Does static make a difference for a const local variable?
Clash Royale CLAN TAG#URR8PPP
Imagine the following declaration:
void foo()
const std::array<int, 80000> arr = /* a lot of different values*/;
//do stuff
And a second one:
void foo()
static const std::array<int, 80000> arr = /* a lot of different values*/;
//do stuff
What are the possible performance differences between these two if any? And is there any danger associated with any of these solutions?
c++ static const
|
show 5 more comments
Imagine the following declaration:
void foo()
const std::array<int, 80000> arr = /* a lot of different values*/;
//do stuff
And a second one:
void foo()
static const std::array<int, 80000> arr = /* a lot of different values*/;
//do stuff
What are the possible performance differences between these two if any? And is there any danger associated with any of these solutions?
c++ static const
3
In thestatic
case they may not be on the stack, but in a read-only section. Probably compiler dependent as well.
– Matthieu Brucher
Mar 1 at 10:31
3
out of curiosity: do you have a real problem at hand, or is this just an academic exercise? (its a valid question in both cases)
– user463035818
Mar 1 at 10:35
1
@user463035818 I am having discussion during code review ;)
– bartop
Mar 1 at 10:37
2
depending on the reviewer that can be a real problem :P
– user463035818
Mar 1 at 10:38
5
@ScheffwithoutStatic
builds the array each times it is invoked from static data (.LC0
).withStatic
uses an array whose construction has been optimized as a constant (withStatic()::arr
).
– YSC
Mar 1 at 10:53
|
show 5 more comments
Imagine the following declaration:
void foo()
const std::array<int, 80000> arr = /* a lot of different values*/;
//do stuff
And a second one:
void foo()
static const std::array<int, 80000> arr = /* a lot of different values*/;
//do stuff
What are the possible performance differences between these two if any? And is there any danger associated with any of these solutions?
c++ static const
Imagine the following declaration:
void foo()
const std::array<int, 80000> arr = /* a lot of different values*/;
//do stuff
And a second one:
void foo()
static const std::array<int, 80000> arr = /* a lot of different values*/;
//do stuff
What are the possible performance differences between these two if any? And is there any danger associated with any of these solutions?
c++ static const
c++ static const
edited Mar 1 at 15:57
Boann
37.4k1290121
37.4k1290121
asked Mar 1 at 10:27
bartopbartop
3,2601031
3,2601031
3
In thestatic
case they may not be on the stack, but in a read-only section. Probably compiler dependent as well.
– Matthieu Brucher
Mar 1 at 10:31
3
out of curiosity: do you have a real problem at hand, or is this just an academic exercise? (its a valid question in both cases)
– user463035818
Mar 1 at 10:35
1
@user463035818 I am having discussion during code review ;)
– bartop
Mar 1 at 10:37
2
depending on the reviewer that can be a real problem :P
– user463035818
Mar 1 at 10:38
5
@ScheffwithoutStatic
builds the array each times it is invoked from static data (.LC0
).withStatic
uses an array whose construction has been optimized as a constant (withStatic()::arr
).
– YSC
Mar 1 at 10:53
|
show 5 more comments
3
In thestatic
case they may not be on the stack, but in a read-only section. Probably compiler dependent as well.
– Matthieu Brucher
Mar 1 at 10:31
3
out of curiosity: do you have a real problem at hand, or is this just an academic exercise? (its a valid question in both cases)
– user463035818
Mar 1 at 10:35
1
@user463035818 I am having discussion during code review ;)
– bartop
Mar 1 at 10:37
2
depending on the reviewer that can be a real problem :P
– user463035818
Mar 1 at 10:38
5
@ScheffwithoutStatic
builds the array each times it is invoked from static data (.LC0
).withStatic
uses an array whose construction has been optimized as a constant (withStatic()::arr
).
