ASCII Triangles

Question

Your task is to write a program or a function that prints an ASCII triangle. They look like this:

|\
| \
|  \
----

Your program will take a single numeric input n, with the constraints 0 <= n <= 1000. The above triangle had a value of n=3.

The ASCII triangle will have n backslashes (\) and vertical bars (|), n+1 lines and dashes (-), and each line will have an amount of spaces equal to the line number (0-based, ie first line is line 0) besides the ultimate line.

Examples:

Input:

4

Output:

|\
| \
|  \
|   \
-----

Input:

0

Output:

In this test case, the output must be empty. No whitespace.

Input:

1

Output:

|\
--

Input & output must be exactly how I specified.

This is code-golf, so aim for the shortest code possible!

Answer 1

Jelly, 14 bytes

’⁶x⁾|\jṄµ€Ṫ”-ṁ

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How it works.

’⁶x⁾|\jṄµ€Ṫ”-ṁ  Main link. Argument: n

        µ       Combine the links to the left into a chain.
         €      Map the chain over [1, ..., n]; for each k:
’                 Decrement; yield k-1.
 ⁶x               Repeat the space character k-1 times, yielding a string.
   ⁾\j            Join the character array ['|', '\'], separating by those spaces.
      Ṅ           Print the result, followed by a linefeed.
         Ṫ      Tail; extract the last line.
                This will yield 0 if the array is empty.
          ⁾-ṁ   Mold the character '-' like that line (or 0), yielding a string
                of an equal amount of hyphen-minus characters.  

Answer 2

C, 58 bytes

i;f(n){for(i=2*n;~i--;printf(i<n?"-":"|%*c\n",2*n-i,92));}

--

Thanks to @Steadybox who's comments on this answer helped me shave a few bytes in my above solution

Answer 3

Javascript (ES6), 97 85 81 75 74 bytes

n=>(g=(n,s)=>n?g(--n,`|${" ".repeat(n)}\\
`+s):s)(n,"")+"-".repeat(n&&n+1)

Turns out I wasn't using nearly enough recursion

f=n=>(g=(n,s)=>n?g(--n,`|${" ".repeat(n)}\\
`+s):s)(n,"")+"-".repeat(n&&n+1)

console.log(f(0))
console.log(f(1))
console.log(f(2))
console.log(f(3))
console.log(f(4))

Answer 4

05AB1E, 16 15 16 bytes

Saved a byte thanks to Adnan

FðN×…|ÿ\}Dg'-×»?

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Explanation

F       }         # for N in range [0 ... input-1]
 ðN×              # push <space> repeated N times
    …|ÿ\          # to the middle of the string "|\"
         Dg       # get length of last string pushed
           '-×    # repeat "-" that many times
              »   # join strings by newline
               ?  # print without newline

Answer 5

V, 18 17 16 bytes

1 byte saved thanks to @nmjcman101 for using another way of outputting nothing if the input is 0

é\é|ÀñÙá ñÒ-xÀ«D

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Hexdump:

00000000: e95c e97c c0f1 d9e1 20f1 d22d 78c0 ab44  .\.|.... ..-x..D

Explanation (outdated)

We first have a loop to check if the argument is 0. If so, the code below executes (|\ is written). Otherwise, nothing is written and the buffer is empty.

Àñ     ñ            " Argument times do:
  é\é|              " Write |\
      h             " Exit loop by creating a breaking error

Now that we got the top of the triangle, we need to create its body.

Àñ   ñ              " Argument times do:
  Ù                 " Duplicate line, the cursor comes down
   à<SPACE>         " Append a space

Now we got one extra line at the bottom of the buffer. This has to be replaced with -s.

Ó-                  " Replace every character with a -
   x                " Delete the extra '-'

---

This answer would be shorter if we could whatever we want for input 0

V, 14 13 bytes

é\é|ÀñÙá ñÒ-x

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Answer 6

C#, 93 bytes

n=>{var s=n>0?new string('-',n+1):"";while(n-->0)s="|"+new string(' ',n)+"\\\n"+s;return s;};

Anonymous function which returns the ASCII triangle as a string.

