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Diferences between "A& operator = (A& a)" and "A& operator =

 
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metalbr@gmail.com
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PostPosted: Wed May 16, 2007 4:21 am    Post subject: Diferences between "A& operator = (A& a)" and "A& operator = Reply with 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!


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John Moeller
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PostPosted: Wed May 16, 2007 7:01 am    Post subject: Re: Diferences between "A& operator = (A& a)" and "A& operat Reply with quote


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.

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John Moeller
fishcorn (AT) gmail (DOT) com

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Abhishek Padmanabh
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PostPosted: Wed May 16, 2007 7:04 am    Post subject: Re: Diferences between "A& operator = (A& a)" and "A& operat Reply with quote


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.


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