const access and non-const access
I have a class which internally owns a vector of foo
class bar {
private:
vector<Foo> foos_;
}
Now I want to design public access to this vector. I am thinking of two versions of the function:
Foo& getFoo(int index) {
// first do size checking, return ref
return foos[index];
}
and
const Foo& getFoo(int index) const {
// first do size checking, return const reference
return foos[index];
}
Any downside of this approach? One obvious downside is I copy the almost identical code simply twice. Is there a better way to do this?
----- edit -----
the second accessor forgets const, updated
c++ design-patterns
asked Nov 24 '18 at 1:43
WhatABeautifulWorldWhatABeautifulWorld
89211123
|
show 9 more comments
I have a class which internally owns a vector of foo
class bar {
private:
vector<Foo> foos_;
}
Now I want to design public access to this vector. I am thinking of two versions of the function:
Foo& getFoo(int index) {
// first do size checking, return ref
return foos[index];
}
and
const Foo& getFoo(int index) const {
// first do size checking, return const reference
return foos[index];
}
Any downside of this approach? One obvious downside is I copy the almost identical code simply twice. Is there a better way to do this?
----- edit -----
the second accessor forgets const, updated
c++ design-patterns
asked Nov 24 '18 at 1:43
WhatABeautifulWorldWhatABeautifulWorld
89211123
2
why not call the non-const function from within the const one?
– dave
Nov 24 '18 at 1:46
2
Because you can't call a non-const method from a const method. C++ does not work this way.
– Sam Varshavchik
Nov 24 '18 at 1:49
Avoid the situation. You have a design flaw.
– Ted Lyngmo
Nov 24 '18 at 1:57
A common technique goes like this:Foo& getFoo(int index) { const bar* constThis = this; return const_cast<Foo&>(constThis->getFoo(index)); }
– Igor Tandetnik
Nov 24 '18 at 1:58
3
@TedLyngmo Then so doesstd::vector. Itsoperatoruses the same approach.
– Igor Tandetnik
Nov 24 '18 at 1:59
|
show 9 more comments
I have a class which internally owns a vector of foo
class bar {
private:
vector<Foo> foos_;
}
Now I want to design public access to this vector. I am thinking of two versions of the function:
Foo& getFoo(int index) {
// first do size checking, return ref
return foos[index];
}
and
const Foo& getFoo(int index) const {
// first do size checking, return const reference
return foos[index];
}
Any downside of this approach? One obvious downside is I copy the almost identical code simply twice. Is there a better way to do this?
----- edit -----
the second accessor forgets const, updated
c++ design-patterns
asked Nov 24 '18 at 1:43
WhatABeautifulWorldWhatABeautifulWorld
89211123
I have a class which internally owns a vector of foo
class bar {
private:
vector<Foo> foos_;
}
Now I want to design public access to this vector. I am thinking of two versions of the function:
Foo& getFoo(int index) {
// first do size checking, return ref
return foos[index];
}
and
const Foo& getFoo(int index) const {
// first do size checking, return const reference
return foos[index];
}
Any downside of this approach? One obvious downside is I copy the almost identical code simply twice. Is there a better way to do this?
----- edit -----
the second accessor forgets const, updated
c++ design-patterns
c++ design-patterns
asked Nov 24 '18 at 1:43
WhatABeautifulWorldWhatABeautifulWorld
89211123
asked Nov 24 '18 at 1:43
WhatABeautifulWorldWhatABeautifulWorld
89211123
edited Nov 24 '18 at 2:03
WhatABeautifulWorld
asked Nov 24 '18 at 1:43
WhatABeautifulWorldWhatABeautifulWorld
89211123
asked Nov 24 '18 at 1:43
WhatABeautifulWorldWhatABeautifulWorld
89211123
asked Nov 24 '18 at 1:43
WhatABeautifulWorldWhatABeautifulWorld
89211123
89211123
2
why not call the non-const function from within the const one?
– dave
Nov 24 '18 at 1:46
2
Because you can't call a non-const method from a const method. C++ does not work this way.
– Sam Varshavchik
Nov 24 '18 at 1:49
Avoid the situation. You have a design flaw.
– Ted Lyngmo
Nov 24 '18 at 1:57
A common technique goes like this:Foo& getFoo(int index) { const bar* constThis = this; return const_cast<Foo&>(constThis->getFoo(index)); }
– Igor Tandetnik
Nov 24 '18 at 1:58
3
@TedLyngmo Then so doesstd::vector. Itsoperatoruses the same approach.
– Igor Tandetnik
Nov 24 '18 at 1:59
|
show 9 more comments
2
why not call the non-const function from within the const one?
– dave
Nov 24 '18 at 1:46
2
Because you can't call a non-const method from a const method. C++ does not work this way.
