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unsigned long long + int

P: n/a
When I add an unsigned long long and an int, what type do each of the
values get promoted to before the addition is performed? What is the
type of the resulting expression? What occurs if the addition overflows
or underflows?

Thanks,

-Peter
Nov 14 '05 #1
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P: n/a
"Peter Ammon" <pe*********@rocketmail.com> wrote in message
news:cj**********@news.apple.com...
When I add an unsigned long long and an int, what type do each of the
values get promoted to before the addition is performed?
The 'unsigned long long' object is not promoted at all.
The 'int' object is promoted to 'unsigned long long'.
What is the
type of the resulting expression?
'unsigned long long'
What occurs if the addition overflows
or underflows?


Undefined.

-Mike
Nov 14 '05 #2

P: n/a

"Mike Wahler" <mk******@mkwahler.net> wrote in message
news:WI****************@newsread3.news.pas.earthli nk.net...
"Peter Ammon" <pe*********@rocketmail.com> wrote in message
news:cj**********@news.apple.com...
When I add an unsigned long long and an int, what type do each of the
values get promoted to before the addition is performed?


The 'unsigned long long' object is not promoted at all.
The 'int' object is promoted to 'unsigned long long'.
What is the
type of the resulting expression?


'unsigned long long'
What occurs if the addition overflows
or underflows?


Undefined.


Sorry, this is not correct. Upon overflow, unsigned integer
types will 'wrap' to zero. The undefined behavior occurs with
signed integer types.

-Mike
Nov 14 '05 #3

P: n/a
Mike Wahler wrote:
"Peter Ammon" <pe*********@rocketmail.com> wrote in message
news:cj**********@news.apple.com...
When I add an unsigned long long and an int, what type do each of the
values get promoted to before the addition is performed?

The 'unsigned long long' object is not promoted at all.
The 'int' object is promoted to 'unsigned long long'.


Thanks! How does this promotion occur? Is it guaranteed to convert the
value modulo ULLONG_MAX + 1 (like I hope), or is it an implementation
defined thing?

[...]

-Peter

--
Pull out a splinter to reply.
Nov 14 '05 #4

P: n/a
kal
Peter Ammon <pe*********@rocketmail.com> wrote in message news:<cj**********@news.apple.com>...
When I add an unsigned long long and an int, what type do each of the
values get promoted to before the addition is performed? What is the
type of the resulting expression? What occurs if the addition overflows
or underflows?


Try it on your machine/compiler and let us know.

On of these days I am going to figure out what the following means.

6.3.1.8 Usual arithmetic conversions

1 Many operators that expect operands of arithmetic type cause
conversions and yield result types in a similar way. The purpose
is to determine a common real type for the operands and result.
For the specified operands, each operand is converted, without
change of type domain, to a type whose corresponding real type
is the common real type. Unless explicitly stated otherwise, the
common real type is also the corresponding real type of the result,
whose type domain is the type domain of the operands if they are
the same, and complex otherwise. This pattern is called the usual
arithmetic conversions:

First, if the corresponding real type of either operand is long
double, the other operand is converted, without change of type
domain, to a type whose corresponding real type is long double.

Otherwise, if the corresponding real type of either operand is
double, the other operand is converted, without change of type
domain, to a type whose corresponding real type is double.

Otherwise, if the corresponding real type of either operand is
float, the other operand is converted, without change of type
domain, to a type whose corresponding real type is float.

Otherwise, the integer promotions are performed on both operands.
Then the following rules are applied to the promoted operands:

If both operands have the same type, then no further
conversion is needed.

Otherwise, if both operands have signed integer types or
both have unsigned integer types, the operand with the type
of lesser integer conversion rank is converted to the type
of the operand with greater rank.

Otherwise, if the operand that has unsigned integer type has
rank greater or equal to the rank of the type of the other
operand, then the operand with signed integer type is
converted to the type of the operand with unsigned integer
type.

Otherwise, if the type of the operand with signed integer
type can represent all of the values of the type of the
operand with unsigned integer type, then the operand with
unsigned integer type is converted to the type of the
operand with signed integer type.