– YSC
Mar 1 at 10:53
3
3
In the
static
case they may not be on the stack, but in a read-only section. Probably compiler dependent as well.– Matthieu Brucher
Mar 1 at 10:31
In the
static
case they may not be on the stack, but in a read-only section. Probably compiler dependent as well.– Matthieu Brucher
Mar 1 at 10:31
3
3
out of curiosity: do you have a real problem at hand, or is this just an academic exercise? (its a valid question in both cases)
– user463035818
Mar 1 at 10:35
out of curiosity: do you have a real problem at hand, or is this just an academic exercise? (its a valid question in both cases)
– user463035818
Mar 1 at 10:35
1
1
@user463035818 I am having discussion during code review ;)
– bartop
Mar 1 at 10:37
@user463035818 I am having discussion during code review ;)
– bartop
Mar 1 at 10:37
2
2
depending on the reviewer that can be a real problem :P
– user463035818
Mar 1 at 10:38
depending on the reviewer that can be a real problem :P
– user463035818
Mar 1 at 10:38
5
5
@Scheff
withoutStatic
builds the array each times it is invoked from static data (.LC0
). withStatic
uses an array whose construction has been optimized as a constant (withStatic()::arr
).– YSC
Mar 1 at 10:53
@Scheff
withoutStatic
builds the array each times it is invoked from static data (.LC0
). withStatic
uses an array whose construction has been optimized as a constant (withStatic()::arr
).– YSC
Mar 1 at 10:53
|
show 5 more comments
4 Answers
4
active
oldest
votes
And is there any danger associated with any of these solutions?
Non-static is dangerous because the array is huge, and the memory reserved for automatic storage is limited. Depending on the system and configuration, that array could use about 30% of the space available for automatic storage. As such, it greatly increases the possibility of stack overflow.
While an optimiser might certainly avoid allocating memory on the stack, there are good reasons why you would want your non-optimised debug build to also not crash.
add a comment |
Forget the array for a moment. That muddles two separate issues. You've got answers that address the lifetime and storage issue. I'll address the initialization issue.
void f()
static const int x = get_x();
// do something with x
void g()
const int x = get_x();
// do something with x
The difference between these two is that the first one will only call get_x()
the first time that f()
is called; x
retains that value through the remainder of the program. The second one will call get_x()
each time that g()
is called.
That matters if get_x()
returns different values on subsequent calls:
int current_x = 0;
int get_x() return current_x++;
add a comment |
What are the possible performance differences between these two if any?And is there any danger associated with any of these solutions?
The difference depends exactly on how you use foo()
.
1st case:(low probability): Your implementation is such that you will call foo()
only once , maybe you have created separate function to divide code logic as practiced. Well in this case declaring as static is very bad, because a static variable or object remains in memory until programs ends . So just imagine that your variable occupying memory unnecessarily.
2nd case:(high probability): Your implementation is such that you will call foo()
again and again . Then non-static object will get allocated and de allocated again and again.This will take huge amount of cpu clock cycles which is not desired .Use static in this case.
add a comment |
In this particular context, one point to consider regarding using static
on a variable with initialization:
From C++17 standard:
6.7.1 Static storage duration [basic.stc.static]
...
2 If a variable with static storage duration has initialization or a destructor with side effects, it shall not be eliminated even if it appears to be unused, except that a class object or its copy/move may be eliminated as specified in 15.8.
Actually, there are more differences: Its lifetime is different, for one....
– CharonX
Mar 1 at 14:19
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
And is there any danger associated with any of these solutions?
Non-static is dangerous because the array is huge, and the memory reserved for automatic storage is limited. Depending on the system and configuration, that array could use about 30% of the space available for automatic storage. As such, it greatly increases the possibility of stack overflow.
While an optimiser might certainly avoid allocating memory on the stack, there are good reasons why you would want your non-optimised debug build to also not crash.
add a comment |
And is there any danger associated with any of these solutions?
Non-static is dangerous because the array is huge, and the memory reserved for automatic storage is limited. Depending on the system and configuration, that array could use about 30% of the space available for automatic storage. As such, it greatly increases the possibility of stack overflow.
While an optimiser might certainly avoid allocating memory on the stack, there are good reasons why you would want your non-optimised debug build to also not crash.
add a comment |
And is there any danger associated with any of these solutions?
Non-static is dangerous because the array is huge, and the memory reserved for automatic storage is limited. Depending on the system and configuration, that array could use about 30% of the space available for automatic storage. As such, it greatly increases the possibility of stack overflow.
