C++ Operator overloaded using friend function. Attempt to add multiple objects failed

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Why compiler shows 'error' in 2nd case? (I have given link to full program)
Why do I have to use const keyword?

1st case:

friend Complex operator + (const Complex &,const Complex &);



Complex c5 = c1+c2+c3+c4; 

2nd case:

friend Complex operator + ( Complex &, Complex &); 



Complex c5 = c1+c2+c3+c4; 

1st case Full Program - I get proper Output

2nd case Full Program - error: no match for 'operator+'

c++ operator-overloading overloading

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edited Nov 19 at 12:16

asked Nov 19 at 12:11

Lelouch Yagami

144

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up vote
0
down vote

favorite

Why compiler shows 'error' in 2nd case? (I have given link to full program)
Why do I have to use const keyword?

1st case:

friend Complex operator + (const Complex &,const Complex &);



Complex c5 = c1+c2+c3+c4; 

2nd case:

friend Complex operator + ( Complex &, Complex &); 



Complex c5 = c1+c2+c3+c4; 

1st case Full Program - I get proper Output

2nd case Full Program - error: no match for 'operator+'

c++ operator-overloading overloading

share|improve this question

edited Nov 19 at 12:16

asked Nov 19 at 12:11

Lelouch Yagami

144

add a comment | 

up vote
0
down vote

favorite

up vote
0
down vote

favorite

Why compiler shows 'error' in 2nd case? (I have given link to full program)
Why do I have to use const keyword?

1st case:

friend Complex operator + (const Complex &,const Complex &);



Complex c5 = c1+c2+c3+c4; 

2nd case:

friend Complex operator + ( Complex &, Complex &); 



Complex c5 = c1+c2+c3+c4; 

1st case Full Program - I get proper Output

2nd case Full Program - error: no match for 'operator+'

c++ operator-overloading overloading

share|improve this question

edited Nov 19 at 12:16

asked Nov 19 at 12:11

Lelouch Yagami

144

Why compiler shows 'error' in 2nd case? (I have given link to full program)
Why do I have to use const keyword?

1st case:

friend Complex operator + (const Complex &,const Complex &);



Complex c5 = c1+c2+c3+c4; 

2nd case:

friend Complex operator + ( Complex &, Complex &); 



Complex c5 = c1+c2+c3+c4; 

1st case Full Program - I get proper Output

2nd case Full Program - error: no match for 'operator+'

c++ operator-overloading overloading

c++ operator-overloading overloading

share|improve this question

edited Nov 19 at 12:16

asked Nov 19 at 12:11

Lelouch Yagami

144

share|improve this question

edited Nov 19 at 12:16

asked Nov 19 at 12:11

Lelouch Yagami

144

share|improve this question

share|improve this question

edited Nov 19 at 12:16

edited Nov 19 at 12:16

edited Nov 19 at 12:16

asked Nov 19 at 12:11

Lelouch Yagami

144

asked Nov 19 at 12:11

Lelouch Yagami

144

asked Nov 19 at 12:11

Lelouch Yagami

144

144

add a comment | 

add a comment | 

3 Answers
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2
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Complex& will not bind to a temporary, Complex const& will.

Each + returns a temporary.

As a general rule, you want:

friend Complex operator + (Complex,const Complex &);

but here two const& will do.

share|improve this answer

answered Nov 19 at 12:23

Yakk - Adam Nevraumont

178k19185363

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up vote
1
down vote

Temporaries dont bind to non-const references. When you write this

auto c3 = c2 + c1 + c0; 

then first c1+c0 will be computed, and the result passed to c2.operator+(). When the operator is declared to take a Complex& then you cannot pass a temporary, when it takes const Complex& then you can. Passing a temporary when a non-const reference is expected is a logic error in 90% of the cases, hence it is forbidden.

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edited Nov 19 at 12:34

answered Nov 19 at 12:23

user463035818

15.7k42561

add a comment | 

up vote
1
down vote

The expression c1+c2+c3+c4 is parsed and evaluated as if

Complex c5 = operator+(c1, operator+(c2, operator+(c3, c4)));

operator+(c3, c4) builds and returns a temporary Complex object: an rvalue.

C++ forbids an rvalue to be bound to a non-const lvalue-reference.

But operator+(Complex&, Complex&) takes a non-const lvalue-reference. Hence the error message.

On the other hand, operator+(Complex const&, Complex const&) takes references to const lvalues.

share|improve this answer

edited 10 hours ago

answered Nov 19 at 12:23

YSC

19.5k34591

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3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
2
down vote

Complex& will not bind to a temporary, Complex const& will.

Each + returns a temporary.

As a general rule, you want:

friend Complex operator + (Complex,const Complex &);

but here two const& will do.

share|improve this answer

answered Nov 19 at 12:23

Yakk - Adam Nevraumont

178k19185363

add a comment | 

up vote
2
down vote

Complex& will not bind to a temporary, Complex const& will.

Each + returns a temporary.

As a general rule, you want:

friend Complex operator + (Complex,const Complex &);

but here two const& will do.

share|improve this answer

answered Nov 19 at 12:23

Yakk - Adam Nevraumont

178k19185363

add a comment | 

up vote
2
down vote

up vote
2
down vote

Complex& will not bind to a temporary, Complex const& will.

Each + returns a temporary.

As a general rule, you want:

friend Complex operator + (Complex,const Complex &);

but here two const& will do.

share|improve this answer

answered Nov 19 at 12:23

Yakk - Adam Nevraumont

178k19185363

Complex& will not bind to a temporary, Complex const& will.

Each + returns a temporary.