Otherwise, both operands are converted to the unsigned
integer type corresponding to the type of the operand
with signed integer type.
Nov 14 '05 #5

P: n/a
Peter Ammon wrote:
Mike Wahler wrote:
"Peter Ammon" <pe*********@rocketmail.com> wrote in message
When I add an unsigned long long and an int, what type do each
of the values get promoted to before the addition is performed?


The 'unsigned long long' object is not promoted at all.
The 'int' object is promoted to 'unsigned long long'.


Thanks! How does this promotion occur? Is it guaranteed to
convert the value modulo ULLONG_MAX + 1 (like I hope), or is it
an implementation defined thing?


Yes, guaranteed.

--
A: Because it fouls the order in which people normally read text.
Q: Why is top-posting such a bad thing?
A: Top-posting.
Q: What is the most annoying thing on usenet and in e-mail?

Nov 14 '05 #6

P: n/a
Mike Wahler <mk******@mkwahler.net> wrote:
"Peter Ammon" <pe*********@rocketmail.com> wrote in message
news:cj**********@news.apple.com...
When I add an unsigned long long and an int, what type do each of the
values get promoted to before the addition is performed?
The 'unsigned long long' object is not promoted at all.
The 'int' object is promoted to 'unsigned long long'.

^^^^^^^^ -> converted

--
Stan Tobias
sed 's/[A-Z]//g' to email
Nov 14 '05 #7

P: n/a
kal
CBFalconer <cb********@yahoo.com> wrote in message news:<41***************@yahoo.com>...
Peter Ammon wrote:

Mike Wahler wrote:
"Peter Ammon" <pe*********@rocketmail.com> wrote in message

When I add an unsigned long long and an int, what type do each
of the values get promoted to before the addition is performed?

The 'unsigned long long' object is not promoted at all.
The 'int' object is promoted to 'unsigned long long'.


Thanks! How does this promotion occur? Is it guaranteed to
convert the value modulo ULLONG_MAX + 1 (like I hope), or is it
an implementation defined thing?


Yes, guaranteed.


Seems so. But I still can't figure out what the standard says.

Say, while converting from long to unsigned long where the value
of long is negative, does this mean the value of unsigned long
is calculated as ULONG_MAX + 1 + <value of long> ?

6.3.1.3 Signed and unsigned integers

2 Otherwise, if the new type is unsigned, the value is converted
by repeatedly adding or subtracting one more than the maximum
value that can be represented in the new type until the value
is in the range of the new type.
Nov 14 '05 #8

P: n/a
kal wrote:

CBFalconer <cb********@yahoo.com> wrote in message news:<41***************@yahoo.com>...
Peter Ammon wrote:

Mike Wahler wrote:
> "Peter Ammon" <pe*********@rocketmail.com> wrote in message
>
>> When I add an unsigned long long and an int, what type do each
>> of the values get promoted to before the addition is performed?
>
> The 'unsigned long long' object is not promoted at all.
> The 'int' object is promoted to 'unsigned long long'.

Thanks! How does this promotion occur? Is it guaranteed to
convert the value modulo ULLONG_MAX + 1 (like I hope), or is it
an implementation defined thing?


Yes, guaranteed.


Seems so. But I still can't figure out what the standard says.

Say, while converting from long to unsigned long where the value
of long is negative, does this mean the value of unsigned long
is calculated as ULONG_MAX + 1 + <value of long> ?

6.3.1.3 Signed and unsigned integers

2 Otherwise, if the new type is unsigned, the value is converted
by repeatedly adding or subtracting one more than the maximum
value that can be represented in the new type until the value
is in the range of the new type.


In most cases it means modulo arithmetic. The verbiage is
necessary to cater to the various allowable integral arithmetic
forms, 2s complement, 1s complement, sign-magnitude.

