add without carry

[Hugh]

I would like to perform an addition without carrying of two integers...
I've got no idea how to do this in python, although I've been using it
for web/cgi/db work for a few years now.

Any help would be great.

Hugh

[Marc 'BlackJack' Rintsch]

In <11**********************@i42g2000cwa.googlegroups .com>, Hugh wrote:

I would like to perform an addition without carrying of two integers...
I've got no idea how to do this in python, although I've been using it
for web/cgi/db work for a few years now.

Your description is a bit vague. What is `without carrying`? Do you want
to limit the result to a given bit length with a "wrap around" behavior if
the number gets too large to fit into those bits? Then this should do the
trick:

a = (b + c) & (2**bits - 1)

Ciao,
Marc 'BlackJack' Rintsch

[John Machin]

Hugh wrote:

I would like to perform an addition without carrying of two integers...
I've got no idea how to do this in python, although I've been using it
for web/cgi/db work for a few years now.

In multiword addition in assembly language, one uses a normal ADD
instruction on the lowest-order pair of words, and an ADC (or ADDC or
whatever) i.e add *with* carry which adds in the carry bit from the
previous operation for each subsequent pair of words. IOW, addition
*without* carrying is normal, and I don't recall hearing the expression
before ......

Do you possibly mean: add two 32-bit integers such the result fits in
32 bits i.e. discard the generated carry bit, keep the low-order
32-bits?

If so, that's simply (a + b) & 0xFFFFFFFF (in Python as well as a few
other languages).

If you meant a different number of bits, well the answer for 8 bits
would be (a + b) & 0xFF.

Otherwise, you need to provide some examples of two integers and what
"addition without carrying" produces.

HTH,
John

[Hugh]

Sorry, here's an example...

5+7=12

added without carrying, 5+7=2

i.e the result is always less than 10

I've been thinking some more about this and my brain is starting to
work something out... I just wondered if there was a function in python
math to do this automatically...

Hugh

[MonkeeSage]

Hugh wrote:

Sorry, here's an example...

5+7=12

added without carrying, 5+7=2

i.e the result is always less than 10

def add(a, b, c=10):
an = a + b
if an >= c:
an -= c
return an

add(5, 7) # = 2

?

Regards,
Jordan

[Bryan Olson]

Hugh wrote:

Sorry, here's an example...

5+7=12

added without carrying, 5+7=2

i.e the result is always less than 10

Are you looking for bitwise exclusive or? In Python it's
the '^' operator. For example:

print 5 ^ 7
--
--Bryan

[Hugh]

Thankyou everyone this gives me something to work with.

Hugh

[Peter Otten]

Hugh wrote:

Sorry, here's an example...

5+7=12

added without carrying, 5+7=2

i.e the result is always less than 10

I've been thinking some more about this and my brain is starting to
work something out... I just wondered if there was a function in python
math to do this automatically...

>>(5 + 7) % 10

2

In this context '%' is called 'modulo operator'. What John and Marc
suggested is basically the same operation for the special case of binary
numbers, i. e.

a % b == a & (b-1)

if a >= 0 and b == 2**N.

Peter

[Hugh]

Peter,

That was what I was thinking along the lines of, It's been two years
since I finished my CS degree and working in mechanical engineering
means I've nearly forgotten it all! :(

Thanks, I'll write a function in my app to handle this...

Hugh

>(5 + 7) % 10

2

In this context '%' is called 'modulo operator'. What John and Marc
suggested is basically the same operation for the special case of binary
numbers, i. e.

a % b == a & (b-1)

if a >= 0 and b == 2**N.

Peter

[Bruno Desthuilliers]

Hugh wrote:

Sorry, here's an example...

5+7=12

added without carrying, 5+7=2

i.e the result is always less than 10

I've been thinking some more about this and my brain is starting to
work something out...

No need to think too long to come up with the most possibly Q&D solution:

res = int(str(5 + 7)[-1])

Like it ?-)

--
bruno desthuilliers
python -c "print '@'.join(['.'.join([w[::-1] for w in p.split('.')]) for
p in 'o****@xiludom.gro'.split('@')])"

[Bruno Desthuilliers]

Bryan Olson wrote:

Hugh wrote:

>Sorry, here's an example...

5+7=12

added without carrying, 5+7=2

i.e the result is always less than 10

Are you looking for bitwise exclusive or? In Python it's
the '^' operator. For example:

print 5 ^ 7

>>10 ^ 21

31

Not really "less than 10"...

--
bruno desthuilliers
python -c "print '@'.join(['.'.join([w[::-1] for w in p.split('.')]) for
p in 'o****@xiludom.gro'.split('@')])"

[Christophe]

Bruno Desthuilliers a écrit :

Bryan Olson wrote:

>Hugh wrote:

>>Sorry, here's an example...