While an optimiser might certainly avoid allocating memory on the stack, there are good reasons why you would want your non-optimised debug build to also not crash.
And is there any danger associated with any of these solutions?
Non-static is dangerous because the array is huge, and the memory reserved for automatic storage is limited. Depending on the system and configuration, that array could use about 30% of the space available for automatic storage. As such, it greatly increases the possibility of stack overflow.
While an optimiser might certainly avoid allocating memory on the stack, there are good reasons why you would want your non-optimised debug build to also not crash.
edited Mar 1 at 11:00
answered Mar 1 at 10:53
eerorikaeerorika
87.9k663134
87.9k663134
add a comment |
add a comment |
Forget the array for a moment. That muddles two separate issues. You've got answers that address the lifetime and storage issue. I'll address the initialization issue.
void f()
static const int x = get_x();
// do something with x
void g()
const int x = get_x();
// do something with x
The difference between these two is that the first one will only call get_x()
the first time that f()
is called; x
retains that value through the remainder of the program. The second one will call get_x()
each time that g()
is called.
That matters if get_x()
returns different values on subsequent calls:
int current_x = 0;
int get_x() return current_x++;
add a comment |
Forget the array for a moment. That muddles two separate issues. You've got answers that address the lifetime and storage issue. I'll address the initialization issue.
void f()
static const int x = get_x();
// do something with x
void g()
const int x = get_x();
// do something with x
The difference between these two is that the first one will only call get_x()
the first time that f()
is called; x
retains that value through the remainder of the program. The second one will call get_x()
each time that g()
is called.
That matters if get_x()
returns different values on subsequent calls:
int current_x = 0;
int get_x() return current_x++;
add a comment |
Forget the array for a moment. That muddles two separate issues. You've got answers that address the lifetime and storage issue. I'll address the initialization issue.
void f()
static const int x = get_x();
// do something with x
void g()
const int x = get_x();
// do something with x
The difference between these two is that the first one will only call get_x()
the first time that f()
is called; x
retains that value through the remainder of the program. The second one will call get_x()
each time that g()
is called.
That matters if get_x()
returns different values on subsequent calls:
int current_x = 0;
int get_x() return current_x++;
Forget the array for a moment. That muddles two separate issues. You've got answers that address the lifetime and storage issue. I'll address the initialization issue.
void f()
static const int x = get_x();
// do something with x
void g()
const int x = get_x();
// do something with x
The difference between these two is that the first one will only call get_x()
the first time that f()
is called; x
retains that value through the remainder of the program. The second one will call get_x()
each time that g()
is called.
That matters if get_x()
returns different values on subsequent calls:
int current_x = 0;
int get_x() return current_x++;
answered Mar 1 at 13:00
Pete BeckerPete Becker
58.7k442122
58.7k442122
add a comment |
add a comment |
What are the possible performance differences between these two if any?And is there any danger associated with any of these solutions?
The difference depends exactly on how you use foo()
.
1st case:(low probability): Your implementation is such that you will call foo()
only once , maybe you have created separate function to divide code logic as practiced. Well in this case declaring as static is very bad, because a static variable or object remains in memory until programs ends . So just imagine that your variable occupying memory unnecessarily.
2nd case:(high probability): Your implementation is such that you will call foo()
again and again . Then non-static object will get allocated and de allocated again and again.This will take huge amount of cpu clock cycles which is not desired .Use static in this case.
add a comment |
What are the possible performance differences between these two if any?And is there any danger associated with any of these solutions?
The difference depends exactly on how you use foo()
.
1st case:(low probability): Your implementation is such that you will call foo()
only once , maybe you have created separate function to divide code logic as practiced. Well in this case declaring as static is very bad, because a static variable or object remains in memory until programs ends . So just imagine that your variable occupying memory unnecessarily.
2nd case:(high probability): Your implementation is such that you will call foo()
again and again . Then non-static object will get allocated and de allocated again and again.This will take huge amount of cpu clock cycles which is not desired .Use static in this case.
add a comment |
What are the possible performance differences between these two if any?And is there any danger associated with any of these solutions?
