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Claudius
Guest
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 Posted: Sat Feb 25, 2006 3:06 am Post subject: Q: overloaded operator[](int)const: influence on what versio |
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Hello,
I have written a class A with the access operator[](int) overloaded by
a
A-const version which returns an int by value:
------------------------------------------------
#include <iostream>
using namespace std;
class A {
public:
A() {}
int operator [] ( int i ) const {
cout << "int operator[]() called" << endl;
return vec[i];
}
int & operator [] ( int i ) {
cout << "int & operator[]() called" << endl;
return vec[i];
}
private:
int vec[10];
};
------------------------------------------------
However, the compiler uses the non const version even if only an
int-value is read:
------------------------------------------------
A aObj;
int a;
a = aObj[5]; //int & operator []() is used
------------------------------------------------
Is it possible to recognize if the compiler can use the const-version?
I'd like to write a sparse data type, where the []-operator is used in
a
cascaded way, e.g. aObj[5][6][7]. If it was possible to recognize that
a
value is only read rather than written, the use of an access operator
could return null immediatedly in the case of a missed index.
Claudius |
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Ian Collins
Guest
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| Posted: Sat Feb 25, 2006 3:06 am Post subject: Re: Q: overloaded operator[](int)const: influence on what ve |
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Claudius wrote:
| Quote: | Hello,
I have written a class A with the access operator[](int) overloaded by
a
A-const version which returns an int by value:
------------------------------------------------
[snip] |
| Quote: |
Is it possible to recognize if the compiler can use the const-version?
The constness applies to the object, so if you have a const A, the const |
operator will be used.
| Quote: | I'd like to write a sparse data type, where the []-operator is used in
a
cascaded way, e.g. aObj[5][6][7]. If it was possible to recognize that
a
value is only read rather than written, the use of an access operator
could return null immediatedly in the case of a missed index.
You can't use const as a means of determining read or write access. You |
have to return a proxy object with assignment and conversion operators
from operator[] and use those.
--
Ian Collins. |
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Bo Persson
Guest
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| Posted: Sat Feb 25, 2006 10:06 am Post subject: Re: overloaded operator[](int)const: influence on what versi |
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"Claudius" <Claudius.Schnoerr (AT) gmx (DOT) de> skrev i meddelandet
news:1140833361.407141.268750 (AT) p10g2000cwp (DOT) googlegroups.com...
| Quote: | Hello,
I have written a class A with the access operator[](int) overloaded
by
a
A-const version which returns an int by value:
------------------------------------------------
#include <iostream
using namespace std;
class A {
public:
A() {}
int operator [] ( int i ) const {
cout << "int operator[]() called" << endl;
return vec[i];
}
int & operator [] ( int i ) {
cout << "int & operator[]() called" << endl;
return vec[i];
}
private:
int vec[10];
};
------------------------------------------------
However, the compiler uses the non const version even if only an
int-value is read:
------------------------------------------------
A aObj;
int a;
a = aObj[5]; //int & operator []() is used
|
The operator used depends on the object. If aObj is const, the const
version is used. Otherwise the non-const.
| Quote: | ------------------------------------------------
Is it possible to recognize if the compiler can use the
const-version?
|
The compiler cannot know what you intend to do with the return value.
You might save the reference, and do the update later.
Consider this:
int& aref = aObj[5]; // saves the reference, not the value
// lots of other code
aref = 42; // the update comes here!
Bo Persson |
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