29 Jan 2013 01:27

## quotRem and divMod

```Hi!

I’ve always thought that `quotRem` is faster than `quot` + `rem`, since
both `quot` and `rem` are just "wrappers" that compute both the quotient
and the remainder and then just throw one out. However, today I looked
into the implementation of `quotRem` for `Int32` and found out that it’s
not true:

quotRem x <at> (I32# x#) y <at> (I32# y#)
| y == 0                     = divZeroError
| x == minBound && y == (-1) = overflowError
| otherwise                  = (I32# (narrow32Int# (x# `quotInt#`
y#)),
I32# (narrow32Int# (x# `remInt#`
y#)))

Why? The `DIV` instruction computes both, doesn’t it? And yet it’s being
performed twice here. Couldn’t one of the experts clarify this bit?

_______________________________________________
```
29 Jan 2013 07:09

### Re: quotRem and divMod

```On Mon, Jan 28, 2013 at 4:27 PM, Artyom Kazak <artyom.kazak <at> gmail.com> wrote:
> Hi!
>
> I’ve always thought that `quotRem` is faster than `quot` + `rem`, since both
> `quot` and `rem` are just "wrappers" that compute both the quotient and the
> remainder and then just throw one out. However, today I looked into the
> implementation of `quotRem` for `Int32` and found out that it’s not true:
>
>     quotRem x <at> (I32# x#) y <at> (I32# y#)
>         | y == 0                     = divZeroError
>         | x == minBound && y == (-1) = overflowError
>         | otherwise                  = (I32# (narrow32Int# (x# `quotInt#`
> y#)),
>                                         I32# (narrow32Int# (x# `remInt#`
> y#)))
>
> Why? The `DIV` instruction computes both, doesn’t it? And yet it’s being
> performed twice here. Couldn’t one of the experts clarify this bit?
>

That code is from base 4.5. Here's base 4.6:

quotRem x <at> (I32# x#) y <at> (I32# y#)
| y == 0                     = divZeroError
-- Note [Order of tests]
| y == (-1) && x == minBound = (overflowError, 0)
| otherwise                  = case x# `quotRemInt#` y# of
(# q, r #) ->
(I32# (narrow32Int# q),
I32# (narrow32Int# r))
```

29 Jan 2013 13:26

### Re: quotRem and divMod

```Shachaf Ben-Kiki <shachaf <at> gmail.com> писал(а) в своём письме Tue, 29 Jan
2013 09:09:37 +0300:

> That code is from base 4.5. Here's base 4.6:
>
>     quotRem x <at> (I32# x#) y <at> (I32# y#)
>         | y == 0                     = divZeroError
>           -- Note [Order of tests]
>         | y == (-1) && x == minBound = (overflowError, 0)
>         | otherwise                  = case x# `quotRemInt#` y# of
>                                        (# q, r #) ->
>                                            (I32# (narrow32Int# q),
>                                             I32# (narrow32Int# r))
>
> So it looks like it was improved in GHC 7.6. In particular, by this
> commit:
>
>     Shacha

Well, I’m glad to know that it has been fixed in the newer GHC (I’m still
on 7.4). Thanks!

_______________________________________________
```
29 Jan 2013 12:39

### Re: quotRem and divMod

```On Tuesday 29 January 2013, 03:27:41, Artyom Kazak wrote:
> Hi!
>
> I’ve always thought that `quotRem` is faster than `quot` + `rem`, since
> both `quot` and `rem` are just "wrappers" that compute both the quotient
> and the remainder and then just throw one out. However, today I looked
> into the implementation of `quotRem` for `Int32` and found out that it’s
> not true:
>
>      quotRem x <at> (I32# x#) y <at> (I32# y#)
>
>          | y == 0                     = divZeroError
>          | x == minBound && y == (-1) = overflowError
>          | otherwise                  = (I32# (narrow32Int# (x# `quotInt#`
>
> y#)),
>                                          I32# (narrow32Int# (x# `remInt#`
> y#)))
>
> Why? The `DIV` instruction computes both, doesn’t it? And yet it’s being
> performed twice here. Couldn’t one of the experts clarify this bit?

It's not necessarily performed twice.

func a b = case a `quotRem` b of
(q, r) -> q+r

produces

idivq 8(%rbp)
```