Turn an integer n into a list containing it n times [closed]

Question

Given an integer n as input, return a list containing n, repeated n times. For example, the program would take 5 and turn it into [5,5,5,5,5]. The elements need to be integers, not strings. No built-in functions that accomplish the task are allowed.

This is code-golf, so standard rules apply.

Answer 1

Just found this lang and wanted to give it a go.

LOLCODE, 164 bytes

HAI 1.2
I HAS A CHZBRGR
GIMMEH CHZBRGR
CHZBRGR IS NOW A NUMBR
IM IN YR CRIB UPPIN YR HMBRGR TIL BOTH SAEM CHZBRGR AN HMBRGR
VISIBLE CHZBRGR
IM OUTTA YR CRIB
KTHXBYE

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Answer 2

Pyt, 3 bytes

Đ⑴*

Explanation:

                Implicit input
Đ               Duplicates input
 ⑴             Pushes an array, [1,1,...,1], with input 1's, onto the stack
   *            Multiply the array element-wise by the input
                Implicit output

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Answer 3

Tcl, 22 bytes

time {lappend L $n} $n

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---

With built-ins:

Tcl, 13 bytes

lrepeat $n $n

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Answer 4

FORTH 21 Bytes

: B 0 DO DUP . LOOP ;

OUTPUT:

10 B
10 B 10 10 10 10 10 10 10 10 10 10 

Answer 5

Gol><>, 5 bytes

:MR:B

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Example full program & How it works

1AGIE;GDlR~
:MR:B

1AG          Register row 1 as function G
   IE;       Take number input; halt on EOF
      G      Call G
       D     Print the stack contents
        lR~  Empty the stack
             Repeat indefinitely

:M           Duplicate top and decrement
  R:         Duplicate n-1 times
    B        Return

Answer 6

Charcoal, 9 5 4 3 bytes

Ｅθθ

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-4 bytes thanks to Neil, -1 byte thanks to Erik the Outgolfer

Explanation

Ｅ  Ｉθ Map with lambda returning first input cast to integer
  Ｎ    Implicit range over next (first) input

Answer 7

Integral, 6 Bytes

Outputs the list as individual stack elements joined by the newline.

^{If built-ins weren't banned, it would be possible via ♫ (1 bytes).}

◙v◙♦▼e

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Explanation

◙      Duplicate input, in order to get around the edge case 1
 v◙    "◙": Create a copy of the operand, when evaluated
   ♦   Swap up a copy of the input
    ▼  Decrement
     e Repeat input-1 times: Duplicate stack

Answer 8

MAWP, 13 bytes

%@!1A[1A~!~]%

-15 bytes after integer input rework (Dion).

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Answer 9

Keg, 5 bytes

:(|:.

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Answer 10

Arn, 7 bytes

|ÅF└¾c)

Explanation

Unpacked: v{_}\1=>

    \ Fold/map (in this case just map) with...
v{ A block with the key of v
  _ Input
}
      1=> Range, [1, in]

Answer 11

Braingolf, 6 bytes

!A.$_=

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Explanation

!A.$_=  Implicit input from commandline args
!A      Read top of stack without popping, run next operator that many times
  .     Duplicate top of stack
   $_   Pop top of stack
     =  Print stack

We have to pop the top of the stack because otherwise we'll have 1 too many copies of n. If we remove the ! and let A pop the top of the stack, duplicate will have nothing to duplicate.

Answer 12

[Swift 3], 55 49 bytes

func a(n:Int)->[Int] {return (0..<n).map{_ in n}}

Answer 13

REXX, 24 bytes

arg n
say copies(n' ',n)

Uses the built-in function copies(), and naturally, this being REXX, the copies are integers just as much as they are strings.

Answer 14

WendyScript, 10 bytes

#:(x)[x]*x

(#:(x)[x]*x)(5) // => [5,5,5,5,5]

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Answer 15

V, 5 bytes

Du@"Ä

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Answer 16

Pyth, 5 bytes

Super simple

.[QQY

Explanation:

.[QQY     Pad Y with eval(input) copies of eval(input)

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Answer 17

R, 7 bytes

0*1:n+n

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Answer 18

Julia 12

~x=fill(x,x)

Also it kinda generalizes to tuples making tensors.

