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1

Define the function iota1(n, m) that takes positive integers n, m with n < m as input, and outputs the list (n,n+1,n+2,...,m)

I've tried switching the code around multiple times but cannot seem to get it to function and display a list the right way

(define (iota1 n m)
 (if (eq? n 0)
 '()
 (append (iota1 (< n m) (+ n 1)) (list n))))

racket

share|improve this question

edited Mar 26 at 12:21

GAD3R

2,32011229

asked Mar 24 at 15:21

detail8891detail8891

61

add a comment | 

1

Define the function iota1(n, m) that takes positive integers n, m with n < m as input, and outputs the list (n,n+1,n+2,...,m)

I've tried switching the code around multiple times but cannot seem to get it to function and display a list the right way

(define (iota1 n m)
 (if (eq? n 0)
 '()
 (append (iota1 (< n m) (+ n 1)) (list n))))

racket

share|improve this question

edited Mar 26 at 12:21

GAD3R

2,32011229

asked Mar 24 at 15:21

detail8891detail8891

61

add a comment | 

Define the function iota1(n, m) that takes positive integers n, m with n < m as input, and outputs the list (n,n+1,n+2,...,m)

I've tried switching the code around multiple times but cannot seem to get it to function and display a list the right way

(define (iota1 n m)
 (if (eq? n 0)
 '()
 (append (iota1 (< n m) (+ n 1)) (list n))))

racket

share|improve this question

edited Mar 26 at 12:21

GAD3R

2,32011229

asked Mar 24 at 15:21

detail8891detail8891

61

(define (iota1 n m)
 (if (eq? n 0)
 '()
 (append (iota1 (< n m) (+ n 1)) (list n))))

share|improve this question

edited Mar 26 at 12:21

GAD3R

2,32011229

asked Mar 24 at 15:21

detail8891detail8891

61

share|improve this question

edited Mar 26 at 12:21

GAD3R

2,32011229

asked Mar 24 at 15:21

detail8891detail8891

61

edited Mar 26 at 12:21

GAD3R

2,32011229

edited Mar 26 at 12:21

GAD3R

2,32011229

asked Mar 24 at 15:21

detail8891detail8891

61

asked Mar 24 at 15:21

detail8891detail8891

61

2 Answers
2

active

oldest

votes

2

There's a few oddities to the code you provided, which I've formatted for readability:

(define (iota1 n m)
 (if (eq? n 0)
 '()
 (append (iota (< n m) (+ n 1))
 (list n))))

The first is that the expression (< n m) evaluates to a boolean value, depending on whether n is less than m or not. When you apply iota to (< n m) in the expression (iota (< n m) (+ n 1)), you are giving iota a boolean value for its first argument instead of a positive integer.

Secondly, the use of append is strange here. When constructing a list in Racket, it's much more common to use the function cons, which takes as arguments a value, and a list, and returns a new list with the value added to the front. For example,

(append '(3) (append '(4) (append '(5) '()))) ==> '(3 4 5)
(cons 3 (cons 4 (cons 5 '()))) ==> '(3 4 5)

It's a good idea to opt for using cons instead of append because it's simpler, and because it is faster, since cons does not traverse the entire list like append does.

Since this sounds a bit like a homework problem, I'll leave you with a "code template" to help you find the answer:

; n : integer
; m : integer
; return value : list of integers
(define (iota1 n m)
 (if (> n m) ; base case; no need to do work when n is greater than m
 ... ; value that goes at the end of the list
 (cons ... ; the value we want to add to the front of the list
 (iota1 ... ...)))) ; the call to iota, generating the rest of the list

share|improve this answer

answered Mar 24 at 19:11

Michael MacLeodMichael MacLeod

6016

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Thanks for the feedback and sorry for the late response but, the only problem I am having is that whenever I put in the second argument("value that goes at the end of the list") it comes back as "expects a real as 2nd argument, given (cons 3 '())", I have the rest of the program complete just not that part, what does it mean by looking for a real argument? I tried googling it but came up with nothing. Thanks again for the help!
  
