Function's const meaning for the return data type
What is the real const meaning for the 2nd declaration foo:B() ?
int foo::A() const {
return m_var;
}
int const foo::B() {
return m_var;
}
For the 1st declaration I know for sure that it "protects" the member variable ie m_var.
But what the whole meaning of the 2nd declaration, which it just returns a constant int to the caller probably non-constant variable ? I mean does this make sense for any reason ?
c++ function const class-design
edited Nov 18 '18 at 11:39
Christophe
39.3k43576
asked Nov 18 '18 at 9:16
MaverickMaverick
396115
|
show 6 more comments
What is the real const meaning for the 2nd declaration foo:B() ?
int foo::A() const {
return m_var;
}
int const foo::B() {
return m_var;
}
For the 1st declaration I know for sure that it "protects" the member variable ie m_var.
But what the whole meaning of the 2nd declaration, which it just returns a constant int to the caller probably non-constant variable ? I mean does this make sense for any reason ?
c++ function const class-design
edited Nov 18 '18 at 11:39
Christophe
39.3k43576
asked Nov 18 '18 at 9:16
MaverickMaverick
396115
4
Returning aconstvalue makes no sense at all.
– Some programmer dude
Nov 18 '18 at 9:19
1
Not only that, the type of the expressionf.B()for a made upfoo f;isint. So I would say it makes negative sense for fundamentals value types. Pre C++11 there was marginal sense to do it for some user defined types. Now there is none.
– StoryTeller
Nov 18 '18 at 9:22
with or without theconst,f.B() = 5;is an error, so its meaningless
– user463035818
Nov 18 '18 at 9:33
that is the wrong dupe. OP has two examples. First is a const method, the question is about the second. Voting to reopen
– user463035818
Nov 18 '18 at 9:44
@user463035818 I seriously disagree. The two dupes perfectly explained what theconstkeyword does for both of the code examples.
– πάντα ῥεῖ
Nov 18 '18 at 9:48
|
show 6 more comments
What is the real const meaning for the 2nd declaration foo:B() ?
int foo::A() const {
return m_var;
}
int const foo::B() {
return m_var;
}
For the 1st declaration I know for sure that it "protects" the member variable ie m_var.
But what the whole meaning of the 2nd declaration, which it just returns a constant int to the caller probably non-constant variable ? I mean does this make sense for any reason ?
c++ function const class-design
edited Nov 18 '18 at 11:39
Christophe
39.3k43576
asked Nov 18 '18 at 9:16
MaverickMaverick
396115
What is the real const meaning for the 2nd declaration foo:B() ?
int foo::A() const {
return m_var;
}
int const foo::B() {
return m_var;
}
For the 1st declaration I know for sure that it "protects" the member variable ie m_var.
But what the whole meaning of the 2nd declaration, which it just returns a constant int to the caller probably non-constant variable ? I mean does this make sense for any reason ?
c++ function const class-design
c++ function const class-design
edited Nov 18 '18 at 11:39
Christophe
39.3k43576
asked Nov 18 '18 at 9:16
MaverickMaverick
396115
edited Nov 18 '18 at 11:39
Christophe
39.3k43576
asked Nov 18 '18 at 9:16
MaverickMaverick
396115
edited Nov 18 '18 at 11:39
Christophe
39.3k43576
edited Nov 18 '18 at 11:39
Christophe
39.3k43576
edited Nov 18 '18 at 11:39
Christophe
39.3k43576
39.3k43576
asked Nov 18 '18 at 9:16
MaverickMaverick
396115
asked Nov 18 '18 at 9:16
MaverickMaverick
396115
asked Nov 18 '18 at 9:16
MaverickMaverick
396115
396115
4
Returning aconstvalue makes no sense at all.
– Some programmer dude
Nov 18 '18 at 9:19
1
Not only that, the type of the expressionf.B()for a made upfoo f;isint. So I would say it makes negative sense for fundamentals value types. Pre C++11 there was marginal sense to do it for some user defined types. Now there is none.
