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overloading of ","

P: n/a
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,(Other 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;
}

Mar 12 '07 #1
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35 Replies


P: n/a
josh a écrit :
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,(Other 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);
Because you have 5.
The expression evaluates:
Other ox = ( (o1 + o1) , (o3 = o2 + o1) );

Or with names
Other ox = o1.operator+(o1).operator,(o3.operator=(o2.operato r+(o1)));

Since o3 is 5, then o1 is also 5 and ox is 5.

The reason is overloaded operator, doesn't have the same precedence as
POD operator,. Is is very confusing.

// ------------------
cout << ox.i << endl;

return 0;
}


Michael
Mar 12 '07 #2

P: n/a
josh wrote:
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,(Other 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;
}
I suspect that this could be undefined behaviour as the + operators have
equal precedence and the order they are evaluated depends on the
compiler. evaluation order is '+', '=', ',' based on precedence and
hence the code is not safe.

JB
Mar 12 '07 #3

P: n/a
Michael DOUBEZ wrote:
josh a écrit :
>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,(Other 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);

Because you have 5.
The expression evaluates:
Other ox = ( (o1 + o1) , (o3 = o2 + o1) );

Or with names
Other ox = o1.operator+(o1).operator,(o3.operator=(o2.operato r+(o1)));

Since o3 is 5, then o1 is also 5 and ox is 5.

The reason is overloaded operator, doesn't have the same precedence as
POD operator,. Is is very confusing.
That is not the only thing confusing! Your use of operator,() is the so
obfuscated! Why on Earth would you want to do something like that?
What possible reason could you have?

On a side note, why isn't operator,() the same precedence as the regular
',' operator? I thought that all user defined operators had the same
precedence as the builtins.
Adrian
--
================================================== ========
Adrian Hawryluk BSc. Computer Science
----------------------------------------------------------
Specialising in: OOD Methodologies in UML
OOP Methodologies in C, C++ and more
RT Embedded Programming
__--------------------------------------------------__
----- [blog: http://adrians-musings.blogspot.com/] -----
'--------------------------------------------------------'
My newsgroup writings are licensed under the Creative
Commons Attribution-Noncommercial-Share Alike 3.0 License
http://creativecommons.org/licenses/by-nc-sa/3.0/
================================================== ========
Mar 12 '07 #4

P: n/a
josh wrote:
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,(Other 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;
}
You are modifying and evaluating the same object (o1) between sequence
points. Your program's behavior is therefore undefined.

Please see the FAQ on ++i,i++, and following (39.15 and 39.16)

http://www.parashift.com/c++-faq-lit...html#faq-39.15

Mar 12 '07 #5

P: n/a
Adrian Hawryluk a écrit :
Michael DOUBEZ wrote:
>josh a écrit :
>>Hi, I coded the following but It does not return what I expect, why?
[snip]
// HERE it returns 5 AND not 6 WHY ???????????????????
Other ox = (o1 + o1, o3 = o2 + o1);

Because you have 5.
The expression evaluates:
Other ox = ( (o1 + o1) , (o3 = o2 + o1) );

Or with names
Other ox = o1.operator+(o1).operator,(o3.operator=(o2.operato r+(o1)));

Since o3 is 5, then o1 is also 5 and ox is 5.

The reason is overloaded operator, doesn't have the same precedence as
POD operator,. Is is very confusing.

That is not the only thing confusing! Your use of operator,() is the so
obfuscated! Why on Earth would you want to do something like that? What
possible reason could you have?
Mostly for learning what not to do I guess.

blitz++ has matrix initalisation with comma overloading.
>
On a side note, why isn't operator,() the same precedence as the regular
',' operator? I thought that all user defined operators had the same
precedence as the builtins.
Concerning the precedence, it is sloppy writing from my part: it doesn't
have the same ordering properties but the precedence is kept (the lowest
if I remember correctly). The point is: in POD, the comma operator is a
sequence point and evaluates from left to right, the lhs of comma being
discarded while when overloaded, it is just another function (no
sequence point).
(++i),(i*=2) is defined behavior with POD while with Operator,(): it is UB.

This was the case up in the OP.

Michael



Mar 12 '07 #6

P: n/a
red floyd wrote:
josh wrote:
>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,(Other 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;
}

You are modifying and evaluating the same object (o1) between sequence
points. Your program's behavior is therefore undefined.

