What is practical difference between `inline` and `template`?
up vote
16
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?
c++
edited Nov 15 at 9:32
VTT
23.1k32345
asked Nov 15 at 9:07
Vitaliy
1,85532344
|
show 11 more comments
up vote
16
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?
c++
edited Nov 15 at 9:32
VTT
23.1k32345
asked Nov 15 at 9:07
Vitaliy
1,85532344
2
I fail to see where the template solution is better than the inline one.
– YSC
Nov 15 at 9:10
3
Thenmy_header_only_function<int>,my_header_only_function<char>, andmy_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 ignoreinlinekeyword 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
16
down vote
favorite
up vote
16
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?
c++
edited Nov 15 at 9:32
VTT
23.1k32345
asked Nov 15 at 9:07
Vitaliy
1,85532344
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?
c++
c++
edited Nov 15 at 9:32
VTT
23.1k32345
asked Nov 15 at 9:07
Vitaliy
1,85532344
edited Nov 15 at 9:32
VTT
23.1k32345
asked Nov 15 at 9:07
Vitaliy
1,85532344
edited Nov 15 at 9:32
VTT
23.1k32345
edited Nov 15 at 9:32
VTT
23.1k32345
edited Nov 15 at 9:32
VTT
23.1k32345
23.1k32345
asked Nov 15 at 9:07
Vitaliy
1,85532344
asked Nov 15 at 9:07
Vitaliy
1,85532344
asked Nov 15 at 9:07
Vitaliy
1,85532344
1,85532344
2
I fail to see where the template solution is better than the inline one.
– YSC
Nov 15 at 9:10
3
Thenmy_header_only_function<int>,my_header_only_function<char>, andmy_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 ignoreinlinekeyword 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
Thenmy_header_only_function<int>,my_header_only_function<char>, andmy_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 ignoreinlinekeyword 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
2
2
I fail to see where the template solution is better than the inline one.
– YSC
Nov 15 at 9:10
I fail to see where the template solution is better than the inline one.
– YSC
Nov 15 at 9:10
3
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
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
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
"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
1
@Vitaliy: it's worse because it allows a fake non-used parameter for no reason.
– geza
Nov 15 at 9:16
@Vitaliy: it's worse because it allows a fake non-used parameter for no reason.
– geza
Nov 15 at 9:16
1
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
@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
8
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
answered Nov 15 at 9:25
VTT
23.1k32345
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 thattemplate <> int foo<void>(int data) {mock();}is an alternative toint 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 isstda "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,stdis a "special" namespace because the standard explicitly makes most cases of putting stuff insidestdUndefined Behaviour.
– Angew
Nov 15 at 10:59
|
show 14 more comments
up vote
7
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 Nov 16 at 2:19
Vinnie Falco
2,4401432
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
8
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
answered Nov 15 at 9:25
VTT
23.1k32345
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 thattemplate <> int foo<void>(int data) {mock();}is an alternative toint 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 isstda "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,stdis a "special" namespace because the standard explicitly makes most cases of putting stuff insidestdUndefined Behaviour.
– Angew
Nov 15 at 10:59
|
show 14 more comments
up vote
8
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
answered Nov 15 at 9:25
VTT
23.1k32345
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 thattemplate <> int foo<void>(int data) {mock();}is an alternative toint 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 isstda "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,stdis a "special" namespace because the standard explicitly makes most cases of putting stuff insidestdUndefined Behaviour.
– Angew
Nov 15 at 10:59
|
show 14 more comments
up vote
8
down vote
accepted
up vote
8
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
answered Nov 15 at 9:25
VTT
23.1k32345
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
answered Nov 15 at 9:25
VTT
23.1k32345
edited Nov 15 at 9:45
answered Nov 15 at 9:25
VTT
23.1k32345
answered Nov 15 at 9:25
VTT
23.1k32345
answered Nov 15 at 9:25
VTT
23.1k32345
23.1k32345
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 thattemplate <> int foo<void>(int data) {mock();}is an alternative toint 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 isstda "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,stdis a "special" namespace because the standard explicitly makes most cases of putting stuff insidestdUndefined 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 thattemplate <> int foo<void>(int data) {mock();}is an alternative toint 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 isstda "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,stdis a "special" namespace because the standard explicitly makes most cases of putting stuff insidestdUndefined Behaviour.
– Angew
Nov 15 at 10:59
2
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
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
This is actually a nice trick.
– YSC
Nov 15 at 9:30
1
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
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
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
@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
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
@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
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
7
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
7
down vote
up vote
7
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
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
answered Nov 15 at 9:23
Angew
130k11245336
answered Nov 15 at 9:23
Angew
130k11245336
answered Nov 15 at 9:23
Angew
130k11245336
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
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
@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
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
@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
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
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up vote
2
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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 Nov 16 at 2:19
Vinnie Falco
2,4401432
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 Nov 16 at 2:19
Vinnie Falco
2,4401432
add a comment |
up vote
2
down vote
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 Nov 16 at 2:19
Vinnie Falco
2,4401432
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 Nov 16 at 2:19
Vinnie Falco
2,4401432
answered Nov 16 at 2:19
Vinnie Falco
2,4401432
answered Nov 16 at 2:19
Vinnie Falco
2,4401432
answered Nov 16 at 2:19
Vinnie Falco
2,4401432
2,4401432
add a comment |
add a comment |
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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>, andmy_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
inlinekeyword 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