– StoryTeller
Nov 18 '18 at 9:22
with or without theconst,f.B() = 5;is an error, so its meaningless
– user463035818
Nov 18 '18 at 9:33
that is the wrong dupe. OP has two examples. First is a const method, the question is about the second. Voting to reopen
– user463035818
Nov 18 '18 at 9:44
@user463035818 I seriously disagree. The two dupes perfectly explained what theconstkeyword does for both of the code examples.
– πάντα ῥεῖ
Nov 18 '18 at 9:48
|
show 6 more comments
4
Returning aconstvalue makes no sense at all.
– Some programmer dude
Nov 18 '18 at 9:19
1
Not only that, the type of the expressionf.B()for a made upfoo f;isint. So I would say it makes negative sense for fundamentals value types. Pre C++11 there was marginal sense to do it for some user defined types. Now there is none.
– StoryTeller
Nov 18 '18 at 9:22
with or without theconst,f.B() = 5;is an error, so its meaningless
– user463035818
Nov 18 '18 at 9:33
that is the wrong dupe. OP has two examples. First is a const method, the question is about the second. Voting to reopen
– user463035818
Nov 18 '18 at 9:44
@user463035818 I seriously disagree. The two dupes perfectly explained what theconstkeyword does for both of the code examples.
– πάντα ῥεῖ
Nov 18 '18 at 9:48
4
4
Returning a
const value makes no sense at all.– Some programmer dude
Nov 18 '18 at 9:19
Returning a
const value makes no sense at all.– Some programmer dude
Nov 18 '18 at 9:19
1
1
Not only that, the type of the expression
f.B() for a made up foo f; is int. So I would say it makes negative sense for fundamentals value types. Pre C++11 there was marginal sense to do it for some user defined types. Now there is none.– StoryTeller
Nov 18 '18 at 9:22
Not only that, the type of the expression
f.B() for a made up foo f; is int. So I would say it makes negative sense for fundamentals value types. Pre C++11 there was marginal sense to do it for some user defined types. Now there is none.– StoryTeller
Nov 18 '18 at 9:22
with or without the
const, f.B() = 5; is an error, so its meaningless– user463035818
Nov 18 '18 at 9:33
with or without the
const, f.B() = 5; is an error, so its meaningless– user463035818
Nov 18 '18 at 9:33
that is the wrong dupe. OP has two examples. First is a const method, the question is about the second. Voting to reopen
– user463035818
Nov 18 '18 at 9:44
that is the wrong dupe. OP has two examples. First is a const method, the question is about the second. Voting to reopen
– user463035818
Nov 18 '18 at 9:44
@user463035818 I seriously disagree. The two dupes perfectly explained what the
const keyword does for both of the code examples.– πάντα ῥεῖ
Nov 18 '18 at 9:48
@user463035818 I seriously disagree. The two dupes perfectly explained what the
const keyword does for both of the code examples.– πάντα ῥεῖ
Nov 18 '18 at 9:48
|
show 6 more comments
1 Answer
1
active
oldest
votes
Case 1: The const after the function signature says that the function will not change the object. So you can use it on const objects (or with pointer to const or a const reference).
Case 2: The const before the function name is indeed about the return type. You are completely right: in practice it doesn't change anything for the object, since the return is in this snippet done by value, and this value in a temp that cannot be changed (e.g. a ++ or a -- would not be valid anyway because there's no lvalue).
Case 3: The const in the return type would make more sense with the return of a pointer to const or a const reference. In this case it would prevent the object state to be changed from outside.
Here a summary:
class foo {
public:
int A() const { // const function
return m_var;
}
int const B() { // non const function, but const return type
return m_var;
}
int const& C() const { // non const function, but const reference return type
return m_var;
}
private:
int m_var;
};
int main() {
const foo x{};
x.A(); // ok
//x.B(); // not ok -> function B() doesn't guarantee to leave x unchanged.
x.C(); // ok
const int& y = x.C(); // ok (y will not alter m_var.