Please see the FAQ on ++i,i++, and following (39.15 and 39.16)

http://www.parashift.com/c++-faq-lit...html#faq-39.15
Further explanation:

Because operator+ modifies its object **AND** you've overloaded
operator, it's equivalent to calling operator,(x++,x) -- undefined behavior

Mar 12 '07 #7

P: n/a
Adrian Hawryluk wrote:
>The reason is overloaded operator, doesn't have the same precedence as
POD operator,. Is is very confusing.

That is not the only thing confusing! Your use of operator,() is the so
obfuscated! Why on Earth would you want to do something like that?
What possible reason could you have?
Actually, all the overloaded operators in the OP's code are doing confusing
things.
Mar 12 '07 #8

P: n/a
On Mar 13, 10:07 am, red floyd <no.s...@here.dudewrote:
// HERE it returns 5 AND not 6 WHY ???????????????????
Other ox = (o1 + o1, o3 = o2 + o1);
You are modifying and evaluating the same object (o1) between sequence
points. Your program's behavior is therefore undefined.
Not true, there is a sequence point before and after every
function call -- in particular, operator+ and operator,
Further explanation:

Because operator+ modifies its object **AND** you've overloaded
operator, it's equivalent to calling operator,(x++,x) -- undefined behavior
Not equivalent at all. It's equivalent to h(x.f(x), g(x))
which is fine (even if f modifies x).

To answer OP's question: the behaviour is unspecified because
overloaded comma operator does not retain the ordering of the
builtin comma operator. What's happening in this case is that
(o3 = o2 + o1) is being executed, and then (o1 + o1), and then
the overloaded comma operator is called with both of those results.
A different compiler might evaluate (o1 + o1) first, resulting in
o5 == 6.
Mar 13 '07 #9

P: n/a
Old Wolf wrote:
On Mar 13, 10:07 am, red floyd <no.s...@here.dudewrote:
>>> // HERE it returns 5 AND not 6 WHY ???????????????????
Other ox = (o1 + o1, o3 = o2 + o1);
You are modifying and evaluating the same object (o1) between sequence
points. Your program's behavior is therefore undefined.

Not true, there is a sequence point before and after every
function call -- in particular, operator+ and operator,
>Further explanation:

Because operator+ modifies its object **AND** you've overloaded
operator, it's equivalent to calling operator,(x++,x) -- undefined behavior

Not equivalent at all. It's equivalent to h(x.f(x), g(x))
which is fine (even if f modifies x).

To answer OP's question: the behaviour is unspecified because
overloaded comma operator does not retain the ordering of the
builtin comma operator. What's happening in this case is that
(o3 = o2 + o1) is being executed, and then (o1 + o1), and then
the overloaded comma operator is called with both of those results.
A different compiler might evaluate (o1 + o1) first, resulting in
o5 == 6.
So does that mean that the comma operator's evaluation order of left to
right is not implemented correctly on the compiler? I thought that it
was defined as left to right.
Adrian

--
================================================== ========
Adrian Hawryluk BSc. Computer Science
----------------------------------------------------------
Specialising in: OOD Methodologies in UML
OOP Methodologies in C, C++ and more
RT Embedded Programming
__--------------------------------------------------__
----- [blog: http://adrians-musings.blogspot.com/] -----
'--------------------------------------------------------'
My newsgroup writings are licensed under the Creative
Commons Attribution-Noncommercial-Share Alike 3.0 License
http://creativecommons.org/licenses/by-nc-sa/3.0/
================================================== ========
Mar 13 '07 #10

P: n/a
Adrian Hawryluk wrote:
Old Wolf wrote:
>>
To answer OP's question: the behaviour is unspecified because
overloaded comma operator does not retain the ordering of the
builtin comma operator. What's happening in this case is that
(o3 = o2 + o1) is being executed, and then (o1 + o1), and then
the overloaded comma operator is called with both of those results.
A different compiler might evaluate (o1 + o1) first, resulting in
o5 == 6.

So does that mean that the comma operator's evaluation order of left to
right is not implemented correctly on the compiler? I thought that it
was defined as left to right.

It is, if you don't overload it. If you overload it, then the
evaluation order is undefined, as I understand it. Similarly, if you
overload && or ||, the "short-circuit" behavior does not occur.
Mar 13 '07 #11

P: n/a
On 12 Mar, 13:16, Michael DOUBEZ <michael.dou...@free.frwrote:
josh a écrit :
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,(Other 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);

Because you have 5.
The expression evaluates:
Other ox = ( (o1 + o1) , (o3 = o2 + o1) );

Or with names
Other ox = o1.operator+(o1).operator,(o3.operator=(o2.operato r+(o1)));

Since o3 is 5, then o1 is also 5 and ox is 5.

