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Does static makes difference for const local variable?

What is the difference between const and readonly?What are the differences between a pointer variable and a reference variable in C++?Are static class variables possible?Difference between static class and singleton pattern?What does “static” mean in C?What is the difference between const int*, const int * const, and int const *?Static variables in JavaScriptWhy are static variables considered evil?Difference between `constexpr` and `const`Replacing a 32-bit loop counter with 64-bit introduces crazy performance deviations

6

1

Imagine 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

share|improve this question

edited 2 hours ago

bartop

asked 2 hours ago

bartopbartop

2,957826

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  In the static case they may not be on the stack, but in a read-only section. Probably compiler dependent as well.
  
  – Matthieu Brucher
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  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
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  @user463035818 I am having discussion during code review ;)
  
  – bartop
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  depending on the reviewer that can be a real problem :P
  
  – user463035818
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  @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
  1 hour ago
  

  
  
  
  

 | 
show 4 more comments

6

1

Imagine 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

share|improve this question

edited 2 hours ago

bartop

asked 2 hours ago

bartopbartop

2,957826

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  In the static case they may not be on the stack, but in a read-only section. Probably compiler dependent as well.
  
  – Matthieu Brucher
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  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
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  @user463035818 I am having discussion during code review ;)
  
  – bartop
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  depending on the reviewer that can be a real problem :P
  
  – user463035818
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  @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
  1 hour ago
  

  
  
  
  

 | 
show 4 more comments

Imagine 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

share|improve this question

edited 2 hours ago

bartop

asked 2 hours ago

bartopbartop

2,957826

void foo(){

    const std::array<int, 80000> arr = {/* a lot of different values*/};

    //do stuff

}

void foo(){

    static const std::array<int, 80000> arr = {/* a lot of different values*/};

    //do stuff

}

share|improve this question

edited 2 hours ago

bartop

asked 2 hours ago

bartopbartop

2,957826

share|improve this question

edited 2 hours ago

bartop

asked 2 hours ago

bartopbartop

2,957826

asked 2 hours ago

bartopbartop

2,957826

asked 2 hours ago

bartopbartop

2,957826

3 Answers
3

active

oldest

votes

4

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.

share|improve this answer

edited 1 hour ago

answered 1 hour ago

eerorikaeerorika

83.8k662128

add a comment | 

2

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.

share|improve this answer

answered 1 hour ago

Abhishek GargAbhishek Garg

1168

add a comment | 

1

There is one difference with 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.

share|improve this answer

answered 54 mins ago

P.WP.W

15.4k31453

add a comment | 

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4

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.

share|improve this answer

edited 1 hour ago

answered 1 hour ago

eerorikaeerorika

83.8k662128

add a comment | 

4

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.

share|improve this answer

edited 1 hour ago

answered 1 hour ago

eerorikaeerorika

83.8k662128

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.

share|improve this answer

edited 1 hour ago

answered 1 hour ago

eerorikaeerorika

83.8k662128

share|improve this answer

edited 1 hour ago

answered 1 hour ago

eerorikaeerorika

83.8k662128

answered 1 hour ago

eerorikaeerorika

83.8k662128

answered 1 hour ago

eerorikaeerorika

83.8k662128

2

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.

share|improve this answer

answered 1 hour ago

Abhishek GargAbhishek Garg

1168

add a comment | 

2

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.

share|improve this answer

answered 1 hour ago

Abhishek GargAbhishek Garg

1168

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.

share|improve this answer

answered 1 hour ago

Abhishek GargAbhishek Garg

1168

share|improve this answer

answered 1 hour ago

Abhishek GargAbhishek Garg

1168

answered 1 hour ago

Abhishek GargAbhishek Garg

1168

answered 1 hour ago

Abhishek GargAbhishek Garg

1168

1

There is one difference with 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.

share|improve this answer

answered 54 mins ago

P.WP.W

15.4k31453

add a comment | 

1

There is one difference with 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.

share|improve this answer

answered 54 mins ago

P.WP.W

15.4k31453

add a comment | 

There is one difference with 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.

share|improve this answer

answered 54 mins ago

P.WP.W

15.4k31453

share|improve this answer

answered 54 mins ago

P.WP.W

15.4k31453

answered 54 mins ago

P.WP.W

15.4k31453

answered 54 mins ago

P.WP.W

15.4k31453