Full program with ungolfed, commented function and test cases:

using System;

class ASCIITriangles
{
    static void Main()
    {
      Func<int, string> f =
      n =>
      {
          // creates the triangle's bottom, made of dashes
          // or an empty string if n == 0
          var s = n > 0 ? new string('-', n + 1) : "";

          // a bottom to top process
          while ( n-- > 0)
          // that creates each precedent line
            s = "|" + new string(' ', n) + "\\\n" + s;

          // and returns the resulting ASCII art
          return s;
      };

      // test cases:
      Console.WriteLine(f(4));
      Console.WriteLine(f(0));
      Console.WriteLine(f(1));
    }
}

Answer 7

CJam, 24 22 21 bytes

Saved 1 byte thanks to Martin Ender

ri_{S*'|\'\N}%\_g+'-*

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Explanation

ri                     e# Take an integer from input
  _                    e# Duplicate it
   {                   e# Map the following to the range from 0 to input-1
    S*                 e#   Put that many spaces
      '|               e#   Put a pipe
        \              e#   Swap the spaces and the pipe
         '\            e#   Put a backslash
           N           e#   Put a newline
            }%         e# (end of map block)
              \        e# Swap the top two stack elements (bring input to the top)
               _g+     e# Add the input's signum to itself. Effectively this increments any 
                       e#  non-zero number and leaves zero as zero.
                  '-*  e# Put that many dashes

Answer 8

Python 2, 69 bytes

lambda x:'\n'.join(['|'+' '*n+'\\'for n in range(x)]+['-'*-~x*(x>0)])

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Answer 9

SmileBASIC, 51 bytes

INPUT N
FOR I=0TO N-1?"|";" "*I;"\
NEXT?"-"*(N+!!N)

Answer 10

PowerShell, 51 67 bytes

param($n)if($n){1..$n|%{"|"+" "*--$_+"\"};write-host -n ('-'*++$n)}

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(Byte increase to account for no trailing newline)

Takes input $n and verifies it is non-zero. Then loops to construct the triangle, and finishes with a line of -. Implicit Write-Output happens at program completion.

Answer 11

C 101 93 75 bytes

f(n){i;for(i=0;i++<n;)printf("|%*c\n",i,92);for(;n--+1;)prin‌​tf("-");}

Ungolfed version

void f(int n)
{
  int i;

  for(i=0;i++<n;)
    printf("|%*c\n",i,92);

  for(;n--+1;)
    printf("-");

}

@Steadybox Thanks for pointing out, makes a lot of sense.

Answer 12

Common Lisp, 89 86 bytes

Creates an anonymous function that takes the n input and prints the triangle to *standard-output* (stdout, by default).

Golfed

(lambda(n)(when(< 0 n)(dotimes(i n)(format t"|~v@t\\~%"i))(format t"~v,,,'-<~>"(1+ n))))

Ungolfed

(lambda (n)
  (when (< 0 n)
    (dotimes (i n)
      (format t "|~v@t\\~%" i))
    (format t "~v,,,'-<~>" (1+ n))))

I'm sure I could make this shorter somehow.

Answer 13

Retina, 39 bytes

.*
$*
*\`(?<=(.*)).
|$.1$* \¶
1
-
-$
--

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Convert decimal input to unary. Replace each 1 with |<N-1 spaces>\¶, print, and undo replace. Replace each 1 with a hyphen, and the last hyphen with 2 hyphens. Tadaa!

Answer 14

Charcoal, 15 bytes

Ｎβ¿β«↓β→⁺¹β↖↖β»

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Breakdown

Ｎβ¿β«↓β→⁺¹β↖↖β»
Ｎβ               assign input to variable β
   ¿β«         »  if β != 0:
      ↓β           draw vertical line β bars long
        →⁺¹β       draw horizontal line β+1 dashes long
            ↖      move cursor up one line and left one character
             ↖β    draw diagonal line β slashes long

Answer 15

Japt, 20 bytes

Saved 2 bytes thanks to @ETHproductions

o@'|+SpX +'\Ãp-pUÄ)·

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Explanation

o@'|+SpX +'\Ãp-pUÄ)·
o                       // Creates a range from 0 to input
 @                      // Iterate through the array
  '|+                   // "|" + 
     SpX +              // S (" ") repeated X (index) times +
          '\Ã            // "\" }
             p-pU       // "-" repeated U (input) +1 times
                 Ä)·    // Join with newlines

Answer 16

J, 20 bytes

-13 bytes thanks to bob

*#' \|-'{~3,~2,.=@i.