The reason is overloaded operator, doesn't have the same precedence as
POD operator,. Is is very confusing.
// ------------------
cout << ox.i << endl;
return 0;
}

Michael
so the compiler is doing:
o1.operator+(01).operator,(03 = 02.operator+(01))
and evaluating it to 5
but when we define overloaded operators the rules "should" be that
they have the same precedence and the same associativity and the same
arity of
the PDO and so really I don't understand why here the rules seems to
be "could"....
Here in the code it seems that it doesn't save the first evaluating
operation on o1...
may be only for the comma operators is there an exception rule????

Mar 13 '07 #12

P: n/a
On 12 Mar, 11:17, "josh" <xdevel2...@yahoo.comwrote:
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,(Other 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;

}
the precedence, associativity and arity of overload operator, as rule,
"should" be the same for the PDO but here for the comma it doesn't
seem so...
if I add to my class an overload of * and than I make i.e.
o1 + o2 * 03 the evaluation order is correct and in fact will be FIRST
o2 * o3 and SECOND
o1 + (the mul result).
So I think that the compiler when meet that expression will do:
o1.operator+(o2.operator*(o3))
but it should do the same with:
(o1.operator+(o1)).operator,(o3 = 02.operator+(o1))
so it should evaluate from left to right as the same PDO rule...
so what's wrong?

Mar 13 '07 #13

P: n/a
On 12 Mar, 13:16, Michael DOUBEZ <michael.dou...@free.frwrote:
josh a écrit :
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,(Other 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);

Because you have 5.
The expression evaluates:
Other ox = ( (o1 + o1) , (o3 = o2 + o1) );

Or with names
Other ox = o1.operator+(o1).operator,(o3.operator=(o2.operato r+(o1)));

Since o3 is 5, then o1 is also 5 and ox is 5.

The reason is overloaded operator, doesn't have the same precedence as
POD operator,. Is is very confusing.
// ------------------
cout << ox.i << endl;
return 0;
}

Michael
the precedence, associativity and arity of overload operator, as rule,
"should" be the same for the PDO but here for the comma it doesn't
seem so...
if I add to my class an overload of * and than I make i.e.
o1 + o2 * 03 the evaluation order is correct and in fact will be FIRST
o2 * o3 and SECOND
o1 + (the mul result).
So I think that the compiler when meet that expression will do:
o1.operator+(o2.operator*(o3))
but it should do the same with:
(o1.operator+(o1)).operator,(o3 = 02.operator+(o1))
so it should evaluate from left to right as the same PDO rule...
so what's wrong?

Mar 13 '07 #14

P: n/a
josh wrote:
On 12 Mar, 11:17, "josh" <xdevel2...@yahoo.comwrote:
>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,(Other 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;

}

the precedence, associativity and arity of overload operator, as rule,
"should" be the same for the PDO but here for the comma it doesn't
seem so...
Precedence and arity do match. I have no idea, what you mean by
associativity. Anyhow, what an overloaded comma operator lacks is the
sequence point that the built-in comma operator gives you.

if I add to my class an overload of * and than I make i.e.
o1 + o2 * 03 the evaluation order is correct and in fact will be FIRST
o2 * o3 and SECOND
This is false: the evaluation order is unspecified. This is unrelated to
overloading. It is unspecified for built-in types, as well. See clause
[5/4].

o1 + (the mul result).
So I think that the compiler when meet that expression will do:
o1.operator+(o2.operator*(o3))
but it should do the same with:
(o1.operator+(o1)).operator,(o3 = 02.operator+(o1))
so it should evaluate from left to right as the same PDO rule...
so what's wrong?
Your code has undefined behavior since it modifies the same object several
times without sequence points. The overloaded comma operator does not give
you the sequence point that forces left-to right evaluation.
Best

Kai-Uwe Bux
Mar 13 '07 #15

P: n/a
On 13 Mar, 04:28, red floyd <no.s...@here.dudewrote:
Adrian Hawryluk wrote:
Old Wolf wrote:
To answer OP's question: the behaviour is unspecified because
overloaded comma operator does not retain the ordering of the
builtin comma operator. What's happening in this case is that
(o3 = o2 + o1) is being executed, and then (o1 + o1), and then
the overloaded comma operator is called with both of those results.
A different compiler might evaluate (o1 + o1) first, resulting in
o5 == 6.
So does that mean that the comma operator's evaluation order of left to
right is not implemented correctly on the compiler? I thought that it
was defined as left to right.

