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Memory usage: #define vs. static const for uint8_t

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3

1

I'm writing an Arduino library to communicate with an I2C device, and I'm wondering what the best way is to declare all the register addresses so as to save memory.

Using #defines:

#define REGISTER_MOTOR_1_MODE 0x44

#define REGISTER_MOTOR_2_MODE 0x47

Or using static const:

static const uint8_t REGISTER_MOTOR_1_MODE = 0x44;

static const uint8_t REGISTER_MOTOR_2_MODE = 0x47;

(Obviously I have more than just two registers I need to declare, but I thought two would illustrate the point just fine)

c++ memory-usage

share|improve this question

asked 19 hours ago

Android DevAndroid Dev

1184

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  define's are just "find and replace"s, and can bit you, if you do anything other than numbers. For example #define LED_MASK 0x01<<2. A safe(r) way to write that would be #define LED_MASK (0x01<<2). See also stackoverflow.com/questions/6542270/…
  
  – Gerben
  18 hours ago
  

  
  
  
  

add a comment | 

3

1

I'm writing an Arduino library to communicate with an I2C device, and I'm wondering what the best way is to declare all the register addresses so as to save memory.

Using #defines:

#define REGISTER_MOTOR_1_MODE 0x44

#define REGISTER_MOTOR_2_MODE 0x47

Or using static const:

static const uint8_t REGISTER_MOTOR_1_MODE = 0x44;

static const uint8_t REGISTER_MOTOR_2_MODE = 0x47;

(Obviously I have more than just two registers I need to declare, but I thought two would illustrate the point just fine)

c++ memory-usage

share|improve this question

asked 19 hours ago

Android DevAndroid Dev

1184

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  define's are just "find and replace"s, and can bit you, if you do anything other than numbers. For example #define LED_MASK 0x01<<2. A safe(r) way to write that would be #define LED_MASK (0x01<<2). See also stackoverflow.com/questions/6542270/…
  
  – Gerben
  18 hours ago
  

  
  
  
  

add a comment | 

I'm writing an Arduino library to communicate with an I2C device, and I'm wondering what the best way is to declare all the register addresses so as to save memory.

Using #defines:

#define REGISTER_MOTOR_1_MODE 0x44

#define REGISTER_MOTOR_2_MODE 0x47

Or using static const:

static const uint8_t REGISTER_MOTOR_1_MODE = 0x44;

static const uint8_t REGISTER_MOTOR_2_MODE = 0x47;

(Obviously I have more than just two registers I need to declare, but I thought two would illustrate the point just fine)

c++ memory-usage

share|improve this question

asked 19 hours ago

Android DevAndroid Dev

1184

#define REGISTER_MOTOR_1_MODE 0x44

#define REGISTER_MOTOR_2_MODE 0x47

static const uint8_t REGISTER_MOTOR_1_MODE = 0x44;

static const uint8_t REGISTER_MOTOR_2_MODE = 0x47;

share|improve this question

asked 19 hours ago

Android DevAndroid Dev

1184

share|improve this question

asked 19 hours ago

Android DevAndroid Dev

1184

asked 19 hours ago

Android DevAndroid Dev

1184

asked 19 hours ago

Android DevAndroid Dev

1184

3 Answers
3

active

oldest

votes

8

You'll find no noticeable difference memory-wise between the two.

The only real difference is that the const method also imposes a type to the value, which can be useful for function overloading or mathematical operations.

share|improve this answer

answered 19 hours ago

Majenko♦Majenko

68.2k43277

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Macros also have types. #define REGISTER_MOTOR_1_MODE 0x44 would cause REGISTER_MOTOR_1_MODE to have a type of int. The difference is more in the explicitness of the type.
  
  – Pharap
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @Pharap Almost... Macros do not have a type. However, the content of the macro can have either an implicit or explicit type. That type is nothing to do with the macro.
  
  – Majenko♦
  14 mins ago
  

  
  
  
  

add a comment | 

5

A #define is a preprocessor macro. As Gerben says in his comment, it's just an automated find-and-replace.

If you use it to hold things like C string constants, e.g. #define ERROR_STRING "You messed up, bud!" it could actually cause your program to take more memory, since that same string literal will be duplicated every time you reference it. In that case a static const would be both type-safe AND reduce memory usage.

share|improve this answer

answered 17 hours ago

Duncan CDuncan C

1,7291617

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Would it be duplicated? I mean, isn't it the job of the compiler to ensure "You messed up, bud!" occurs exactly once in the string constant table?
  
  – juhist
  15 hours ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  @juhist: C++ explicitly states that the compiler can choose to duplicate identical strings, or not. Most compilers have a flag to either do this, or not. I think, by default, most compilers do not merge strings, in case the app relies on them being unique.
  
