Overloading resolution with numbers

j.romildo＠gmail.com

5 Apr 2012 5 Apr '12

9:16 a.m.

Hello. Consider the following ghci session: Prelude> :t read "2" + 1 read "2" + 1 :: (Num a, Read a) => a Prelude> :t read "2.3" + 1 read "2.3" + 1 :: (Num a, Read a) => a Prelude> read "2" + 1 3 Prelude> read "2.3" + 1 *** Exception: Prelude.read: no parse Why does (read "2" + 1) works, but (read "2.3" + 1) fail at runtime? Romildo

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Amy de Buitléir

5 Apr 5 Apr

9:18 a.m.

writes:

...

Why does (read "2" + 1) works, but (read "2.3" + 1) fail at runtime?

Try this: read "2.3" + 1 :: Float Or this: read "2.3" + 1.0 The reason that your version didn't work is because GHCi is guessing that you want the read operation to parse an Integer, since you're adding it to 1.

j.romildo＠gmail.com

9:54 a.m.

On Thu, Apr 05, 2012 at 01:18:39PM +0000, Amy de Buitléir wrote:

...

writes:

...
Why does (read "2" + 1) works, but (read "2.3" + 1) fail at runtime?

Try this:

read "2.3" + 1 :: Float

Or this:

read "2.3" + 1.0

The reason that your version didn't work is because GHCi is guessing that you want the read operation to parse an Integer, since you're adding it to 1.

This is explanation does not seem to be enough once we consider the type of the literal 1: Prelude> :t 1 1 :: Num a => a That is, the literal 1 is overloaded to any numeric type. It is not necessarily an Integer. Romildo

Jason Dusek

10:53 a.m.

The 1 is not necessarily an Integer nor is it necessarily a Float, Double or Rational. GHC applies so called type-defaulting rules to make a best guess in this scenario. Informally, if it could be an Integer (due to lack of static evidence to the contrary), then you can expect the defaulting rules to assign it that type. -- Jason Dusek On Apr 5, 2012 7:45 AM, wrote:

...

...
writes:

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Why does (read "2" + 1) works, but (read "2.3" + 1) fail at runtime?

Try this:

read "2.3" + 1 :: Float

Or this:

read "2.3" + 1.0

The reason that your version didn't work is because GHCi is guessing

On Thu, Apr 05, 2012 at 01:18:39PM +0000, Amy de Buitléir wrote: that you

...
want the read operation to parse an Integer, since you're adding it to 1.

This is explanation does not seem to be enough once we consider the type of the literal 1:

Prelude> :t 1 1 :: Num a => a

That is, the literal 1 is overloaded to any numeric type. It is not necessarily an Integer.

Romildo

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damodar kulkarni

10:59 a.m.

See this: Prelude> :t 2.3 2.3 :: Fractional a => a Hope this helps. On Thu, Apr 5, 2012 at 7:24 PM, wrote:

...

...
writes:

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Why does (read "2" + 1) works, but (read "2.3" + 1) fail at runtime?

Try this:

read "2.3" + 1 :: Float

Or this:

read "2.3" + 1.0

The reason that your version didn't work is because GHCi is guessing

On Thu, Apr 05, 2012 at 01:18:39PM +0000, Amy de Buitléir wrote: that you

...
want the read operation to parse an Integer, since you're adding it to 1.

This is explanation does not seem to be enough once we consider the type of the literal 1:

Prelude> :t 1 1 :: Num a => a

That is, the literal 1 is overloaded to any numeric type. It is not necessarily an Integer.

Romildo

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-- Thanks and regards, -Damodar Kulkarni

damodar kulkarni

11 a.m.

On Thu, Apr 5, 2012 at 8:29 PM, damodar kulkarni wrote:

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See this: Prelude> :t 2.3 2.3 :: Fractional a => a

Hope this helps.

On Thu, Apr 5, 2012 at 7:24 PM, wrote:

...
...
writes:

...
Why does (read "2" + 1) works, but (read "2.3" + 1) fail at runtime?

Try this:

read "2.3" + 1 :: Float

Or this:

read "2.3" + 1.0

The reason that your version didn't work is because GHCi is guessing

...
want the read operation to parse an Integer, since you're adding it to

On Thu, Apr 05, 2012 at 01:18:39PM +0000, Amy de Buitléir wrote: that you 1.

This is explanation does not seem to be enough once we consider the type of the literal 1:

Prelude> :t 1 1 :: Num a => a

That is, the literal 1 is overloaded to any numeric type. It is not necessarily an Integer.

Romildo

_______________________________________________ Beginners mailing list Beginners@haskell.org http://www.haskell.org/mailman/listinfo/beginners

-- Thanks and regards, -Damodar Kulkarni

-- Thanks and regards, -Damodar Kulkarni

Marius Ghita

9:19 a.m.

Because this works Prelude> read "2.3" + 1.0 3.3 Seriously I can only assume that via the inference it decides based on the one that the string must be of type Int and uses the Read instance of Int to parse it; whereas that string is obviously a float. That is what* I think* it does. On Thu, Apr 5, 2012 at 4:16 PM, wrote:

...

Hello.

Consider the following ghci session:

Prelude> :t read "2" + 1 read "2" + 1 :: (Num a, Read a) => a

Prelude> :t read "2.3" + 1 read "2.3" + 1 :: (Num a, Read a) => a

Prelude> read "2" + 1 3

Prelude> read "2.3" + 1 *** Exception: Prelude.read: no parse

Why does (read "2" + 1) works, but (read "2.3" + 1) fail at runtime?

Romildo

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Amy de Buitléir
damodar kulkarni
j.romildo＠gmail.com
Jason Dusek
Marius Ghita