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original: 33 bytes

(#&'| \'@(1,1,~])"0 i.),('-'#~>:)

ungolfed

(#&'| \' @ (1,1,~])"0 i.) , ('-'#~>:)

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Answer 17

QBIC, 41 bytes

:~a>0|[a|?@|`+space$(b-1)+@\`][a+1|Z=Z+@-

Explanation

:~a>0|  Gets a, and checks if a > 0
        If it isn't the program quits without printing anything
[a|     For b=1; b <= a; b++
?@|`+   Print "|"
space$  and a number of spaces
(b-1)   euqal to our current 1-based line - 1
+@\`    and a "\"
]       NEXT
[a+1|   FOR c=1; c <= a+1; c++
Z=Z+@-  Add a dash to Z$
        Z$ gets printed implicitly at the end of the program, if it holds anything
        The last string literal, IF and second FOR loop are closed implicitly.

Answer 18

Pyke, 18 17 bytes

I Fd*\|R\\s)Qh\-*

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Answer 19

MATL, 19 bytes

?'\|- '2GXyYc!3Yc!)

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?         % Implicit input. If non-zero
  '\|- '  %   Push this string
  2       %   Push 2
  G       %   Push input
  Xy      %   Identity matrix of that size
  Yc      %   Prepend a column of 2's to that matrix
  !       %   Transpose
  3       %   Push 3
  Yc      %   Postpend a column of 3's to the matrix
  !       %   Transpose
  )       %   Index into string
          % Implicit end. Implicit display

Answer 20

Python2, 73 bytes

n=input()
w=0
exec'print"|"+" "*w+"\\\\"+("\\n"+"-"*-~n)*(w>n-2);w+=1;'*n

A full program. I also tried string interpolation for the last line, but it turned out be a couple bytes longer :/

exec'print"|%s\\\\%s"%(" "*w,("\\n"+"-"*-~n)*(w>n-2));w+=1;'*n

Another solution at 73 bytes:

n=j=input()
exec'print"|"+" "*(n-j)+"\\\\"+("\\n"+"-"*-~n)*(j<2);j-=1;'*n

Test cases

0:

1:
|\
--

2:
|\
| \
---

3:
|\
| \
|  \
----

6:
|\
| \
|  \
|   \
|    \
|     \
-------

Answer 21

Pyth, 23 18 bytes

VQ++\|*dN\\)IQ*\-h

Test suite available online.
Thanks to Ven for golfing off 5 bytes.

Explanation

VQ++\|*dN\\)IQ*\-h
 Q           Q    Q  [Q is implicitly appended, initializes to eval(input)]
       d             [d initializes to ' ' (space)]
VQ         )         For N in range(0, eval(input)):
      *dN             Repeat space N times
   +\|                Prepend |
  +      \\           Append \
                      Implicitly print on new line
            IQ       If (input): [0 is falsy, all other valid inputs are truthy]
                 hQ   Increment input by 1
              *\-     Repeat - that many times
                      Implicitly print on new line

Answer 22

R, 101 bytes

for(i in 1:(n=scan())){stopifnot(n>0);cat("|",rep(" ",i-1),"\\\n",sep="")};cat("-",rep("-",n),sep="")

This code complies with the n=0 test-case if you only consider STDOUT !
Indeed, the stopifnot(n>0) part stops the script execution, displays nothing to STDOUT but writes Error: n > 0 is not TRUE to SDTERR.

Ungolfed :

for(i in 1:(n=scan()))
    {
    stopifnot(n>0)
    cat("|", rep(" ", i-1), "\\\n", sep = "")
    }

cat("-", rep("-", n), sep = "")

Answer 23

Python 2, 62 bytes

n=input();s='\\'
exec"print'|'+s;s=' '+s;"*n
if n:print'-'*-~n

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Prints line by line, each time adding another space before the backslash. If a function that doesn't print would be allowed, that would likely be shorter.