– Sam Varshavchik
Nov 24 '18 at 1:49
Avoid the situation. You have a design flaw.
– Ted Lyngmo
Nov 24 '18 at 1:57
A common technique goes like this:Foo& getFoo(int index) { const bar* constThis = this; return const_cast<Foo&>(constThis->getFoo(index)); }
– Igor Tandetnik
Nov 24 '18 at 1:58
3
@TedLyngmo Then so doesstd::vector. Itsoperatoruses the same approach.
– Igor Tandetnik
Nov 24 '18 at 1:59
2
2
why not call the non-const function from within the const one?
– dave
Nov 24 '18 at 1:46
why not call the non-const function from within the const one?
– dave
Nov 24 '18 at 1:46
2
2
Because you can't call a non-const method from a const method. C++ does not work this way.
– Sam Varshavchik
Nov 24 '18 at 1:49
Because you can't call a non-const method from a const method. C++ does not work this way.
– Sam Varshavchik
Nov 24 '18 at 1:49
Avoid the situation. You have a design flaw.
– Ted Lyngmo
Nov 24 '18 at 1:57
Avoid the situation. You have a design flaw.
– Ted Lyngmo
Nov 24 '18 at 1:57
A common technique goes like this:
Foo& getFoo(int index) { const bar* constThis = this; return const_cast<Foo&>(constThis->getFoo(index)); }– Igor Tandetnik
Nov 24 '18 at 1:58
A common technique goes like this:
Foo& getFoo(int index) { const bar* constThis = this; return const_cast<Foo&>(constThis->getFoo(index)); }– Igor Tandetnik
Nov 24 '18 at 1:58
3
3
@TedLyngmo Then so does
std::vector. Its operator uses the same approach.– Igor Tandetnik
Nov 24 '18 at 1:59
@TedLyngmo Then so does
std::vector. Its operator uses the same approach.– Igor Tandetnik
Nov 24 '18 at 1:59
|
show 9 more comments
1 Answer
1
active
oldest
votes
Having both const and non-const accessors is somewhat common in C++. There is no language feature to combine the code for both--you really do need to write it twice.
By the way, you don't need to do bounds checking yourself, you can use foos_.at(index) instead of foos_[index] and then you'll have automatic bounds checking.
answered Nov 24 '18 at 3:53
John ZwinckJohn Zwinck
153k17177294
add a comment |
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1 Answer
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votes
Having both const and non-const accessors is somewhat common in C++. There is no language feature to combine the code for both--you really do need to write it twice.
By the way, you don't need to do bounds checking yourself, you can use foos_.at(index) instead of foos_[index] and then you'll have automatic bounds checking.
answered Nov 24 '18 at 3:53
John ZwinckJohn Zwinck
153k17177294
add a comment |
Having both const and non-const accessors is somewhat common in C++. There is no language feature to combine the code for both--you really do need to write it twice.
By the way, you don't need to do bounds checking yourself, you can use foos_.at(index) instead of foos_[index] and then you'll have automatic bounds checking.
answered Nov 24 '18 at 3:53
John ZwinckJohn Zwinck
153k17177294
add a comment |
Having both const and non-const accessors is somewhat common in C++. There is no language feature to combine the code for both--you really do need to write it twice.
By the way, you don't need to do bounds checking yourself, you can use foos_.at(index) instead of foos_[index] and then you'll have automatic bounds checking.
answered Nov 24 '18 at 3:53
John ZwinckJohn Zwinck
153k17177294
Having both const and non-const accessors is somewhat common in C++. There is no language feature to combine the code for both--you really do need to write it twice.
By the way, you don't need to do bounds checking yourself, you can use foos_.at(index) instead of foos_[index] and then you'll have automatic bounds checking.
answered Nov 24 '18 at 3:53
John ZwinckJohn Zwinck
153k17177294
answered Nov 24 '18 at 3:53
John ZwinckJohn Zwinck
153k17177294
answered Nov 24 '18 at 3:53
John ZwinckJohn Zwinck
153k17177294
answered Nov 24 '18 at 3:53
John ZwinckJohn Zwinck
153k17177294
153k17177294
add a comment |
add a comment |
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2
why not call the non-const function from within the const one?
– dave
Nov 24 '18 at 1:46
2
Because you can't call a non-const method from a const method. C++ does not work this way.
– Sam Varshavchik
Nov 24 '18 at 1:49
Avoid the situation. You have a design flaw.
– Ted Lyngmo
Nov 24 '18 at 1:57
A common technique goes like this:
Foo& getFoo(int index) { const bar* constThis = this; return const_cast<Foo&>(constThis->getFoo(index)); }– Igor Tandetnik
Nov 24 '18 at 1:58
3
@TedLyngmo Then so does
std::vector. Itsoperatoruses the same approach.– Igor Tandetnik
Nov 24 '18 at 1:59