  – Mooing Duck
  14 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @MooingDuck I compiled a program having many std::printf("%pn", "foo"); lines in it. All of them print the same address.
  
  – juhist
  13 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @MooingDuck: While few compilers will guarantee that strings will always be merged, I think the normal reason is that upholding such a guarantee may impose extra build-time costs and complexity, and the savings in storage from merging strings may or may not be worthwhile.
  
  – supercat
  13 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  @juhist: Sure. In your compiler and your compiler flags. With my compiler and my flags, I get different addresses.
  
  – Mooing Duck
  12 hours ago
  

  
  
  
  

 | 
show 2 more comments

3

Theoretically both approaches should consume the same amount of space and will probably result in the same code.

---

However, rather than that meaning "pick any one", what that actually means is that you should turn your attention to other factors when deciding which one to use.

As a rule of thumb you should generally avoid macros because they don't respect scoping rules due to how they work.

This can lead to some serious problems in certain cases.

See Why are preprocessor macros evil and what are the alternatives?.

---

However, I'd like to point out a third option: constexpr.

constexpr uint8_t motorMode1Register = 0x44;

constexpr implies const and is a better expression of the intent that the value should be capable of being evaluated at compile time.
constexpr literally means "this is a constant expression and thus can be evaluated at compile time".

As a result of this, constexpr variables can be used in situations where other variables cannot, such as template parameters.

See Constexpr vs Macros.

share|improve this answer

answered 4 hours ago

PharapPharap

1498

add a comment | 

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8

You'll find no noticeable difference memory-wise between the two.

The only real difference is that the const method also imposes a type to the value, which can be useful for function overloading or mathematical operations.

share|improve this answer

answered 19 hours ago

Majenko♦Majenko

68.2k43277

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Macros also have types. #define REGISTER_MOTOR_1_MODE 0x44 would cause REGISTER_MOTOR_1_MODE to have a type of int. The difference is more in the explicitness of the type.
  
  – Pharap
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @Pharap Almost... Macros do not have a type. However, the content of the macro can have either an implicit or explicit type. That type is nothing to do with the macro.
  
  – Majenko♦
  14 mins ago
  

  
  
  
  

add a comment | 

8

You'll find no noticeable difference memory-wise between the two.

The only real difference is that the const method also imposes a type to the value, which can be useful for function overloading or mathematical operations.

share|improve this answer

answered 19 hours ago

Majenko♦Majenko

68.2k43277

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Macros also have types. #define REGISTER_MOTOR_1_MODE 0x44 would cause REGISTER_MOTOR_1_MODE to have a type of int. The difference is more in the explicitness of the type.
  
  – Pharap
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @Pharap Almost... Macros do not have a type. However, the content of the macro can have either an implicit or explicit type. That type is nothing to do with the macro.
  
  – Majenko♦
  14 mins ago
  

  
  
  
  

add a comment | 

You'll find no noticeable difference memory-wise between the two.

The only real difference is that the const method also imposes a type to the value, which can be useful for function overloading or mathematical operations.

share|improve this answer

answered 19 hours ago

Majenko♦Majenko

68.2k43277

share|improve this answer

answered 19 hours ago

Majenko♦Majenko

68.2k43277

answered 19 hours ago

Majenko♦Majenko

68.2k43277

answered 19 hours ago

Majenko♦Majenko

68.2k43277

5

A #define is a preprocessor macro. As Gerben says in his comment, it's just an automated find-and-replace.

If you use it to hold things like C string constants, e.g. #define ERROR_STRING "You messed up, bud!" it could actually cause your program to take more memory, since that same string literal will be duplicated every time you reference it. In that case a static const would be both type-safe AND reduce memory usage.

share|improve this answer

answered 17 hours ago

Duncan CDuncan C

1,7291617

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Would it be duplicated? I mean, isn't it the job of the compiler to ensure "You messed up, bud!" occurs exactly once in the string constant table?
  
  – juhist
  15 hours ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  @juhist: C++ explicitly states that the compiler can choose to duplicate identical strings, or not. Most compilers have a flag to either do this, or not. I think, by default, most compilers do not merge strings, in case the app relies on them being unique.
  
  – Mooing Duck
  14 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @MooingDuck I compiled a program having many std::printf("%pn", "foo"); lines in it. All of them print the same address.
  
  – juhist
  13 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @MooingDuck: While few compilers will guarantee that strings will always be merged, I think the normal reason is that upholding such a guarantee may impose extra build-time costs and complexity, and the savings in storage from merging strings may or may not be worthwhile.
  
  – supercat
  13 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  @juhist: Sure. In your compiler and your compiler flags. With my compiler and my flags, I get different addresses.
  
  – Mooing Duck
  12 hours ago
  

  
  
  
  

 | 
show 2 more comments

5

A #define is a preprocessor macro. As Gerben says in his comment, it's just an automated find-and-replace.