Answer 24

JavaScript (ES6), 71 bytes

f=
n=>console.log(' '.repeat(n).replace(/./g,'|$`\\\n')+'-'.repeat(n+!!n))

<form onsubmit=f(+i.value);return!true><input id=i type=number><input type=submit value=Go!>

Outputs to the console. Save 6 bytes if printing to the SpiderMonkey JavaScript shell is acceptable. Save 13 bytes if returning the output is acceptable.

Answer 25

Haskell, 82 65 bytes

g 0=""
g n=((take n$iterate(' ':)"\\\n")>>=('|':))++([0..n]>>"-")

Try it online! Usage:

Prelude> g 4
"|\\\n| \\\n|  \\\n|   \\\n-----"

Or more nicely:

Prelude> putStr $ g 4
|\
| \
|  \
|   \
-----

Answer 26

Python 2, 67 bytes

Another function in Python 2, using rjust.

lambda n:('|'.join(map('\\\n'.rjust,range(n+2)))+'-'*-~n)[4:]*(n>0)

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Answer 27

Python 3, 60 bytes

f=lambda n,k=0:k<n and'|'+' '*k+'\\\n'+f(n,k+1)or'-'[:n]*-~n

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Two more solutions with the same byte count.

f=lambda n,k=0:n and'|'+' '*k+'\\\n'+f(n-1,k+1)or-~k*'-'[:k]
f=lambda n,s='|':-~n*'-'[:n]if s[n:]else s+'\\\n'+f(n,s+' ')

Answer 28

Perl, 63 bytes

$n=shift;print'|',$"x--$_,"\\\n"for 1..$n;print'-'x++$n,$/if$n

Ungolfed:

$ perl -MO=Deparse triangle.pl
$n = shift @ARGV;
print '|', $" x --$_, "\\\n" foreach (1 .. $n);
print '-' x ++$n, $/ if $n;

$" is the list separator, which defaults to " ". $/ is the output record separator, which defaults to "\n". $_ is the implicit loop variable.

Answer 29

brainf**k, 256 bytes

[-]>[-]+[[-]>[-],[+[-----------[>[-]++++++[<------>-]<--<<[->>++++++++++<<]>>[-<<+>>]<+>]]]<]<>++++++++++[>+>+++>++++>+++++++++>++++++++++++<<<<<-]>>++>+++++>++>++++<<<<<<[>[-]+>>>>>.>[>+<<<<<.>>>>-]>[-<+>]<<<.>>+<<<<<.<<-]>>>>>>>[<<<.>>>-]<<<<<<[>>>.<<<-]

Try it online!

How it works

# Read input and convert to number (https://esolangs.org/wiki/Brainfuck_algorithms#Input_a_decimal_number)
[-]>[-]+[[-]>[-],[+[-----------[>[-]++++++[<------>-]<--<<[->>++++++++++<<]>>[-<<+>>]<+>]]]<]<

# We need the characters {LF} (10); {space} (32); {minus} (45); \(92) | (120)

P1 > ++++++++++
   [
p0  < 
P1  >
P2  >  +                     
P3  >  +++
P4  >  ++++
P5  >  +++++++++
P6  >  ++++++++++++
P5  <
P4  <
P3  <
P2  <
P1  < -
   ]
P0 <           # Contains n
P1 >           # is zero
P2 >           # is {LF}
P3 > ++        # is {space}
P4 > +++++     # is {minus}
P5 > ++        # is \
P6 > ++++      # is |
p5 <
p4 <
p3 <
p2 <
p1 <
p0 <
   [           # Loop through n
P1 >[-]+       # p1 will be 1 if input was not 0; used to output the additional {minus}
p2 >
p3 >
p4 >
p5 >
p6 > .
## output p7 spaces; copy p7 to p8
p7 >
   [
p8 > +
p7 < 
p6 <
p5 <
p4 <
p3 < .  # output {space}
p4 >
p5 >
p6 >
p7 > -
   ]
## copy p8 back to p7
p8 >
   [
   -
p7 < +   
p8 >
   ]      
p7 <
p6 <
p5 < . # output \
p6 >
p7 > + # increase number of spaces
p6 <
p5 <
p4 <
p3 <
p2 < . # output {LF}
p1 <
p0 < -
   ]
p1 >
p2 >
p3 >
p4 >
p5 >
p6 >
## output p7 minuses
p7 >
   [
p6 <
p5 <
p4 < .
p5 >
p6 >
p7 > -
   ]
p6 <
p5 <
p4 <
p3 <
p2 <
## output additional minus
p1 <
  [
p2 >  
p3 >
p4 > .
p3 <
p2 <
p1 < -
  ]

