unsigned long long + int


P

Peter Ammon

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
 
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M

Mike Wahler

Peter Ammon said:
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
 
M

Mike Wahler

Mike Wahler said:
The 'unsigned long long' object is not promoted at all.
The 'int' object is promoted to 'unsigned long long'.


'unsigned long long'


Undefined.

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

-Mike
 
P

Peter Ammon

Mike said:
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
 
K

kal

Peter Ammon said:
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.
 
C

CBFalconer

Peter said:
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.
 
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K

kal

CBFalconer said:
Peter Ammon wrote:

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.
 
C

CBFalconer

kal said:
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.
 
S

S.Tobias

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.
 
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D

Dan Pop

In said:
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
 
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J

Joe Wright

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




Chapter and verse.




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.
 

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