//int& z = x.C(); // not ok since z is not const
return 0;
}
online demo
Case 4: (thanks to HolyBlackCat for pointing it out). What make no difference for scalars in case 2 could perfectly make sense for classes. Suppose m_var would be of class bar :
class bar {
public:
void change_it() {}
void read_it() const {}
};
Then the const return value would make a difference:
foo u{};
u.B(); // ok
u.B().read_it(); // ok
u.B().change_it(); // not ok because of constness of B().
online demo
answered Nov 18 '18 at 10:39
ChristopheChristophe
39.3k43576
If I remember correctly,constindeed has no effect when returning arithmetic types. But that's different for classes (e.g. const class instances can't be meaningfully moved from). Though I don't remember the details.
– HolyBlackCat
Nov 18 '18 at 10:45
@HolyBlackCat Thanks for pointing it out ! Indeed, it would make sense for a class. I've edited my answer and gave you credit for this additional case :-)
– Christophe
Nov 18 '18 at 11:08
Maybe it's worth mentioning that const class return values preventsmoveing from them, so their usage is discouraged.
– geza
Nov 18 '18 at 11:17
add a comment |
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
Case 1: The const after the function signature says that the function will not change the object. So you can use it on const objects (or with pointer to const or a const reference).
Case 2: The const before the function name is indeed about the return type. You are completely right: in practice it doesn't change anything for the object, since the return is in this snippet done by value, and this value in a temp that cannot be changed (e.g. a ++ or a -- would not be valid anyway because there's no lvalue).
Case 3: The const in the return type would make more sense with the return of a pointer to const or a const reference. In this case it would prevent the object state to be changed from outside.
Here a summary:
class foo {
public:
int A() const { // const function
return m_var;
}
int const B() { // non const function, but const return type
return m_var;
}
int const& C() const { // non const function, but const reference return type
return m_var;
}
private:
int m_var;
};
int main() {
const foo x{};
x.A(); // ok
//x.B(); // not ok -> function B() doesn't guarantee to leave x unchanged.
x.C(); // ok
const int& y = x.C(); // ok (y will not alter m_var.
//int& z = x.C(); // not ok since z is not const
return 0;
}
online demo
Case 4: (thanks to HolyBlackCat for pointing it out). What make no difference for scalars in case 2 could perfectly make sense for classes. Suppose m_var would be of class bar :
class bar {
public:
void change_it() {}
void read_it() const {}
};
Then the const return value would make a difference:
foo u{};
u.B(); // ok
u.B().read_it(); // ok
u.B().change_it(); // not ok because of constness of B().
online demo
answered Nov 18 '18 at 10:39
ChristopheChristophe
39.3k43576
If I remember correctly,constindeed has no effect when returning arithmetic types. But that's different for classes (e.g. const class instances can't be meaningfully moved from). Though I don't remember the details.
– HolyBlackCat
Nov 18 '18 at 10:45
@HolyBlackCat Thanks for pointing it out ! Indeed, it would make sense for a class. I've edited my answer and gave you credit for this additional case :-)
– Christophe
Nov 18 '18 at 11:08
Maybe it's worth mentioning that const class return values preventsmoveing from them, so their usage is discouraged.
– geza
Nov 18 '18 at 11:17
add a comment |
Case 1: The const after the function signature says that the function will not change the object. So you can use it on const objects (or with pointer to const or a const reference).
Case 2: The const before the function name is indeed about the return type. You are completely right: in practice it doesn't change anything for the object, since the return is in this snippet done by value, and this value in a temp that cannot be changed (e.g. a ++ or a -- would not be valid anyway because there's no lvalue).
Case 3: The const in the return type would make more sense with the return of a pointer to const or a const reference. In this case it would prevent the object state to be changed from outside.