This is my first contribution and I'm not sure whether it is allowed to re-use code from others. My code above uses the code published in the Wiki to read the input characters and convert them into a number. Hope that's okay.

Answer 30

Python 3, 72 bytes

x=int(input())
for i in range(x):
    print('|'+' '*i+'\\')
print('-'*x)

First submission. Very basic and entry level.

Try it online!

Answer 31

Javascript 101(Full Program), 94(Function Output), 79(Return) bytes

Full Program

Will not run in Chrome (as process doesn't exist apparently)
Will not run in TIO (as prompt apparently isn't allowed)

x=prompt();s='';for(i=0;i<x;i++)s+='|'+' '.repeat(i)+`\\
`;process.stdout.write(s+'-'.repeat(x&&x+1))

Function with EXACT print

x=>{s='';for(i=0;i<x;)s+='|'+' '.repeat(i++)+`\\
`;process.stdout.write(s+'-'.repeat(x&&x+1))}

Try it Online

Function with return string

x=>{s='';for(i=0;i<x;)s+='|'+' '.repeat(i++)+`\\
`;return s+'-'.repeat(x&&x+1)}

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Repeating characters in Javascript is dumb and so is suppressing newlines on output

Answer 32

Python 2, 82 bytes

def f(i,c=0):
 if c<i:print'|'+' '*c+'\\';f(i,c+1)
 print'-'*((c+1,c)[c<1]);exit()

Try it online!

Longer that the other Python answers but a recursive function just to be different.

It feels wasteful using two print statements but I can't find a shorter way round it. Also the exit() wastes 7 to stop it printing decreasing number of - under the triangle.

Answer 33

Javascript (ES6), 84 bytes

Creates an anonymous function that presents the result on an alert box. No output is present if 0.

i=>{for(X=i,a='';i--;)a='|'+(' '.repeat(i))+'\\\n'+a;X&&alert(a+('-'.repeat(++X)))}

---

If printing is really required (137 bytes):

i=>for(X=i,d=document.body;i--;)d.innerHTML='|'+('&nbsp;'.repeat(i))+'\\<br>'+d.innerHTML;X&&window.print(d.innerHTML+=('-'.repeat(++X)))

Changes the HTML code and prints the page into a paper sheet. The page stays white if 0.

Answer 34

Befunge-98, 68 bytes

&:!#@_000pv<
:kg00 ',|'<|`g00:p00+1g00,a,\'$,kg00
00:k+1g00-'<@,k+1g

Answer 35

Clojure, 104 bytes

(fn[h](let[r #(apply str(repeat % %2))](doseq[n(range h)](println(str\|(r n\ )\\)))(print(r(inc h)\-))))

At least I should have lots of room to golf from here!

Loops over the range from 0 to h-1, printing a \|, then h-many spaces, then a \\. Finally, at the end, it prints h+1 many dashes.

(defn triangle [hypot-n]
  (let [r #(apply str (repeat % %2))] ; Create a shortcut string repeat function, since I need it twice.
    (doseq [line-n (range hypot-n)] ; Loop over the range 0 to hypot...
      (println (str \| (r line-n \ ) \\))) ; Printing a bar, then line-n many spaces, then a slash
    (print (r (inc hypot-n) \-)))) ; Then print the trailing dashes

Answer 36

Batch, 110 bytes

@set s=
@for /l %%i in (1,1,%1)do @call:c
@if %1 gtr 0 echo -%s: =-%
@exit/b
:c
@echo ^|%s%\
@set s= %s%

Special-casing 0 is no hardship as %s: =-% would fail anyway.