  – detail8891
  Mar 30 at 0:32
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  It's referring to the contract real?, which returns true when given a real number. See docs.racket-lang.org/reference/… for more information
  
  – Michael MacLeod
  Mar 30 at 5:48
  

  
  
  
  

add a comment | 

1

Welcome to the racket world, my version is here:

#lang racket

(define (iota1 n m)
 (let loop ([loop_n n]
 [result_list '()])
 (if (<= loop_n m)
 (loop
 (add1 loop_n)
 (cons loop_n result_list))
 (reverse result_list))))

share|improve this answer

answered Mar 26 at 7:13

simmonesimmone

274110

add a comment | 

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2

There's a few oddities to the code you provided, which I've formatted for readability:

(define (iota1 n m)
 (if (eq? n 0)
 '()
 (append (iota (< n m) (+ n 1))
 (list n))))

The first is that the expression (< n m) evaluates to a boolean value, depending on whether n is less than m or not. When you apply iota to (< n m) in the expression (iota (< n m) (+ n 1)), you are giving iota a boolean value for its first argument instead of a positive integer.

Secondly, the use of append is strange here. When constructing a list in Racket, it's much more common to use the function cons, which takes as arguments a value, and a list, and returns a new list with the value added to the front. For example,

(append '(3) (append '(4) (append '(5) '()))) ==> '(3 4 5)
(cons 3 (cons 4 (cons 5 '()))) ==> '(3 4 5)

It's a good idea to opt for using cons instead of append because it's simpler, and because it is faster, since cons does not traverse the entire list like append does.

Since this sounds a bit like a homework problem, I'll leave you with a "code template" to help you find the answer:

; n : integer
; m : integer
; return value : list of integers
(define (iota1 n m)
 (if (> n m) ; base case; no need to do work when n is greater than m
 ... ; value that goes at the end of the list
 (cons ... ; the value we want to add to the front of the list
 (iota1 ... ...)))) ; the call to iota, generating the rest of the list

share|improve this answer

answered Mar 24 at 19:11

Michael MacLeodMichael MacLeod

6016

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Thanks for the feedback and sorry for the late response but, the only problem I am having is that whenever I put in the second argument("value that goes at the end of the list") it comes back as "expects a real as 2nd argument, given (cons 3 '())", I have the rest of the program complete just not that part, what does it mean by looking for a real argument? I tried googling it but came up with nothing. Thanks again for the help!
  
  – detail8891
  Mar 30 at 0:32
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  It's referring to the contract real?, which returns true when given a real number. See docs.racket-lang.org/reference/… for more information
  
  – Michael MacLeod
  Mar 30 at 5:48
  

  
  
  
  

add a comment | 

2

There's a few oddities to the code you provided, which I've formatted for readability:

(define (iota1 n m)
 (if (eq? n 0)
 '()
 (append (iota (< n m) (+ n 1))
 (list n))))

The first is that the expression (< n m) evaluates to a boolean value, depending on whether n is less than m or not. When you apply iota to (< n m) in the expression (iota (< n m) (+ n 1)), you are giving iota a boolean value for its first argument instead of a positive integer.

Secondly, the use of append is strange here. When constructing a list in Racket, it's much more common to use the function cons, which takes as arguments a value, and a list, and returns a new list with the value added to the front. For example,

(append '(3) (append '(4) (append '(5) '()))) ==> '(3 4 5)
(cons 3 (cons 4 (cons 5 '()))) ==> '(3 4 5)

It's a good idea to opt for using cons instead of append because it's simpler, and because it is faster, since cons does not traverse the entire list like append does.

Since this sounds a bit like a homework problem, I'll leave you with a "code template" to help you find the answer:

; n : integer
; m : integer
; return value : list of integers
(define (iota1 n m)
 (if (> n m) ; base case; no need to do work when n is greater than m
 ... ; value that goes at the end of the list
 (cons ... ; the value we want to add to the front of the list
 (iota1 ... ...)))) ; the call to iota, generating the rest of the list

share|improve this answer

answered Mar 24 at 19:11

Michael MacLeodMichael MacLeod

6016

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Thanks for the feedback and sorry for the late response but, the only problem I am having is that whenever I put in the second argument("value that goes at the end of the list") it comes back as "expects a real as 2nd argument, given (cons 3 '())", I have the rest of the program complete just not that part, what does it mean by looking for a real argument? I tried googling it but came up with nothing. Thanks again for the help!
  