5+7=12

added without carrying, 5+7=2

i.e the result is always less than 10

Are you looking for bitwise exclusive or? In Python it's
the '^' operator. For example:

print 5 ^ 7

>>>10 ^ 21

31

Not really "less than 10"...

But you must use numbers smaller than 10 as input! Still :

>>8 ^ 2

10

:D

[Bryan Olson]

Bruno Desthuilliers wrote:

Bryan Olson wrote:

>Hugh wrote:

>>Sorry, here's an example...

5+7=12

added without carrying, 5+7=2

i.e the result is always less than 10

Are you looking for bitwise exclusive or? In Python it's
the '^' operator. For example:

print 5 ^ 7

>>>10 ^ 21

31

Not really "less than 10"...

I had little idea what he meant by that. My guess was that
no bit clear in an operand could be set in the result.
XOR is also known as mod-2 addition. It's what you get
if you add each bit independently, dropping carries, so
I thought it might be what he was looking for.
Unfortunately the bitwise mod-2 addition of 5 and 7 gives
the same result as mod-10 addition. Oh well.
--
--Bryan

[Jon Ribbens]

In article <45***********************@news.free.fr>, Bruno Desthuilliers wrote:

Hugh wrote:

>Sorry, here's an example...

5+7=12

added without carrying, 5+7=2

i.e the result is always less than 10

>I've been thinking some more about this and my brain is starting to
work something out...

No need to think too long to come up with the most possibly Q&D solution:

res = int(str(5 + 7)[-1])

Am I missing something subtle in the question or is there some reason
that nobody has posted the correct solution:

(a + b) % 10

?

[Bruno Desthuilliers]

Christophe wrote:

Bruno Desthuilliers a écrit :

>Bryan Olson wrote:

>>Hugh wrote:
Sorry, here's an example...

5+7=12

added without carrying, 5+7=2

i.e the result is always less than 10
Are you looking for bitwise exclusive or? In Python it's
the '^' operator. For example:

print 5 ^ 7
10 ^ 21

31

Not really "less than 10"...

But you must use numbers smaller than 10 as input! Still :

>>>8 ^ 2

10
:D

Still fails:

>>8 ^ 7

15

Sorry !-p

--
bruno desthuilliers
python -c "print '@'.join(['.'.join([w[::-1] for w in p.split('.')]) for
p in 'o****@xiludom.gro'.split('@')])"

[Jason]

Jon Ribbens wrote:

In article <45***********************@news.free.fr>, Bruno Desthuilliers wrote:

No need to think too long to come up with the most possibly Q&D solution:

res = int(str(5 + 7)[-1])

Am I missing something subtle in the question or is there some reason
that nobody has posted the correct solution:

(a + b) % 10

?

You're not missing anything. That's the obvious solution. We're just
celebrating our non-Dutchness at the moment.

>>def bound(value, maxBound):

.... while value maxBound:
.... value -= maxBound
.... return value
....

>>bound(5 + 6, 10)

1

>>bound(8 + 7, 10)

5

See? I'm definitely not Dutch.

--Jason

[John Machin]

Jason wrote:

Jon Ribbens wrote:

In article <45***********************@news.free.fr>, Bruno Desthuilliers wrote:

>
No need to think too long to come up with the most possibly Q&D solution:
>
res = int(str(5 + 7)[-1])

Am I missing something subtle in the question or is there some reason
that nobody has posted the correct solution:

(a + b) % 10

?

You're not missing anything. That's the obvious solution. We're just
celebrating our non-Dutchness at the moment.

>def bound(value, maxBound):

... while value maxBound:

Does non-Dutchness require or permit bugs like instead of >= or are
you merely having a braino?

... value -= maxBound
... return value
...

>bound(5 + 6, 10)

1

>bound(8 + 7, 10)

5

See? I'm definitely not Dutch.

--Jason

[Bruno Desthuilliers]

Jon Ribbens a écrit :

In article <45***********************@news.free.fr>, Bruno Desthuilliers wrote:

>>Hugh wrote:

>>>Sorry, here's an example...

5+7=12

added without carrying, 5+7=2

i.e the result is always less than 10

>>>I've been thinking some more about this and my brain is starting to
work something out...

No need to think too long to come up with the most possibly Q&D solution:

res = int(str(5 + 7)[-1])

Am I missing something subtle in the question or is there some reason
that nobody has posted the correct solution:

(a + b) % 10

I'm afraid Peter Otten did a couple hours before you. But please note
that my solution *is* actually 100% correct (base 10 assumed). It may be
totally inefficiant, but that's another problem !-)