Answer 37

JavaScript, 66 bytes

f=(x,i=0)=>i<x?`|${" ".repeat(i)}\\\n`+f(x,i+1):"-".repeat(x&&i+1)

Answer 38

Mathematica, 57 bytes

Array[{"|"," "~Table~#,"\\
"}&,#,0]<>Table["-",#+Sign@#]&

Unnamed function taking a nonnegative integer as input and returning a string-with-newlines. It deals with the special case of input 0 by noting that the number of dashes is equal to the input plus the sign of the input.

Answer 39

Underload, 58 39 bytes

:()~(|)~(~!((-):S~^S)~:S( )*(\
)S)~^^!^

Assumes the input is a Church numeral on the stack.

Explanation

          stack (assume input N = 2):
            (:*)
:         keep a copy of the input on the bottom
            (:*)(:*)
()~       insert the finalization snippet
            (:*)()(:*)
(|)~      insert the accumulator
            (:*)()(|)(:*)
(...)~^^  run (...) N times:


    ~!(...)~  set finalization snippet
                (:*)((-):S~^S)(|)
    :S        output copy of accumulator
                (:*)((-):S~^S)(|)           output = |
    ( )*      append space to accumulator
                (:*)((-):S~^S)(| )          output = |
    (\;)S     output backslash and newline
               (:*)((-):S~^S)(| )          output = |\;


    ~!(...)~  set finalization snippet
                (:*)((-):S~^S)(| )
    :S        output copy of accumulator
                (:*)((-):S~^S)(| )          output = |\;| 
    ( )*      append space to accumulator
                (:*)((-):S~^S)(|  )         output = |\;| 
    (\;)S     output backslash and newline
                (:*)((-):S~^S)(|  )         output = |\;| \;

!         delete accumulator
            (:*)((-):S~^S)                  output = |\;| \;
^         execute finalization snippet:
            (:*)                            output = |\;| \;

    (-)       push hyphen
                (:*)(-)                     output = |\;| \;
    :S        output hyphen
                (:*)(-)                     output = |\;| \;-
    ~^        repeat N times
                (--)                        output = |\;| \;-
    S         print
                                            output = |\;| \;---

Answer 40

8th, 94 89 bytes

Code

: f 0; dup ( "|" . ( " " . ) swap times "\\" . cr ) 0 rot n:1- loop n:1+ "-" swap s:* . ;

Explanation

The word f requires an integer input

: f \ n -- 
   0;                      \ Exit if input equals to 0
   dup                     \ Duplicate input on the stack
   ( "|" .                 \ Print a piece of the vertical side
   ( " " . ) swap times    \ Print space between sides
   "\\" . cr )             \ Print a piece of the slanted side
   0 rot n:1- loop         \ Repeat the above operation for the size required 
   n:1+ "-" swap s:* .     \ Print the bottom side
;

Usage

ok> 4 f
|\
| \
|  \
|   \
-----

Answer 41

Ruby, 48 bytes

->n{puts ?-*(n.times{|x|puts ?|+' '*x+?\\}+(n<=>0))}

Answer 42

Jelly, 19 bytes

Ḷ×⁶j@€⁾|\;‘×”-$$YxṠ

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Explanation:

Ḷ×⁶j@€⁾|\;‘×”-$$YxṠ Link 0. Arguments: z.
Ḷ                   Range from 0 to z - 1. [arg-z]
  ⁶                 Space (' ').
 ×                  Product of x and y. [multi-char-string-warning]
      ⁾|\           Two char string ("|\").
   j                Join x on y.
    @               Reverse arguments.
     €              Convert this link to a map on the left argument.
          ‘         Increment z. [arg-z]
            ”-      Character ('-').
           ×        Product of x and y. [multi-char-string-warning]
              $     Merge last two links to a monadic link.
               $    Merge last two links to a monadic link.
         ;          Concatenate x and y.
                Y   Join z on newlines ('\n').
                  Ṡ Sign of z. [arg-z]
                 x  Repeat x y times.