--
A: Because it fouls the order in which people normally read text.
Q: Why is top-posting such a bad thing?
A: Top-posting.
Q: What is the most annoying thing on usenet and in e-mail?
Nov 14 '05 #9

P: n/a
S.Tobias wrote:
Mike Wahler <mk******@mkwahler.net> wrote:
"Peter Ammon" <pe*********@rocketmail.com> wrote in message
news:cj**********@news.apple.com...
When I add an unsigned long long and an int, what type do each of the
values get promoted to before the addition is performed?


The 'unsigned long long' object is not promoted at all.
The 'int' object is promoted to 'unsigned long long'.


^^^^^^^^ -> converted


Objects are not converted, values are. Conversion to higher rank is
called promotion. Like in the Army. :=)
--
Joe Wright mailto:jo********@comcast.net
"Everything should be made as simple as possible, but not simpler."
--- Albert Einstein ---
Nov 14 '05 #10

P: n/a
Joe Wright <jo********@comcast.net> wrote:
S.Tobias wrote:
The 'int' object is promoted to 'unsigned long long'. ^^^^^^^^ -> converted

Objects are not converted, values are. Conversion to higher rank is
called promotion. Like in the Army. :=)


Are you sure (about integer promotions, not the Army :-) )?
I thought the word "promotions" was reserved only for _automatic_
conversions (rank_less_than_int)->(int); or (float)->(double)
in argument promotions.

--
Stan Tobias
sed 's/[A-Z]//g' to email
Nov 14 '05 #11

P: n/a
In <Y4********************@comcast.com> Joe Wright <jo********@comcast.net> writes:
S.Tobias wrote:
Mike Wahler <mk******@mkwahler.net> wrote:
"Peter Ammon" <pe*********@rocketmail.com> wrote in message
news:cj**********@news.apple.com...

When I add an unsigned long long and an int, what type do each of the
values get promoted to before the addition is performed?
The 'unsigned long long' object is not promoted at all.
The 'int' object is promoted to 'unsigned long long'.


^^^^^^^^ -> converted


Objects are not converted, values are.


Entirely gratuitous nit picking: the purpose of objects is to hold values.
Conversion to higher rank is called promotion.
Chapter and verse.
Like in the Army. :=)


In C, there are four strictly delimited categories of automatic
conversions: the integral promotions, the default argument promotions,
the usual arithmetic conversions and the assignment operator conversions.
To avoid misunderstanding, use the word "promotion" only in one of the
two contexts where the standard uses it, even if the usual arithmetic
conversions are often army-like promotions...

Dan
--
Dan Pop
DESY Zeuthen, RZ group
Email: Da*****@ifh.de
Currently looking for a job in the European Union
Nov 14 '05 #12

P: n/a
Dan Pop wrote:
In <Y4********************@comcast.com> Joe Wright <jo********@comcast.net> writes:

S.Tobias wrote:

Mike Wahler <mk******@mkwahler.net> wrote:
"Peter Ammon" <pe*********@rocketmail.com> wrote in message
news:cj**********@news.apple.com...
>When I add an unsigned long long and an int, what type do each of the
>values get promoted to before the addition is performed?
The 'unsigned long long' object is not promoted at all.
The 'int' object is promoted to 'unsigned long long'.

^^^^^^^^ -> converted


Objects are not converted, values are.

Entirely gratuitous nit picking: the purpose of objects is to hold values.

Conversion to higher rank is called promotion.

Chapter and verse.

Like in the Army. :=)

In C, there are four strictly delimited categories of automatic
conversions: the integral promotions, the default argument promotions,
the usual arithmetic conversions and the assignment operator conversions.
To avoid misunderstanding, use the word "promotion" only in one of the
two contexts where the standard uses it, even if the usual arithmetic
conversions are often army-like promotions...

Dan


What got my attention first was Stan's need to correct Mike's use of
'promoted' to 'converted'. That was gratuitous I think. Then, in
context I see Mike promoting objects. That is simply wrong and that
'the purpose of objects is to hold values' does not change the fact
that objects are not converted or promoted. Values are.

I do appreciate your advice. Thank you.
--
Joe Wright mailto:jo********@comcast.net
"Everything should be made as simple as possible, but not simpler."
--- Albert Einstein ---
Nov 14 '05 #13

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