As a general rule, you want:

friend Complex operator + (Complex,const Complex &);

but here two const& will do.

share|improve this answer

answered Nov 19 at 12:23

Yakk - Adam Nevraumont

178k19185363

share|improve this answer

share|improve this answer

answered Nov 19 at 12:23

Yakk - Adam Nevraumont

178k19185363

answered Nov 19 at 12:23

Yakk - Adam Nevraumont

178k19185363

answered Nov 19 at 12:23

Yakk - Adam Nevraumont

178k19185363

178k19185363

add a comment | 

add a comment | 

up vote
1
down vote

Temporaries dont bind to non-const references. When you write this

auto c3 = c2 + c1 + c0; 

then first c1+c0 will be computed, and the result passed to c2.operator+(). When the operator is declared to take a Complex& then you cannot pass a temporary, when it takes const Complex& then you can. Passing a temporary when a non-const reference is expected is a logic error in 90% of the cases, hence it is forbidden.

share|improve this answer

edited Nov 19 at 12:34

answered Nov 19 at 12:23

user463035818

15.7k42561

add a comment | 

up vote
1
down vote

Temporaries dont bind to non-const references. When you write this

auto c3 = c2 + c1 + c0; 

then first c1+c0 will be computed, and the result passed to c2.operator+(). When the operator is declared to take a Complex& then you cannot pass a temporary, when it takes const Complex& then you can. Passing a temporary when a non-const reference is expected is a logic error in 90% of the cases, hence it is forbidden.

share|improve this answer

edited Nov 19 at 12:34

answered Nov 19 at 12:23

user463035818

15.7k42561

add a comment | 

up vote
1
down vote

up vote
1
down vote

Temporaries dont bind to non-const references. When you write this

auto c3 = c2 + c1 + c0; 

then first c1+c0 will be computed, and the result passed to c2.operator+(). When the operator is declared to take a Complex& then you cannot pass a temporary, when it takes const Complex& then you can. Passing a temporary when a non-const reference is expected is a logic error in 90% of the cases, hence it is forbidden.

share|improve this answer

edited Nov 19 at 12:34

answered Nov 19 at 12:23

user463035818

15.7k42561

Temporaries dont bind to non-const references. When you write this

auto c3 = c2 + c1 + c0; 

then first c1+c0 will be computed, and the result passed to c2.operator+(). When the operator is declared to take a Complex& then you cannot pass a temporary, when it takes const Complex& then you can. Passing a temporary when a non-const reference is expected is a logic error in 90% of the cases, hence it is forbidden.

share|improve this answer

edited Nov 19 at 12:34

answered Nov 19 at 12:23

user463035818

15.7k42561

share|improve this answer

share|improve this answer

edited Nov 19 at 12:34

edited Nov 19 at 12:34

edited Nov 19 at 12:34

answered Nov 19 at 12:23

user463035818

15.7k42561

answered Nov 19 at 12:23

user463035818

15.7k42561

answered Nov 19 at 12:23

user463035818

15.7k42561

15.7k42561

add a comment | 

add a comment | 

up vote
1
down vote

The expression c1+c2+c3+c4 is parsed and evaluated as if

Complex c5 = operator+(c1, operator+(c2, operator+(c3, c4)));

operator+(c3, c4) builds and returns a temporary Complex object: an rvalue.

C++ forbids an rvalue to be bound to a non-const lvalue-reference.

But operator+(Complex&, Complex&) takes a non-const lvalue-reference. Hence the error message.

On the other hand, operator+(Complex const&, Complex const&) takes references to const lvalues.

share|improve this answer

edited 10 hours ago

answered Nov 19 at 12:23

YSC

19.5k34591

add a comment | 

up vote
1
down vote

The expression c1+c2+c3+c4 is parsed and evaluated as if

Complex c5 = operator+(c1, operator+(c2, operator+(c3, c4)));

operator+(c3, c4) builds and returns a temporary Complex object: an rvalue.

C++ forbids an rvalue to be bound to a non-const lvalue-reference.

But operator+(Complex&, Complex&) takes a non-const lvalue-reference. Hence the error message.

On the other hand, operator+(Complex const&, Complex const&) takes references to const lvalues.

share|improve this answer

edited 10 hours ago

answered Nov 19 at 12:23

YSC

19.5k34591

add a comment | 

up vote
1
down vote

up vote
1
down vote

The expression c1+c2+c3+c4 is parsed and evaluated as if

Complex c5 = operator+(c1, operator+(c2, operator+(c3, c4)));

operator+(c3, c4) builds and returns a temporary Complex object: an rvalue.

C++ forbids an rvalue to be bound to a non-const lvalue-reference.

But operator+(Complex&, Complex&) takes a non-const lvalue-reference. Hence the error message.

On the other hand, operator+(Complex const&, Complex const&) takes references to const lvalues.

share|improve this answer

edited 10 hours ago

answered Nov 19 at 12:23

YSC

19.5k34591

The expression c1+c2+c3+c4 is parsed and evaluated as if

Complex c5 = operator+(c1, operator+(c2, operator+(c3, c4)));

operator+(c3, c4) builds and returns a temporary Complex object: an rvalue.

C++ forbids an rvalue to be bound to a non-const lvalue-reference.

But operator+(Complex&, Complex&) takes a non-const lvalue-reference. Hence the error message.

On the other hand, operator+(Complex const&, Complex const&) takes references to const lvalues.

share|improve this answer

edited 10 hours ago

answered Nov 19 at 12:23

YSC

19.5k34591

share|improve this answer

share|improve this answer

edited 10 hours ago

edited 10 hours ago

edited 10 hours ago

answered Nov 19 at 12:23

YSC

19.5k34591

answered Nov 19 at 12:23

YSC

19.5k34591

answered Nov 19 at 12:23

YSC

19.5k34591

19.5k34591

add a comment | 

add a comment | 

 

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