Consider the following three function definitions in Haskell: first x y = x ident x = x omega x = omega x We could also define these in ML, although with a slightly different syntax: fun Ident x = x; etc. How would the evaluations of the following expression in Haskell and ML compare= first (ident 3) (omega 1) (First (Ident 3) (Omega 1). in ML}
Consider the following three function definitions in Haskell: first x y = x ident x = x omega x = omega x We could also define these in ML, although with a slightly different syntax: fun Ident x = x; etc. How would the evaluations of the following expression in Haskell and ML compare= first (ident 3) (omega 1) (First (Ident 3) (Omega 1). in ML}
Database System Concepts
7th Edition
ISBN:9780078022159
Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Chapter1: Introduction
Section: Chapter Questions
Problem 1PE
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![### Function Definitions in Haskell and ML
**Consider the following three function definitions in Haskell:**
```haskell
first x y = x
ident x = x
omega x = omega x
```
- `first x y = x`: This function takes two arguments `x` and `y`, and returns `x`.
- `ident x = x`: This function takes a single argument `x` and returns `x`.
- `omega x = omega x`: This function is recursive and will cause an infinite loop when called, as it is defined in terms of itself with the same argument.
**The same functions can also be defined in ML, although with a slightly different syntax:**
```ml
fun Ident x = x;
```
- `fun Ident x = x;`: This is equivalent to `ident x = x` in Haskell, taking a single argument `x` and returning `x`.
**Comparison of Evaluations in Haskell and ML**
Consider the evaluation of the following expression in Haskell and ML:
```haskell
first (ident 3) (omega 1)
```
```ml
{First (Ident 3) (Omega 1). in ML}
```
- In Haskell:
- `ident 3` evaluates to `3` since `ident x = x`.
- `omega 1` does not terminate as it initiates an infinite loop (`omega 1 = omega 1`).
- `first` takes two arguments but will return the first one without evaluating the second, so `first (ident 3) (omega 1)` will effectively return `3` without getting stuck in the infinite loop caused by `omega 1`.
- In ML:
- `Ident 3` evaluates to `3`, analogous to Haskell's `ident 3`.
- However, `Omega` would initiate an infinite loop in ML just as in Haskell with the `omega` function.
- In ML, depending on the language's evaluation strategy, the overall result may differ from Haskell. In strict evaluation (which most ML variants follow), `Omega 1` would be evaluated before being passed into `First`, causing an infinite loop and preventing a result.
By comparing these evaluations, one can observe the differences in how Haskell and ML approach function evaluation, particularly in the handling of non-terminating functions and short-circuiting behavior.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F4a9bf715-7941-4859-ad33-64f7ce422f38%2F8077f046-8902-462f-b887-bdfa4fc46a26%2Fzp7fn4o_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Function Definitions in Haskell and ML
**Consider the following three function definitions in Haskell:**
```haskell
first x y = x
ident x = x
omega x = omega x
```
- `first x y = x`: This function takes two arguments `x` and `y`, and returns `x`.
- `ident x = x`: This function takes a single argument `x` and returns `x`.
- `omega x = omega x`: This function is recursive and will cause an infinite loop when called, as it is defined in terms of itself with the same argument.
**The same functions can also be defined in ML, although with a slightly different syntax:**
```ml
fun Ident x = x;
```
- `fun Ident x = x;`: This is equivalent to `ident x = x` in Haskell, taking a single argument `x` and returning `x`.
**Comparison of Evaluations in Haskell and ML**
Consider the evaluation of the following expression in Haskell and ML:
```haskell
first (ident 3) (omega 1)
```
```ml
{First (Ident 3) (Omega 1). in ML}
```
- In Haskell:
- `ident 3` evaluates to `3` since `ident x = x`.
- `omega 1` does not terminate as it initiates an infinite loop (`omega 1 = omega 1`).
- `first` takes two arguments but will return the first one without evaluating the second, so `first (ident 3) (omega 1)` will effectively return `3` without getting stuck in the infinite loop caused by `omega 1`.
- In ML:
- `Ident 3` evaluates to `3`, analogous to Haskell's `ident 3`.
- However, `Omega` would initiate an infinite loop in ML just as in Haskell with the `omega` function.
- In ML, depending on the language's evaluation strategy, the overall result may differ from Haskell. In strict evaluation (which most ML variants follow), `Omega 1` would be evaluated before being passed into `First`, causing an infinite loop and preventing a result.
By comparing these evaluations, one can observe the differences in how Haskell and ML approach function evaluation, particularly in the handling of non-terminating functions and short-circuiting behavior.
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