  – detail8891
  Mar 30 at 0:32
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  It's referring to the contract real?, which returns true when given a real number. See docs.racket-lang.org/reference/… for more information
  
  – Michael MacLeod
  Mar 30 at 5:48
  

  
  
  
  

add a comment | 

There's a few oddities to the code you provided, which I've formatted for readability:

(define (iota1 n m)
 (if (eq? n 0)
 '()
 (append (iota (< n m) (+ n 1))
 (list n))))

The first is that the expression (< n m) evaluates to a boolean value, depending on whether n is less than m or not. When you apply iota to (< n m) in the expression (iota (< n m) (+ n 1)), you are giving iota a boolean value for its first argument instead of a positive integer.

Secondly, the use of append is strange here. When constructing a list in Racket, it's much more common to use the function cons, which takes as arguments a value, and a list, and returns a new list with the value added to the front. For example,

(append '(3) (append '(4) (append '(5) '()))) ==> '(3 4 5)
(cons 3 (cons 4 (cons 5 '()))) ==> '(3 4 5)

It's a good idea to opt for using cons instead of append because it's simpler, and because it is faster, since cons does not traverse the entire list like append does.

Since this sounds a bit like a homework problem, I'll leave you with a "code template" to help you find the answer:

; n : integer
; m : integer
; return value : list of integers
(define (iota1 n m)
 (if (> n m) ; base case; no need to do work when n is greater than m
 ... ; value that goes at the end of the list
 (cons ... ; the value we want to add to the front of the list
 (iota1 ... ...)))) ; the call to iota, generating the rest of the list

share|improve this answer

answered Mar 24 at 19:11

Michael MacLeodMichael MacLeod

6016

(define (iota1 n m)
 (if (eq? n 0)
 '()
 (append (iota (< n m) (+ n 1))
 (list n))))

(append '(3) (append '(4) (append '(5) '()))) ==> '(3 4 5)
(cons 3 (cons 4 (cons 5 '()))) ==> '(3 4 5)

; n : integer
; m : integer
; return value : list of integers
(define (iota1 n m)
 (if (> n m) ; base case; no need to do work when n is greater than m
 ... ; value that goes at the end of the list
 (cons ... ; the value we want to add to the front of the list
 (iota1 ... ...)))) ; the call to iota, generating the rest of the list

share|improve this answer

answered Mar 24 at 19:11

Michael MacLeodMichael MacLeod

6016

answered Mar 24 at 19:11

Michael MacLeodMichael MacLeod

6016

answered Mar 24 at 19:11

Michael MacLeodMichael MacLeod

6016

1

Welcome to the racket world, my version is here:

#lang racket

(define (iota1 n m)
 (let loop ([loop_n n]
 [result_list '()])
 (if (<= loop_n m)
 (loop
 (add1 loop_n)
 (cons loop_n result_list))
 (reverse result_list))))

share|improve this answer

answered Mar 26 at 7:13

simmonesimmone

274110

add a comment | 

1

Welcome to the racket world, my version is here:

#lang racket

(define (iota1 n m)
 (let loop ([loop_n n]
 [result_list '()])
 (if (<= loop_n m)
 (loop
 (add1 loop_n)
 (cons loop_n result_list))
 (reverse result_list))))

share|improve this answer

answered Mar 26 at 7:13

simmonesimmone

274110

add a comment | 

Welcome to the racket world, my version is here:

#lang racket

(define (iota1 n m)
 (let loop ([loop_n n]
 [result_list '()])
 (if (<= loop_n m)
 (loop
 (add1 loop_n)
 (cons loop_n result_list))
 (reverse result_list))))

share|improve this answer

answered Mar 26 at 7:13

simmonesimmone

274110

#lang racket

(define (iota1 n m)
 (let loop ([loop_n n]
 [result_list '()])
 (if (<= loop_n m)
 (loop
 (add1 loop_n)
 (cons loop_n result_list))
 (reverse result_list))))

share|improve this answer

answered Mar 26 at 7:13

simmonesimmone

274110

answered Mar 26 at 7:13

simmonesimmone

274110

answered Mar 26 at 7:13

simmonesimmone

274110