It is, if you don't overload it. If you overload it, then the
evaluation order is undefined, as I understand it. Similarly, if you
overload && or ||, the "short-circuit" behavior does not occur.
have you read it on the 14482 iso spec.?

Mar 13 '07 #16

P: n/a
BTW - What is POD?

Mar 13 '07 #17

P: n/a
liam_herron wrote:
BTW - What is POD?
Plain Old Data.

Mar 13 '07 #18

P: n/a
josh wrote:
On 13 Mar, 04:28, red floyd <no.s...@here.dudewrote:
>Adrian Hawryluk wrote:
>>Old Wolf wrote:
To answer OP's question: the behaviour is unspecified because
overloaded comma operator does not retain the ordering of the
builtin comma operator. What's happening in this case is that
(o3 = o2 + o1) is being executed, and then (o1 + o1), and then
the overloaded comma operator is called with both of those results.
A different compiler might evaluate (o1 + o1) first, resulting in
o5 == 6.
So does that mean that the comma operator's evaluation order of left to
right is not implemented correctly on the compiler? I thought that it
was defined as left to right.
It is, if you don't overload it. If you overload it, then the
evaluation order is undefined, as I understand it. Similarly, if you
overload && or ||, the "short-circuit" behavior does not occur.

have you read it on the 14482 iso spec.?
I believe so. I'm at home right now, and my copy of 14882 is at work,
so it'll have to wait, unless other posters wish to comment on that.
Mar 14 '07 #19

P: n/a
Precedence and arity do match. I have no idea, what you mean by
associativity.
when I say about associativity I mean that if we have in an expression
more operators
of equal precedence i.e. * / % than it will be evaluated from left-to-
rigth
int a = 4 / 2 * 8 -first 4/2 and SECOND 2 * 8

if I add to my class an overload of * and than I make i.e.
o1 + o2 * 03 the evaluation order is correct and in fact will be FIRST
o2 * o3 and SECOND

This is false: the evaluation order is unspecified. This is unrelated to
overloading. It is unspecified for built-in types, as well. See clause
[5/4].
This is not false! the evaluation order is the same of PDO. It is
unspecified
only if I have in an expression the same operators i.e.
int a = 4 + 6 + b + f -here is not sure that the compiler goes from
left to right
it is guaranted only for && !! , ?: operators

Best

Mar 14 '07 #20

P: n/a
red floyd a écrit :
josh wrote:
>On 13 Mar, 04:28, red floyd <no.s...@here.dudewrote:
>>Adrian Hawryluk wrote:
Old Wolf wrote:
To answer OP's question: the behaviour is unspecified because
overloaded comma operator does not retain the ordering of the
builtin comma operator. What's happening in this case is that
(o3 = o2 + o1) is being executed, and then (o1 + o1), and then
the overloaded comma operator is called with both of those results.
A different compiler might evaluate (o1 + o1) first, resulting in
o5 == 6.
So does that mean that the comma operator's evaluation order of left to
right is not implemented correctly on the compiler? I thought that it
was defined as left to right.
It is, if you don't overload it. If you overload it, then the
evaluation order is undefined, as I understand it. Similarly, if you
overload && or ||, the "short-circuit" behavior does not occur.
have you read it on the 14482 iso spec.?

I believe so. I'm at home right now, and my copy of 14882 is at work,
so it'll have to wait, unless other posters wish to comment on that.
[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".
Michael
Mar 14 '07 #21

P: n/a
On 14 Mar, 10:16, Michael DOUBEZ <michael.dou...@free.frwrote:
red floyd a écrit :
josh wrote:
On 13 Mar, 04:28, red floyd <no.s...@here.dudewrote:
Adrian Hawryluk wrote:
Old Wolf wrote:
To answer OP's question: the behaviour is unspecified because
overloaded comma operator does not retain the ordering of the
builtin comma operator. What's happening in this case is that
(o3 = o2 + o1) is being executed, and then (o1 + o1), and then
the overloaded comma operator is called with both of those results.
A different compiler might evaluate (o1 + o1) first, resulting in
o5 == 6.
So does that mean that the comma operator's evaluation order of leftto
right is not implemented correctly on the compiler? I thought that it
was defined as left to right.
It is, if you don't overload it. If you overload it, then the
evaluation order is undefined, as I understand it. Similarly, if you
overload && or ||, the "short-circuit" behavior does not occur.
have you read it on the 14482 iso spec.?
I believe so. I'm at home right now, and my copy of 14882 is at work,
so it'll have to wait, unless other posters wish to comment on that.

