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Adding extra constness causes compiler error

Sell me on const correctnessHow come a non-const reference cannot bind to a temporary object?Why does flowing off the end of a non-void function without returning a value not produce a compiler error?Defining static const integer members in class definitionextra qualification error in C++int a[] = 1,2,; Weird comma allowed. Any particular reason?Why is it impossible to build a compiler that can determine if a C++ function will change the value of a particular variable?Why “const” is not causing compile errorReplacing a 32-bit loop counter with 64-bit introduces crazy performance deviationsWhy is a public const method not called when the non-const one is private?

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7

This compiles fine on GCC 8.2:

class M

public:
 const Pointer* getPointer() const return _ptr;
private:
 Pointer* _ptrnullptr;
;

but when I add another const to the function:

class M

public:
 const Pointer* const getPointer() const return _ptr;
private:
 Pointer* _ptrnullptr;
;

I get the compiler error:

error: type qualifiers ignored on function return type [-Werror=ignored-qualifiers]

Why would it not let me add additional const-ness? Since when was extra const bad?

c++

share|improve this question

asked May 31 at 13:42

user997112user997112

10.3k30109227

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  It's not bad but it is unneeded. You return by value so the const is superfluous.
  
  – NathanOliver
  May 31 at 13:46
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  How are you calling getPointer ()?
  
  – Steve
  May 31 at 13:48
  

  
  
  
  

add a comment | 

7

This compiles fine on GCC 8.2:

class M

public:
 const Pointer* getPointer() const return _ptr;
private:
 Pointer* _ptrnullptr;
;

but when I add another const to the function:

class M

public:
 const Pointer* const getPointer() const return _ptr;
private:
 Pointer* _ptrnullptr;
;

I get the compiler error:

error: type qualifiers ignored on function return type [-Werror=ignored-qualifiers]

Why would it not let me add additional const-ness? Since when was extra const bad?

c++

share|improve this question

asked May 31 at 13:42

user997112user997112

10.3k30109227

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  It's not bad but it is unneeded. You return by value so the const is superfluous.
  
  – NathanOliver
  May 31 at 13:46
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  How are you calling getPointer ()?
  
  – Steve
  May 31 at 13:48
  

  
  
  
  

add a comment | 

This compiles fine on GCC 8.2:

class M

public:
 const Pointer* getPointer() const return _ptr;
private:
 Pointer* _ptrnullptr;
;

but when I add another const to the function:

class M

public:
 const Pointer* const getPointer() const return _ptr;
private:
 Pointer* _ptrnullptr;
;

I get the compiler error:

error: type qualifiers ignored on function return type [-Werror=ignored-qualifiers]

Why would it not let me add additional const-ness? Since when was extra const bad?

c++

share|improve this question

asked May 31 at 13:42

user997112user997112

10.3k30109227

class M

public:
 const Pointer* getPointer() const return _ptr;
private:
 Pointer* _ptrnullptr;
;

class M

public:
 const Pointer* const getPointer() const return _ptr;
private:
 Pointer* _ptrnullptr;
;

error: type qualifiers ignored on function return type [-Werror=ignored-qualifiers]

share|improve this question

asked May 31 at 13:42

user997112user997112

10.3k30109227

share|improve this question

asked May 31 at 13:42

user997112user997112

10.3k30109227

asked May 31 at 13:42

user997112user997112

10.3k30109227

asked May 31 at 13:42

user997112user997112

10.3k30109227

2 Answers
2

active

oldest

votes

7

Because returning a const something by value like here makes no difference with or without.

For example:

const int GetMyInt()

 int k = 42;
 return k;


//later..
int ret = GetMyInt();
// modify ret.

Because the returned value from GetMyInt will be copied into ret anyway (not taking (N)RVO into account), having GetMyInt return const makes no difference.

Normally this is a warning because it's superfluous code but -Werror turns every warning into an error so there's that.

share|improve this answer

edited May 31 at 13:49

answered May 31 at 13:45

Sombrero ChickenSombrero Chicken

26.1k33483

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  I understand your example, but in my example I want to return a pointer which cannot be modified and can only call const methods?
  
  – user997112
  May 31 at 13:47
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  
  @user997112 Your pointer is returned by value so it is copied on the calling side, so saying which cannot be modified can't be done from the return type. The caller decides to where this return value is copied.
  
  – Sombrero Chicken
  May 31 at 13:47
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  @user997112 But nothing can ever modify the pointer returned even when it's not const. You cannot do e.g. getPointer() += 4. And at the same time, nothing can prevent you from assigning that pointer to a modifiable one, even if it was returned as const.
  
  – Angew
  May 31 at 13:48
  
  

  
  
  
  

add a comment | 

5

The const qualifier has no effect in this position, since the returned value is a prvalue of non-class type and therefore cannot be modified anyway.