Answer 43

bash + Unix utilities, 71 bytes

(($1))&&(echo '|\';(($1-1))&&$0 $[$1-1]|sed 's/|/| /;s/-/--/'||echo --)

Test program:

for n in 0 1 2 3 4 5; do echo $n; ./triangle $n; echo; done

Test output:

0

1
|\
--

2
|\
| \
---

3
|\
| \
|  \
----

4
|\
| \
|  \
|   \
-----

5
|\
| \
|  \
|   \
|    \
------

Answer 44

dc, 98 bytes

256?dse1+d[q]st1=t^124*23562+dsaP2sk[[lkd256r^32*la+dsaPd1+skle>y]srle1<r]dsyx[45Ple1-dse0!>q]dsqx

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Explanation

This takes due advantage of dc's P command, which utilizes conversion to base 256 on most systems. Therefore, for any input n, the program first raises 256 to the n + 1th power, multiplies the result by 124 (ASCII character |), and then adds 256*92+10=23562 to the product (where 92 is equivalent to the character \ and 10 is the decimal value of the new-line (\n)). This results in a decimal number that when converted to base 256 with P results in the output |\\n where \n is the literal new-line character. A duplicate of this decimal number is also stored on top of register a.

Then, a "macro-loop" is invoked, as long as n > 1, in which a counter is incremented until n, beginning from 2, and, as the 3rd through nth base 256 digits are unset, 256 is raised to each of those increments, a result which is then multiplied by 32 (the ASCII single space character). Then the value on top of register a is incremented by the resulting product, thus, on each iteration, setting each one of the unset base 256 digits in the between the | and the \ characters to a single space.

Finally, after all n-1 lines have been output, another "macro-loop" is invoked in which all the n+1 dashes are output through the feeding of 45 to P on each iteration.

Note: The [q]st1=t segment makes sure that nothing is output for the input 0 by checking if the incremented input is equal to one, and if it is, simply executes the macro [q] which exits the program.

Answer 45

Python 3, 75 bytes

def x(n):
 if n:
  for i in range(n):print('|'+' '*i+'\\')
  print('-'*-~n)

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Here's my Python 3 answer, and it's pretty long. I wonder how I can shorten it...

Thanks to Eric The Outgolfer for helping in the comments...

Answer 46

PHP, 75 bytes

(70 bytes if I used extended ascii)

for($p=str_pad;+$i<$argv[1];)echo$p("|",++$i),"\\
";echo$p("",$i+!!$i,"-");

takes input from command line argument. Run with -nr.

Answer 47

Swift 3 96 bytes

var a=5,b="";for c in 0..<a{var d="|";for _ in 0..<c{d+=" "};print(d+"\\");b+="-"};print(b+"-")

Answer 48

bash + printf, 68 bytes

for((;i<$1;))
{
a=$a-
printf "|%$[i++]s\\\\\n"
}
[ $a ] && echo $a-

Use "bash program_name number" to run. Sample run:

bash-4.1$ bash triangle 0
bash-4.1$ bash triangle 1
|\
--
bash-4.1$ bash triangle 2
|\
| \
---
bash-4.1$ bash triangle 3
|\
| \
|  \
----

Answer 49

VBA Excel, 68 65 bytes

Using Immediate Window and [a1] as input

for x=0to[a1]-1:?"|"string(x,32)"\":next:if x then?string(x+1,45)

Answer 50

Python 3, 61 bytes

n=5
q="".join(map(lambda x:"|%s\\\n"%(x*" "),range(n)))+"-"*n

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Answer 51

Perl 5, 50 +2 (-ap) = 52 bytes

$_&&='|\\';say,s/\\/ $&/ while$F[0]--;s/./-/g;chop

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Answer 52

Acc!!, 179 bytes

N
Count d while _%60-10 {
(_/60*10+_%60-48)*60+N
}
_/60
Count r while _-r {
Write 124
Count s while r-s {
Write 32
}
Write 92
Write 10
}
Count h while _+3*_/(3*_-1)-h {
Write 45
}

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Explanation

Load a number into the accumulator, print that many rows of | ... \, and print that many (plus 1) hyphens. Unless the number is 0, in which case print 0 hyphens. No trailing newline in any case.