The difference depends exactly on how you use foo()
.
1st case:(low probability): Your implementation is such that you will call foo()
only once , maybe you have created separate function to divide code logic as practiced. Well in this case declaring as static is very bad, because a static variable or object remains in memory until programs ends . So just imagine that your variable occupying memory unnecessarily.
2nd case:(high probability): Your implementation is such that you will call foo()
again and again . Then non-static object will get allocated and de allocated again and again.This will take huge amount of cpu clock cycles which is not desired .Use static in this case.
What are the possible performance differences between these two if any?And is there any danger associated with any of these solutions?
The difference depends exactly on how you use foo()
.
1st case:(low probability): Your implementation is such that you will call foo()
only once , maybe you have created separate function to divide code logic as practiced. Well in this case declaring as static is very bad, because a static variable or object remains in memory until programs ends . So just imagine that your variable occupying memory unnecessarily.
2nd case:(high probability): Your implementation is such that you will call foo()
again and again . Then non-static object will get allocated and de allocated again and again.This will take huge amount of cpu clock cycles which is not desired .Use static in this case.
answered Mar 1 at 11:15
Abhishek GargAbhishek Garg
311110
311110
add a comment |
add a comment |
In this particular context, one point to consider regarding using static
on a variable with initialization:
From C++17 standard:
6.7.1 Static storage duration [basic.stc.static]
...
2 If a variable with static storage duration has initialization or a destructor with side effects, it shall not be eliminated even if it appears to be unused, except that a class object or its copy/move may be eliminated as specified in 15.8.
Actually, there are more differences: Its lifetime is different, for one....
– CharonX
Mar 1 at 14:19
add a comment |
In this particular context, one point to consider regarding using static
on a variable with initialization:
From C++17 standard:
6.7.1 Static storage duration [basic.stc.static]
...
2 If a variable with static storage duration has initialization or a destructor with side effects, it shall not be eliminated even if it appears to be unused, except that a class object or its copy/move may be eliminated as specified in 15.8.
Actually, there are more differences: Its lifetime is different, for one....
– CharonX
Mar 1 at 14:19
add a comment |
In this particular context, one point to consider regarding using static
on a variable with initialization:
From C++17 standard:
6.7.1 Static storage duration [basic.stc.static]
...
2 If a variable with static storage duration has initialization or a destructor with side effects, it shall not be eliminated even if it appears to be unused, except that a class object or its copy/move may be eliminated as specified in 15.8.
In this particular context, one point to consider regarding using static
on a variable with initialization:
From C++17 standard:
6.7.1 Static storage duration [basic.stc.static]
...
2 If a variable with static storage duration has initialization or a destructor with side effects, it shall not be eliminated even if it appears to be unused, except that a class object or its copy/move may be eliminated as specified in 15.8.
edited Mar 1 at 14:23
answered Mar 1 at 11:49
P.WP.W
17.5k41556
17.5k41556
Actually, there are more differences: Its lifetime is different, for one....
– CharonX
Mar 1 at 14:19
add a comment |
Actually, there are more differences: Its lifetime is different, for one....
– CharonX
Mar 1 at 14:19
Actually, there are more differences: Its lifetime is different, for one....
– CharonX
Mar 1 at 14:19
Actually, there are more differences: Its lifetime is different, for one....
– CharonX
Mar 1 at 14:19
add a comment |
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3
In the
static
case they may not be on the stack, but in a read-only section. Probably compiler dependent as well.– Matthieu Brucher
Mar 1 at 10:31
3
out of curiosity: do you have a real problem at hand, or is this just an academic exercise? (its a valid question in both cases)
– user463035818
Mar 1 at 10:35
1
@user463035818 I am having discussion during code review ;)
– bartop
Mar 1 at 10:37
2
depending on the reviewer that can be a real problem :P
– user463035818
Mar 1 at 10:38
5
@Scheff
withoutStatic
builds the array each times it is invoked from static data (.LC0
).withStatic
uses an array whose construction has been optimized as a constant (withStatic()::arr
).– YSC
Mar 1 at 10:53