Here a summary:
class foo {
public:
int A() const { // const function
return m_var;
}
int const B() { // non const function, but const return type
return m_var;
}
int const& C() const { // non const function, but const reference return type
return m_var;
}
private:
int m_var;
};
int main() {
const foo x{};
x.A(); // ok
//x.B(); // not ok -> function B() doesn't guarantee to leave x unchanged.
x.C(); // ok
const int& y = x.C(); // ok (y will not alter m_var.
//int& z = x.C(); // not ok since z is not const
return 0;
}
online demo
Case 4: (thanks to HolyBlackCat for pointing it out). What make no difference for scalars in case 2 could perfectly make sense for classes. Suppose m_var would be of class bar :
class bar {
public:
void change_it() {}
void read_it() const {}
};
Then the const return value would make a difference:
foo u{};
u.B(); // ok
u.B().read_it(); // ok
u.B().change_it(); // not ok because of constness of B().
online demo
answered Nov 18 '18 at 10:39
ChristopheChristophe
39.3k43576
If I remember correctly,constindeed has no effect when returning arithmetic types. But that's different for classes (e.g. const class instances can't be meaningfully moved from). Though I don't remember the details.
– HolyBlackCat
Nov 18 '18 at 10:45
@HolyBlackCat Thanks for pointing it out ! Indeed, it would make sense for a class. I've edited my answer and gave you credit for this additional case :-)
– Christophe
Nov 18 '18 at 11:08
Maybe it's worth mentioning that const class return values preventsmoveing from them, so their usage is discouraged.
– geza
Nov 18 '18 at 11:17
add a comment |
Case 1: The const after the function signature says that the function will not change the object. So you can use it on const objects (or with pointer to const or a const reference).
Case 2: The const before the function name is indeed about the return type. You are completely right: in practice it doesn't change anything for the object, since the return is in this snippet done by value, and this value in a temp that cannot be changed (e.g. a ++ or a -- would not be valid anyway because there's no lvalue).
Case 3: The const in the return type would make more sense with the return of a pointer to const or a const reference. In this case it would prevent the object state to be changed from outside.
Here a summary:
class foo {
public:
int A() const { // const function
return m_var;
}
int const B() { // non const function, but const return type
return m_var;
}
int const& C() const { // non const function, but const reference return type
return m_var;
}
private:
int m_var;
};
int main() {
const foo x{};
x.A(); // ok
//x.B(); // not ok -> function B() doesn't guarantee to leave x unchanged.
x.C(); // ok
const int& y = x.C(); // ok (y will not alter m_var.
//int& z = x.C(); // not ok since z is not const
return 0;
}
online demo
Case 4: (thanks to HolyBlackCat for pointing it out). What make no difference for scalars in case 2 could perfectly make sense for classes. Suppose m_var would be of class bar :
class bar {
public:
void change_it() {}
void read_it() const {}
};
Then the const return value would make a difference:
foo u{};
u.B(); // ok
u.B().read_it(); // ok
u.B().change_it(); // not ok because of constness of B().
online demo
answered Nov 18 '18 at 10:39
ChristopheChristophe
39.3k43576
Case 1: The const after the function signature says that the function will not change the object. So you can use it on const objects (or with pointer to const or a const reference).
Case 2: The const before the function name is indeed about the return type. You are completely right: in practice it doesn't change anything for the object, since the return is in this snippet done by value, and this value in a temp that cannot be changed (e.g. a ++ or a -- would not be valid anyway because there's no lvalue).
Case 3: The const in the return type would make more sense with the return of a pointer to const or a const reference. In this case it would prevent the object state to be changed from outside.
Here a summary:
class foo {
public:
int A() const { // const function
return m_var;
}
int const B() { // non const function, but const return type
return m_var;
}
int const& C() const { // non const function, but const reference return type
return m_var;
}
private:
int m_var;
};
int main() {
const foo x{};
x.A(); // ok
//x.B(); // not ok -> function B() doesn't guarantee to leave x unchanged.
x.C(); // ok
const int& y = x.C(); // ok (y will not alter m_var.