julia> ~3
3-element Array{Int64,1}:
 3
 3
 3

julia> ~tuple(2,4)
2×4 Array{Tuple{Int64,Int64},2}:
 (2, 4)  (2, 4)  (2, 4)  (2, 4)
 (2, 4)  (2, 4)  (2, 4)  (2, 4)

julia> ~tuple(2,4,3)
2×4×3 Array{Tuple{Int64,Int64,Int64},3}:
[:, :, 1] =
 (2, 4, 3)  (2, 4, 3)  (2, 4, 3)  (2, 4, 3)
 (2, 4, 3)  (2, 4, 3)  (2, 4, 3)  (2, 4, 3)

[:, :, 2] =
 (2, 4, 3)  (2, 4, 3)  (2, 4, 3)  (2, 4, 3)
 (2, 4, 3)  (2, 4, 3)  (2, 4, 3)  (2, 4, 3)

[:, :, 3] =
 (2, 4, 3)  (2, 4, 3)  (2, 4, 3)  (2, 4, 3)
 (2, 4, 3)  (2, 4, 3)  (2, 4, 3)  (2, 4, 3)

Answer 19

TI-BASIC, 12 bytes

Ans→dim(L₁:Ans(L₁+1

Assumes an empty list.

Answer 20

Objective-C 107 Bytes

#define a NSMutableArray
-(a*)f:(int)n{a*m=[a arrayWithCapacity:n];for(int i=0;i<n;i++)m[i]=@(n);return m;}

A little macro abuse and a basic loop, everyone else took the good languages...

Usage

[self f:5];

returns an array of number objects of the given value

Answer 21

VBA, 75 Bytes

Defined Function that takes input n of expected type Integer and returns an array of length n filled with n to STDOUT

Function l(n)
Dim o
ReDim o(1To n)
For i=1To n
o(i)=n
Next
l=o
End Function

Answer 22

Microscript II, 14 bytes

$s{o`s`+s}sNv*

Answer 23

C++, 86 56 45 bytes

Thanks to @ceilingcat for saving 30 41 bytes!

#import<list>
#define f(n)std::list<int>(n,n)

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Answer 24

Ly, 13 10 bytes

ns[>l<1-]>

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Prints space-separated numbers. If outputting without a separator is allowed, ns[lu1-]; works for 9 bytes.

Explanation:

ns[>l<1-]>

ns         # take input and save it
  [        # loop
   >l<     # load save and push it to another stack
      1-   # decrement
        ]  # end loop
         > # go to other stack
           # implicitly print

Answer 25

F#, 32 bytes

let m n=List.map(fun x->n)[1..n]

Creates a list 1 to n, then mapping a function to each element to replace it with n.

Answer 26

Axiom, 23 bytes

f(x)==[x for i in 1..x]

This 15 bytes would be better

f(x)⊳[x|i∈1..x]

f(x) return the list contains x element x but Axiom is a good approximation to that. Goes under 15 bytes it seems too much golfed

Answer 27

Ruby, 12 bytes

->(n){[n]*n}

13 bytes

Array.new n,n

Answer 28

QBIC, 17 bytes

dim g(:)[a|g(b)=a

This is a direct port of my QBasic answer (48 bytes), and QBIC does what it does best: compressing QBasic.

Explanation

dim g(:) Defines an array of x elements, where x is a cmd line param
[a|      FOR b = 1 TO a (the : above creates 'a' and assigns the cmd line param to it)
g(b)=a   Set each array element to 'a'

Answer 29

dc, 21

?dsn[lnpr1-d0!=F]dsFx

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Ungolfed:

# Read a number from standard input (?), save it to register n (sn) and leave a copy on the stack (d)
?dsn

# Function F: print register n as many times as the top of the stack shows
[            # start of macro
  lnp        # print register n
  r          # swap the counter to the top
  1-         # decrement counter
  d          # duplicate counter for the comparison
  0 !=F      # if counter is not zero run F again
]dsFx        # end of macro, save it to F (sF) and execute (d x)

Run on the command line like this:

echo 5 | dc dcscript

Output:

5
5
5
5
5

This shorter script works for n>1 (15 bytes):

? [d d z 1- <F]dsFx f

This uses the depth of the stack to determine when to stop.

Answer 30

8th, 42 37 bytes

Code

>r a:new ( r@ a:push ) r@ times rdrop

SED (Stack Effect Diagram) is: n -- a

Usage

: f >r a:new ( r@ a:push ) r@ times rdrop ;

ok> 5 f .
[5,5,5,5,5]

We could save another 6 bytes if we accept the idea of leaving numbers on r-stack after processing.