable) even check that any functions which are defined are never used, let alone remove unused functions. Whereas the compiler does not emit code for a template UNLESS the template is either used (resulting in implicit instantiation) or explicitly instantiated.
– Peter
Nov 15 at 9:35

@Peter: compilers don't emit code for non-used inline functions either. At least, I've never see one.
– geza
Nov 15 at 9:37

In that particular boost header all those template functions are always immediately instantiated to be used in non-template functions.
– VTT
Nov 15 at 9:47

add a comment |

up vote
6
down vote

answered Nov 15 at 9:23

Angew

130k11245336

Can you please write any reference that proofs that any compiler will not discard unused function from binary file on optimization stage?
– Vitaliy
Nov 15 at 9:25

@Vitaliy Not really, that's why I wrote "may become", not "will become."
– Angew
Nov 15 at 9:26

It is reasonably common that compilers don't remove unused functions - it sort of breaks the separate compilation model, which means a function defined in one compilation unit needs to be called from another. Very few linkers (the program that combine objects into an executable) even check that any functions which are defined are never used, let alone remove unused functions. Whereas the compiler does not emit code for a template UNLESS the template is either used (resulting in implicit instantiation) or explicitly instantiated.
– Peter
Nov 15 at 9:35

@Peter: compilers don't emit code for non-used inline functions either. At least, I've never see one.
– geza
Nov 15 at 9:37

In that particular boost header all those template functions are always immediately instantiated to be used in non-template functions.
– VTT
Nov 15 at 9:47

add a comment |

up vote
6
down vote

answered Nov 15 at 9:23

Angew

130k11245336

answered Nov 15 at 9:23

Angew

130k11245336

answered Nov 15 at 9:23

Angew

130k11245336

answered Nov 15 at 9:23

Angew

130k11245336

answered Nov 15 at 9:23

Angew

130k11245336

Can you please write any reference that proofs that any compiler will not discard unused function from binary file on optimization stage?
– Vitaliy
Nov 15 at 9:25

@Vitaliy Not really, that's why I wrote "may become", not "will become."
– Angew
Nov 15 at 9:26

It is reasonably common that compilers don't remove unused functions - it sort of breaks the separate compilation model, which means a function defined in one compilation unit needs to be called from another. Very few linkers (the program that combine objects into an executable) even check that any functions which are defined are never used, let alone remove unused functions. Whereas the compiler does not emit code for a template UNLESS the template is either used (resulting in implicit instantiation) or explicitly instantiated.
– Peter
Nov 15 at 9:35

@Peter: compilers don't emit code for non-used inline functions either. At least, I've never see one.
– geza
Nov 15 at 9:37

In that particular boost header all those template functions are always immediately instantiated to be used in non-template functions.
– VTT
Nov 15 at 9:47

add a comment |

Can you please write any reference that proofs that any compiler will not discard unused function from binary file on optimization stage?
– Vitaliy
Nov 15 at 9:25

@Vitaliy Not really, that's why I wrote "may become", not "will become."
– Angew
Nov 15 at 9:26

It is reasonably common that compilers don't remove unused functions - it sort of breaks the separate compilation model, which means a function defined in one compilation unit needs to be called from another. Very few linkers (the program that combine objects into an executable) even check that any functions which are defined are never used, let alone remove unused functions. Whereas the compiler does not emit code for a template UNLESS the template is either used (resulting in implicit instantiation) or explicitly instantiated.
– Peter
Nov 15 at 9:35

@Peter: compilers don't emit code for non-used inline functions either. At least, I've never see one.
– geza
Nov 15 at 9:37

In that particular boost header all those template functions are always immediately instantiated to be used in non-template functions.
– VTT
Nov 15 at 9:47

Can you please write any reference that proofs that any compiler will not discard unused function from binary file on optimization stage?
– Vitaliy
Nov 15 at 9:25

@Vitaliy Not really, that's why I wrote "may become", not "will become."
– Angew
Nov 15 at 9:26

It is reasonably common that compilers don't remove unused functions - it sort of breaks the separate compilation model, which means a function defined in one compilation unit needs to be called from another. Very few linkers (the program that combine objects into an executable) even check that any functions which are defined are never used, let alone remove unused functions. Whereas the compiler does not emit code for a template UNLESS the template is either used (resulting in implicit instantiation) or explicitly instantiated.
– Peter
Nov 15 at 9:35

@Peter: compilers don't emit code for non-used inline functions either. At least, I've never see one.
– geza
Nov 15 at 9:37

In that particular boost header all those template functions are always immediately instantiated to be used in non-template functions.
– VTT
Nov 15 at 9:47

add a comment |

up vote
2
down vote

answered 2 days ago

Vinnie Falco

2,4101432

add a comment |

up vote
2
down vote

answered 2 days ago

Vinnie Falco

2,4101432

add a comment |

up vote
2
down vote

answered 2 days ago

Vinnie Falco

2,4101432

answered 2 days ago

Vinnie Falco

2,4101432

answered 2 days ago

Vinnie Falco

2,4101432

answered 2 days ago

Vinnie Falco

2,4101432

answered 2 days ago

Vinnie Falco

2,4101432

add a comment |

draft saved

draft discarded

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

Post as a guest

Name

Required, but never shown

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

Name

Required, but never shown

Name

Required, but never shown

This page is only for reference, If you need detailed information, please check here

搜尋此網誌

Vrftjkry