//int& z = x.C(); // not ok since z is not const
return 0;
}
online demo
Case 4: (thanks to HolyBlackCat for pointing it out). What make no difference for scalars in case 2 could perfectly make sense for classes. Suppose m_var would be of class bar :
class bar {
public:
void change_it() {}
void read_it() const {}
};
Then the const return value would make a difference:
foo u{};
u.B(); // ok
u.B().read_it(); // ok
u.B().change_it(); // not ok because of constness of B().
online demo
answered Nov 18 '18 at 10:39
ChristopheChristophe
39.3k43576
edited Nov 18 '18 at 11:06
answered Nov 18 '18 at 10:39
ChristopheChristophe
39.3k43576
answered Nov 18 '18 at 10:39
ChristopheChristophe
39.3k43576
answered Nov 18 '18 at 10:39
ChristopheChristophe
39.3k43576
39.3k43576
If I remember correctly,constindeed has no effect when returning arithmetic types. But that's different for classes (e.g. const class instances can't be meaningfully moved from). Though I don't remember the details.
– HolyBlackCat
Nov 18 '18 at 10:45
@HolyBlackCat Thanks for pointing it out ! Indeed, it would make sense for a class. I've edited my answer and gave you credit for this additional case :-)
– Christophe
Nov 18 '18 at 11:08
Maybe it's worth mentioning that const class return values preventsmoveing from them, so their usage is discouraged.
– geza
Nov 18 '18 at 11:17
add a comment |
If I remember correctly,constindeed has no effect when returning arithmetic types. But that's different for classes (e.g. const class instances can't be meaningfully moved from). Though I don't remember the details.
– HolyBlackCat
Nov 18 '18 at 10:45
@HolyBlackCat Thanks for pointing it out ! Indeed, it would make sense for a class. I've edited my answer and gave you credit for this additional case :-)
– Christophe
Nov 18 '18 at 11:08
Maybe it's worth mentioning that const class return values preventsmoveing from them, so their usage is discouraged.
– geza
Nov 18 '18 at 11:17
If I remember correctly,
const indeed has no effect when returning arithmetic types. But that's different for classes (e.g. const class instances can't be meaningfully moved from). Though I don't remember the details.– HolyBlackCat
Nov 18 '18 at 10:45
If I remember correctly,
const indeed has no effect when returning arithmetic types. But that's different for classes (e.g. const class instances can't be meaningfully moved from). Though I don't remember the details.– HolyBlackCat
Nov 18 '18 at 10:45
@HolyBlackCat Thanks for pointing it out ! Indeed, it would make sense for a class. I've edited my answer and gave you credit for this additional case :-)
– Christophe
Nov 18 '18 at 11:08
@HolyBlackCat Thanks for pointing it out ! Indeed, it would make sense for a class. I've edited my answer and gave you credit for this additional case :-)
– Christophe
Nov 18 '18 at 11:08
Maybe it's worth mentioning that const class return values prevents
moveing from them, so their usage is discouraged.– geza
Nov 18 '18 at 11:17
Maybe it's worth mentioning that const class return values prevents
moveing from them, so their usage is discouraged.– geza
Nov 18 '18 at 11:17
add a comment |
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4
Returning a
constvalue makes no sense at all.– Some programmer dude
Nov 18 '18 at 9:19
1
Not only that, the type of the expression
f.B()for a made upfoo f;isint. So I would say it makes negative sense for fundamentals value types. Pre C++11 there was marginal sense to do it for some user defined types. Now there is none.– StoryTeller
Nov 18 '18 at 9:22
with or without the
const,f.B() = 5;is an error, so its meaningless– user463035818
Nov 18 '18 at 9:33
that is the wrong dupe. OP has two examples. First is a const method, the question is about the second. Voting to reopen
– user463035818
Nov 18 '18 at 9:44
@user463035818 I seriously disagree. The two dupes perfectly explained what the
constkeyword does for both of the code examples.– πάντα ῥεῖ
Nov 18 '18 at 9:48