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metalbr@gmail.com
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 Posted: Wed May 16, 2007 4:21 am Post subject: Diferences between "A& operator = (A& a)" and "A& operator = |
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class A{
public:
A operator - (int i){
return A();
}
A& operator = (A& a){ // if A& operator = (const A& a) works!!!
return *this;
}
};
int main(){
A a, b;
a = b; // ok
a = b + 1; // error, why? why const is necessary?
}
Thanks!
--
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John Moeller
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| Posted: Wed May 16, 2007 7:01 am Post subject: Re: Diferences between "A& operator = (A& a)" and "A& operat |
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metalbr (AT) gmail (DOT) com wrote:
| Quote: | class A{
public:
A operator - (int i){
return A();
}
A& operator = (A& a){ // if A& operator = (const A& a) works!!!
return *this;
}
};
int main(){
A a, b;
a = b; // ok
a = b + 1; // error, why? why const is necessary?
}
Thanks!
|
I assume you mean "b - 1", and not "b + 1". The result of b - 1 is a
temporary object. You can't pass a temporary to a function with a
non-const reference, because you can only bind an rvalue to a const
reference. That's why your operator=(const A&) works and operator=(A&)
doesn't.
--
John Moeller
fishcorn (AT) gmail (DOT) com
[ See http://www.gotw.ca/resources/clcm.htm for info about ]
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Abhishek Padmanabh
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| Posted: Wed May 16, 2007 7:04 am Post subject: Re: Diferences between "A& operator = (A& a)" and "A& operat |
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On 16 May, 09:21, "meta...@gmail.com" <meta...@gmail.com> wrote:
| Quote: | int main(){
A a, b;
a = b; // ok
a = b + 1; // error, why? why const is necessary?
|
Because b+1 results in a temporary A to be created, with the result of
addition and then invoke operator= with that temporary as the second
argument to it. That (the temporary) cannot be bound to a reference to
non-const.
--
[ See http://www.gotw.ca/resources/clcm.htm for info about ]
[ comp.lang.c++.moderated. First time posters: Do this! ] |
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