Notice that the compiler message says -Werror=, meaning that it's normally a warning (so the code is not wrong, but warning-worthy). It has been turned into an error by your compilation settings.

share|improve this answer

answered May 31 at 13:46

AngewAngew

138k11270362

add a comment | 

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7

Because returning a const something by value like here makes no difference with or without.

For example:

const int GetMyInt()

 int k = 42;
 return k;


//later..
int ret = GetMyInt();
// modify ret.

Because the returned value from GetMyInt will be copied into ret anyway (not taking (N)RVO into account), having GetMyInt return const makes no difference.

Normally this is a warning because it's superfluous code but -Werror turns every warning into an error so there's that.

share|improve this answer

edited May 31 at 13:49

answered May 31 at 13:45

Sombrero ChickenSombrero Chicken

26.1k33483

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  I understand your example, but in my example I want to return a pointer which cannot be modified and can only call const methods?
  
  – user997112
  May 31 at 13:47
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  
  @user997112 Your pointer is returned by value so it is copied on the calling side, so saying which cannot be modified can't be done from the return type. The caller decides to where this return value is copied.
  
  – Sombrero Chicken
  May 31 at 13:47
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  @user997112 But nothing can ever modify the pointer returned even when it's not const. You cannot do e.g. getPointer() += 4. And at the same time, nothing can prevent you from assigning that pointer to a modifiable one, even if it was returned as const.
  
  – Angew
  May 31 at 13:48
  
  

  
  
  
  

add a comment | 

7

Because returning a const something by value like here makes no difference with or without.

For example:

const int GetMyInt()

 int k = 42;
 return k;


//later..
int ret = GetMyInt();
// modify ret.

Because the returned value from GetMyInt will be copied into ret anyway (not taking (N)RVO into account), having GetMyInt return const makes no difference.

Normally this is a warning because it's superfluous code but -Werror turns every warning into an error so there's that.

share|improve this answer

edited May 31 at 13:49

answered May 31 at 13:45

Sombrero ChickenSombrero Chicken

26.1k33483

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  I understand your example, but in my example I want to return a pointer which cannot be modified and can only call const methods?
  
  – user997112
  May 31 at 13:47
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  
  @user997112 Your pointer is returned by value so it is copied on the calling side, so saying which cannot be modified can't be done from the return type. The caller decides to where this return value is copied.
  
  – Sombrero Chicken
  May 31 at 13:47
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  @user997112 But nothing can ever modify the pointer returned even when it's not const. You cannot do e.g. getPointer() += 4. And at the same time, nothing can prevent you from assigning that pointer to a modifiable one, even if it was returned as const.
  
  – Angew
  May 31 at 13:48
  
  

  
  
  
  

add a comment | 

Because returning a const something by value like here makes no difference with or without.

For example:

const int GetMyInt()

 int k = 42;
 return k;


//later..
int ret = GetMyInt();
// modify ret.

Because the returned value from GetMyInt will be copied into ret anyway (not taking (N)RVO into account), having GetMyInt return const makes no difference.

Normally this is a warning because it's superfluous code but -Werror turns every warning into an error so there's that.

share|improve this answer

edited May 31 at 13:49

answered May 31 at 13:45

Sombrero ChickenSombrero Chicken

26.1k33483

const int GetMyInt()

 int k = 42;
 return k;


//later..
int ret = GetMyInt();
// modify ret.

share|improve this answer

edited May 31 at 13:49

answered May 31 at 13:45

Sombrero ChickenSombrero Chicken

26.1k33483

answered May 31 at 13:45

Sombrero ChickenSombrero Chicken

26.1k33483

answered May 31 at 13:45

Sombrero ChickenSombrero Chicken

26.1k33483

5

The const qualifier has no effect in this position, since the returned value is a prvalue of non-class type and therefore cannot be modified anyway.

Notice that the compiler message says -Werror=, meaning that it's normally a warning (so the code is not wrong, but warning-worthy). It has been turned into an error by your compilation settings.

share|improve this answer

answered May 31 at 13:46

AngewAngew

138k11270362

add a comment | 

5

The const qualifier has no effect in this position, since the returned value is a prvalue of non-class type and therefore cannot be modified anyway.

Notice that the compiler message says -Werror=, meaning that it's normally a warning (so the code is not wrong, but warning-worthy). It has been turned into an error by your compilation settings.

share|improve this answer

answered May 31 at 13:46

AngewAngew

138k11270362

add a comment | 

The const qualifier has no effect in this position, since the returned value is a prvalue of non-class type and therefore cannot be modified anyway.

Notice that the compiler message says -Werror=, meaning that it's normally a warning (so the code is not wrong, but warning-worthy). It has been turned into an error by your compilation settings.

share|improve this answer

answered May 31 at 13:46

AngewAngew

138k11270362

share|improve this answer

answered May 31 at 13:46

AngewAngew

138k11270362

answered May 31 at 13:46

AngewAngew

138k11270362

answered May 31 at 13:46

AngewAngew

138k11270362