Most of the code is quite straightforward (handy loop variable mnemonics: digit, row, space, and hyphen). The two interesting parts are inputting a multi-digit number and handling the 0 case.

Inputting a number

N
Count d while _%60-10 {
(_/60*10+_%60-48)*60+N
}
_/60

We divide the accumulator into two sections: _%60 stores the ASCII code of the character we just read, and _/60 stores the number we're constructing. We loop until _%60 is 10--that is, when we encounter a newline. Inside the loop, _%60 is the ASCII code of the new least-significant digit. _%60-48 is the digit itself. _/60 is the previous value of the number, so we multiply that by 10 and add the new digit to get the new value of the number: _/60*10+_%60-48. Multiply that quantity by 60 and add the next input character N, and we're ready to loop. After the loop finishes, we set the accumulator to _/60, the stored number itself.

Handling the 0 case

Count h while _+3*_/(3*_-1)-h {

My original loop condition was _+1-h: loop (accumulator + 1) times. But if the accumulator is 0, we want to loop 0 times, not 1. So instead of adding 1, we add 3*_/(3*_-1). When _ is 0, this quantity is 0. When _ is greater than 0, this quantity is 1 (since 3*_ is less than twice 3*_-1).

Answer 53

Java 8, 109 bytes

n->{String r="|\\\n";int i=0;for(;++i<n;r+=r.format("|%"+i+"s\\%n",""));for(;i-->=0;r+="-");return n<1?"":r;}

Can most likely be golfed some more.

Explanation:

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n->{                 // Method with integer parameter and String return-type
  String r="|\\\n";  //  Result-string, starting at "|\" + new-line
  int i=0;           //  Index-integer, starting at 0
  for(;++i<n;        //  Loop (1) from 1 to `n` (exclusive)
    r+=              //   Append to the result-String:
       r.format("|   //    A literal "|"
        %"+i+"s      //    + `i` amount of spaces
        \\           //    + a literal "\"
        %n","")      //    + a new-line
  );                 //  End of loop (1)
  for(;i-->=0;       //  Loop (2) from `n` down to 0 (inclusive)
    r+="-"           //   Append that many literal "-" to the result-String
  );                 //  End of loop (2)
  return n<1?        //  If the input was 0:
    ""               //   Return an empty String
   :                 //  Else:
    r;               //   Return the result-String
}                    // End of method

Answer 54

SNOBOL4 (CSNOBOL4), 98 bytes

	N =INPUT
T	OUTPUT =LT(X,N) '|' DUPL(' ',X) '\'	:F(E)
	X =X + 1	:(T)
E	OUTPUT =DUPL('-',X + 1)
END

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Answer 55

Japt -R, 14 bytes

°Æ'\i|úXÄÃpUç-

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°Æ'\i|úXÄÃpUç-     :Implicit input of integer U
°                  :Postfix increment U
 Æ                 :Map each X in the range [0,U)
  '\               :  Literal "\"
    i              :  Prepend
     |ú            :    "|" right padded with spaces to length
       XÄ          :   X+1
         Ã         :End map
          p        :Push
           Uç-     :  "-" repeated U (now incremented) times
                   :Implicit output, joined with newlines

Answer 56

Commodore BASIC (C64/128, TheC64 & Mini, VIC-20, PET, C16/+4) - byte count later; non-competing as this is PETSCII and not ASCII

0inputn:ifnthenfori=1ton:print"{shift+B}{left}"spc(i)"{shift+M}":next:fori=0ton:print"-";:next

There is a sanity check for zero, but any other positive integer will work.

Note, a C64 has 25 rows and 40 columns text (some PETs and the C128 in VDC mode has 80 columns, the VIC-20 is 22 columns by 23 rows) so the output will be skewed if you exceed the screen limit in either direction.

I've added a screenshot so that you may see the PETSCII characters that I can't make happen in this listing.