overloading of ","

[josh]

Hi, I coded the following but It does not return what I expect, why?

#include <iostream>

using namespace std;

class Other
{
public:
int i;

Other(int x=1)
{
i = x;
}

Other *operator-() { return this;}

Other &operator+ (Other t)
{
i += t.i;

return *this;
}

Other &operator,(Othe r oth)
{
i = oth.i;

return *this;
}

};

int main()
{
Other o0, o1, o2(4), o3(5);

o0->i = 100;

cout << o0.i << "\n" << o0->i << "\n";

// HERE it returns 5 AND not 6 WHY ??????????????? ????
Other ox = (o1 + o1, o3 = o2 + o1);
// ------------------
cout << ox.i << endl;

return 0;
}

[josh]

Precedence and associativity don't define order of operations.

Overloading operators changes neither of these.

No? and how the compiler could know which operations came fisrt????
it is the same algebra principles!
May be I don't have understand what you want to say?

Best

[Victor Bazarov]

josh wrote:

>Precedence and associativity don't define order of operations.
Overloading operators changes neither of these.

No? and how the compiler could know which operations came fisrt????
it is the same algebra principles!
May be I don't have understand what you want to say?

My guess [as to what ?? wants to say] follows.

In an expression a*b + c*d both multiplications have to be done
before the sum can be calculated, but it is unspecified in the
language whether 'a' is multiplied by 'b' first and then 'c' by
'd' or vice versa. Precedence only says that the expression
a*b + c*d cannot be interpreted as a * (b+c) * d.

Same with overloaded operators: the expression a*b + c*d can be
rewritten as operator+(opera tor*(a,b), operator*(c,d)) and the
order in which the arguments of the operator+ function (the return
values of the two calls to operator*) are calculated is not set by
the standard. The generated code can call the first operator*
before the second or vice versa.

Precedence cannot be changed for overloaded ops, that is, you
cannot expect the expression a*b + c*d to evaluate as
operator*(a, operator+(b, operator*(c,d)) ).

[BTW, please give proper attributions when you quote. It is not
clear to whom you replied and whom you meant when you wrote in
your reply "... what *you* want to say", emphasis mine]

V
--
Please remove capital 'A's when replying by e-mail
I do not respond to top-posted replies, please don't ask

[josh]

In an expression a*b + c*d both multiplications have to be done

before the sum can be calculated, but it is unspecified in the
language whether 'a' is multiplied by 'b' first and then 'c' by
'd' or vice versa. Precedence only says that the expression
a*b + c*d cannot be interpreted as a * (b+c) * d.

I think your guess is ok...

[Clark Cox]

On 03/14/07, "josh" <xd********@yah oo.comsaid:

On 14 Mar, 10:51, Michael DOUBEZ <michael.dou... @free.frwrote:

>josh a écrit :

>>On 14 Mar, 10:16, Michael DOUBEZ <michael.dou... @free.frwrote:
[snip]
[1.9/18] The POD &&, ||, ?: and , define a sequence.
In footnote (12) of the paragraph, it is specified that when those
operators are overloaded, they designate a function and "the operand
form an argument list without an implied sequence point between them".

>>so when am I overloading that operators the left-to-right sequence is
not implied (guaranted) and
if I have i.e. a && b && c (where a,b,c are object of a Type class
with that operators overloaded) is not evaluated first a && b and then
the result with c ???

No. Suposing the operator&& resolves to binary function, your expression
will be
operator&&(ope rator&&(a,b),c) ;

Then a,b and c will be evaluated in any order.

Michael

yes but the calling function operator () came before any other
operator and in that case
will be logic if the compiler first called operator&&(a,b) and then
comapared the result with c

Not necessarily. There is nothing stopping the compiler from evaluating
c first, *then* evaluating operator&&(a,b) . The only guarantee that we
have as to the ordering of the evaluations of a, b and c is that c
cannot come between a and b. These are the possible orderings:

a, b, c
b, a, c
(b and a simultaneously) , c
c, a, b
c, b, a
c, (b and a simultaneously)
--
Clark S. Cox III
cl*******@gmail .com

[Ron Natalie]

josh wrote:

>Precedence and associativity don't define order of operations.
Overloading operators changes neither of these.

No? and how the compiler could know which operations came fisrt????
it is the same algebra principles!
May be I don't have understand what you want to say?

Precedence and associativity affect the meaning of the operators as
to which operator binds to which operands.

The ordering however in C++ is UNSPECIFIED other than obviously the
values that are needed for later stages in the calculation have to
be done earlier. However, the ones that don't matter can be done
in any order. Side-effects can be applied at any time up until
the sequence point.

a() * b() * c()

for example could execute the functions a, b, and c before
it does nay of the multiples, or it could do them immediately
before the value is needed...

[Old Wolf]

On Mar 16, 4:48 am, Clark Cox <clarkc...@gmai l.comwrote:

operator&&(oper ator&&(a,b),c);

There is nothing stopping the compiler from evaluating
c first, *then* evaluating operator&&(a,b) . The only guarantee that we
have as to the ordering of the evaluations of a, b and c is that c
cannot come between a and b.

There is no such guarantee. For example, the order could be a,c,b.