[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".

Michael
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 ???

Mar 14 '07 #22

P: n/a
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&&(operator&&(a,b),c);

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

Michael
Mar 14 '07 #23

P: n/a
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&&(operator&&(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

Josh

Mar 14 '07 #24

P: n/a
josh a écrit :
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&&(operator&&(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
Google for Sequence Point.

Michael
Mar 14 '07 #25

P: n/a
josh wrote:
>Precedence and arity do match. I have no idea, what you mean by
associativity.

when I say about associativity I mean that if we have in an expression
more operators
of equal precedence i.e. * / % than it will be evaluated from left-to-
rigth
int a = 4 / 2 * 8 -first 4/2 and SECOND 2 * 8
I see. Note, however, that this provision does not reflect the evaluation
order in terms of which subexpressions get evaluated first. E.g.,

a + b - c * e

by left-to-right grouping for + and - is equivalent to

( a + b ) - ( c * e )

The compiler, however, is free to have c*e evaluated before a+b.

if I add to my class an overload of * and than I make i.e.
o1 + o2 * 03 the evaluation order is correct and in fact will be FIRST
o2 * o3 and SECOND

This is false: the evaluation order is unspecified. This is unrelated to
overloading. It is unspecified for built-in types, as well. See clause
[5/4].
This is not false! the evaluation order is the same of PDO. It is
unspecified
only if I have in an expression the same operators i.e.
Nope: in a*b-d/c it is unspecified whether a*b is evaluated before or after
d/c.
int a = 4 + 6 + b + f -here is not sure that the compiler goes from
left to right
it is guaranted only for && !! , ?: operators

Best

Kai-Uwe Bux
Mar 14 '07 #26

P: n/a
Nope: in a*b-d/c it is unspecified whether a*b is evaluated before or after
d/c.

yes for a*b-d/c but we are speaking about more complex expressions
where
the order is: precedence and associativity but however I think that
the
comma operator when we overload it don't respect the PDO rules where
is guaranted the left-to-right evaluation (as && || and ?:)

Josh

Mar 14 '07 #27

P: n/a
Michael DOUBEZ wrote:
red floyd a écrit :
[discussion on overloaded &&, ||, and ',' as sequence points redacted]
>>>

I believe so. I'm at home right now, and my copy of 14882 is at work,
so it'll have to wait, unless other posters wish to comment on that.

[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".
Thanks, Michael. I missed that one. I was looking at 13.6 and couldn't
find it.
Mar 14 '07 #28

P: n/a
josh wrote:
>
>Nope: in a*b-d/c it is unspecified whether a*b is evaluated before or
after d/c.


yes for a*b-d/c but we are speaking about more complex expressions
where
the order is: precedence and associativity but however I think that
the
comma operator when we overload it don't respect the PDO rules where
is guaranted the left-to-right evaluation (as && || and ?:)
Exactly, the overloaded comma-operator differs from the comma-operator for
POD types in this regard. And, I already said that:
Anyhow, what an overloaded comma operator lacks is the
sequence point that the built-in comma operator gives you.

Best

Kai-Uwe Bux
Mar 14 '07 #29

P: n/a
josh wrote:
>Nope: in a*b-d/c it is unspecified whether a*b is evaluated before or after
d/c.


yes for a*b-d/c but we are speaking about more complex expressions
where
the order is: precedence and associativity but however I think that
the
comma operator when we overload it don't respect the PDO rules where
is guaranted the left-to-right evaluation (as && || and ?:)

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

Overloading operators ||, &&, and , does alter their LEFT
SIDE FIRST rule as well as in the case of the two formers
the REQUIRED omission of the evaluation of the second
operand if the first is true (for or) or false (for and).

Mar 15 '07 #30

P: n/a
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
Mar 15 '07 #31

P: n/a
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+(operator*(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
Mar 15 '07 #32

P: n/a
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...


Mar 15 '07 #33

P: n/a
On 03/14/07, "josh" <xd********@yahoo.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&&(operator&&(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

Mar 15 '07 #34

P: n/a
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...
Mar 16 '07 #35

P: n/a
On Mar 16, 4:48 am, Clark Cox <clarkc...@gmail.comwrote:
operator&&(operator&&(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.

Mar 16 '07 #36

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