Question

Rather than making a parameterised type into instances of the Functor, Applicative and Monad classes in this order, in practice it is sometimes simpler to define the functor and applicative instances in terms of the monad instance, relying on the fact that the order in which declarations are made is not important in Haskell. Complete the missing parts in the following declarations for the ST type using the do notation. instance Functor ST where - fmap :: (a $\rightarrow$ b) $\rightarrow$ ST a $\rightarrow$ ST b fmap $g$ st $=$ do $\ldots$ instance Applicative ST where - pure :: a $\rightarrow$ ST a pure $x=s(\backslash s \rightarrow(x, s))$ $-(\langle *\rangle):$ ST $(\mathrm{a} \rightarrow$ b) $\rightarrow$ ST a $\rightarrow$ ST b stf $\langle *\rangle \operatorname{stx}=$ do ... instance Monad ST where $-(\gg=)::$ ST a $\rightarrow$ (a $\rightarrow$ ST b) $\rightarrow$ ST b $s t>=f=S(\backslash s \rightarrow$ let $\left(x, s^{\prime}\right)=\operatorname{app}$ st $s$ in app (f $x$ ) $s^{\prime}$ ) Solutions to exercises 1-4 are given in appendix A.

   Rather than making a parameterised type into instances of the Functor, Applicative and Monad classes in this order, in practice it is sometimes simpler to define the functor and applicative instances in terms of the monad instance, relying on the fact that the order in which declarations are made is not important in Haskell. Complete the missing parts in the following declarations for the ST type using the do notation.
instance Functor ST where
- fmap :: (a $\rightarrow$ b) $\rightarrow$ ST a $\rightarrow$ ST b
fmap $g$ st $=$ do $\ldots$
instance Applicative ST where
- pure :: a $\rightarrow$ ST a
pure $x=s(\backslash s \rightarrow(x, s))$
$-(\langle *\rangle):$ ST $(\mathrm{a} \rightarrow$ b) $\rightarrow$ ST a $\rightarrow$ ST b
stf $\langle *\rangle \operatorname{stx}=$ do ...
instance Monad ST where
$-(\gg=)::$ ST a $\rightarrow$ (a $\rightarrow$ ST b) $\rightarrow$ ST b
$s t>=f=S(\backslash s \rightarrow$
let $\left(x, s^{\prime}\right)=\operatorname{app}$ st $s$ in app (f $x$ ) $s^{\prime}$ )
Solutions to exercises 1-4 are given in appendix A.
Show more…
Programming in Haskell
Programming in Haskell
Graham Hutton 2nd Edition
Chapter 12, Problem 8 ↓

Instant Answer

verified

Step 1

The Monad instance for ST is already provided, which allows us to sequence computations that involve state. The bind operation `(>>=)` is defined for the ST type, which is crucial for defining the Functor and Applicative instances.  Show more…

Show all steps

lock
AceChat toggle button
Close icon
Ace pointing down

Please give Ace some feedback

Your feedback will help us improve your experience

Thumb up icon Thumb down icon
Thanks for your feedback!
Profile picture
Rather than making a parameterised type into instances of the Functor, Applicative and Monad classes in this order, in practice it is sometimes simpler to define the functor and applicative instances in terms of the monad instance, relying on the fact that the order in which declarations are made is not important in Haskell. Complete the missing parts in the following declarations for the ST type using the do notation. instance Functor ST where - fmap :: (a $\rightarrow$ b) $\rightarrow$ ST a $\rightarrow$ ST b fmap $g$ st $=$ do $\ldots$ instance Applicative ST where - pure :: a $\rightarrow$ ST a pure $x=s(\backslash s \rightarrow(x, s))$ $-(\langle *\rangle):$ ST $(\mathrm{a} \rightarrow$ b) $\rightarrow$ ST a $\rightarrow$ ST b stf $\langle *\rangle \operatorname{stx}=$ do ... instance Monad ST where $-(\gg=)::$ ST a $\rightarrow$ (a $\rightarrow$ ST b) $\rightarrow$ ST b $s t>=f=S(\backslash s \rightarrow$ let $\left(x, s^{\prime}\right)=\operatorname{app}$ st $s$ in app (f $x$ ) $s^{\prime}$ ) Solutions to exercises 1-4 are given in appendix A.
Close icon
Play audio
Feedback
Powered by NumerAI
Need help? Use Ace
Ace is your personal tutor. It breaks down any question with clear steps so you can learn.
Start Using Ace
Ace is your personal tutor for learning
Step-by-step explanations
Instant summaries
Summarize YouTube videos
Understand textbook images or PDFs
Study tools like quizzes and flashcards
Listen to your notes as a podcast
Continue solving this problem
Create a free account to:
  • View full step-by-step solution
  • Ask follow-up questions with Ace AI
  • Save progress and study later
Continue Free
Numerade

Get step-by-step video solution
from top educators

Continue with Clever
or



By creating an account, you agree to the Terms of Service and Privacy Policy
Already have an account? Log In

A free answer
just for you

Watch the video solution with this free unlock.

Numerade

Log in to watch this video
...and 100,000,000 more!


EMAIL

PASSWORD

OR
Continue with Clever