If you use it to hold things like C string constants, e.g. #define ERROR_STRING "You messed up, bud!" it could actually cause your program to take more memory, since that same string literal will be duplicated every time you reference it. In that case a static const would be both type-safe AND reduce memory usage.

share|improve this answer

answered 17 hours ago

Duncan CDuncan C

1,7291617

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Would it be duplicated? I mean, isn't it the job of the compiler to ensure "You messed up, bud!" occurs exactly once in the string constant table?
  
  – juhist
  15 hours ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  @juhist: C++ explicitly states that the compiler can choose to duplicate identical strings, or not. Most compilers have a flag to either do this, or not. I think, by default, most compilers do not merge strings, in case the app relies on them being unique.
  
  – Mooing Duck
  14 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @MooingDuck I compiled a program having many std::printf("%pn", "foo"); lines in it. All of them print the same address.
  
  – juhist
  13 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @MooingDuck: While few compilers will guarantee that strings will always be merged, I think the normal reason is that upholding such a guarantee may impose extra build-time costs and complexity, and the savings in storage from merging strings may or may not be worthwhile.
  
  – supercat
  13 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  @juhist: Sure. In your compiler and your compiler flags. With my compiler and my flags, I get different addresses.
  
  – Mooing Duck
  12 hours ago
  

  
  
  
  

 | 
show 2 more comments

A #define is a preprocessor macro. As Gerben says in his comment, it's just an automated find-and-replace.

If you use it to hold things like C string constants, e.g. #define ERROR_STRING "You messed up, bud!" it could actually cause your program to take more memory, since that same string literal will be duplicated every time you reference it. In that case a static const would be both type-safe AND reduce memory usage.

share|improve this answer

answered 17 hours ago

Duncan CDuncan C

1,7291617

share|improve this answer

answered 17 hours ago

Duncan CDuncan C

1,7291617

answered 17 hours ago

Duncan CDuncan C

1,7291617

answered 17 hours ago

Duncan CDuncan C

1,7291617

3

Theoretically both approaches should consume the same amount of space and will probably result in the same code.

---

However, rather than that meaning "pick any one", what that actually means is that you should turn your attention to other factors when deciding which one to use.

As a rule of thumb you should generally avoid macros because they don't respect scoping rules due to how they work.

This can lead to some serious problems in certain cases.

See Why are preprocessor macros evil and what are the alternatives?.

---

However, I'd like to point out a third option: constexpr.

constexpr uint8_t motorMode1Register = 0x44;

constexpr implies const and is a better expression of the intent that the value should be capable of being evaluated at compile time.
constexpr literally means "this is a constant expression and thus can be evaluated at compile time".

As a result of this, constexpr variables can be used in situations where other variables cannot, such as template parameters.

See Constexpr vs Macros.

share|improve this answer

answered 4 hours ago

PharapPharap

1498

add a comment | 

3

Theoretically both approaches should consume the same amount of space and will probably result in the same code.

---

However, rather than that meaning "pick any one", what that actually means is that you should turn your attention to other factors when deciding which one to use.

As a rule of thumb you should generally avoid macros because they don't respect scoping rules due to how they work.

This can lead to some serious problems in certain cases.

See Why are preprocessor macros evil and what are the alternatives?.

---

However, I'd like to point out a third option: constexpr.

constexpr uint8_t motorMode1Register = 0x44;

constexpr implies const and is a better expression of the intent that the value should be capable of being evaluated at compile time.
constexpr literally means "this is a constant expression and thus can be evaluated at compile time".

As a result of this, constexpr variables can be used in situations where other variables cannot, such as template parameters.

See Constexpr vs Macros.

share|improve this answer

answered 4 hours ago

PharapPharap

1498

add a comment | 

Theoretically both approaches should consume the same amount of space and will probably result in the same code.

---

However, rather than that meaning "pick any one", what that actually means is that you should turn your attention to other factors when deciding which one to use.

As a rule of thumb you should generally avoid macros because they don't respect scoping rules due to how they work.

This can lead to some serious problems in certain cases.

See Why are preprocessor macros evil and what are the alternatives?.

---

However, I'd like to point out a third option: constexpr.

constexpr uint8_t motorMode1Register = 0x44;

constexpr implies const and is a better expression of the intent that the value should be capable of being evaluated at compile time.
constexpr literally means "this is a constant expression and thus can be evaluated at compile time".

As a result of this, constexpr variables can be used in situations where other variables cannot, such as template parameters.

See Constexpr vs Macros.

share|improve this answer

answered 4 hours ago

PharapPharap

1498

constexpr uint8_t motorMode1Register = 0x44;

share|improve this answer

answered 4 hours ago

PharapPharap

1498

answered 4 hours ago

PharapPharap

1498

answered 4 hours ago

PharapPharap

1498