# Specifications

Given a number n, output an ASCII "meme arrow" (greater-than symbol, >) of size n.

n will always be a positive integer, greater than 0.

## Examples

n = 2

\
\
/
/


n = 5

\
\
\
\
\
/
/
/
/
/


# Sample code

Here is a sample program, written in Crystal, that returns the correct results. Run it as ./arrow 10.

arrow.cr:

def f(i)
i.times { |j|
j.times { print ' ' }
puts "\\"
}
i.times { |j|
(i-j-1).times { print ' ' }
puts '/'
}
end

f(ARGV[0].to_i)


# Rules

• This is . The shortest answer wins. However, I will not select an answer, because the shortest answer may change over time.
• Standard loopholes are not allowed.
• Are you sure this is not a duplicate (that is, did you sandbox this)? I think this is likely to be, but it's slightly nearly impossible to search for. – my pronoun is monicareinstate Jun 9 at 4:46
• Can n be zero? – xnor Jun 9 at 6:05
• I think it's pretty silly to call this common symbol a "meme arrow". They're obviously comedy chevrons. – scatter Jun 10 at 18:04
• @Christian They're actually amusing angles – dkudriavtsev Jun 10 at 20:39
• @ArtemisFowl I thought they were interesting increases – dkudriavtsev Jun 12 at 22:17

# Java 8, 80 bytes

n->{for(int i=0;i<2*n;)System.out.printf("%"+(i++<n?i:n-~n-i)+"c\n",i>n?47:92);}


Try it online.

Explanation:

n->{                    // Method with integer parameter and no return-type
for(int i=0;i<2*n;)   //  Loop i in the range [0, 2 * input n):
System.out.printf(  //   Print with format:
"%"+(i++<n?       //    If i is smaller than input n:
//    (and increase i by 1 right after this check with i++)
i           //     Use i leading spaces
:            //    Else:
n-~n-i)     //     Use 2 * n - i + 1 leading spaces
+"c\n",i>n?       //    If i is larger than input n:
47        //     Append a trailing '/' and newline
:          //    Else:
92);}     //     Append a trailing '\' and newline


# Japt-R, 15 bytes

"\\/"¬c@õ!ùX zY


Try it

õ!ù'\ cUÆ'/ ùU´


Try it

# Pyth, 23 bytes

V2VQ
+*?N--QH1H\ ?N\/\\


Try it online!

Note that the program must not have a terminating newline! If saving it on linux, you can remove the last byte in a file using head -c -1 <file>.

I think this is the first time I have used for loops in pyth? Anyways, this is a rather direct port of the example source code in the question (though the 2 loops are combined, and the innermost loop is converted into string multiplication, and the remaining 2 prints are joined into one).

# Pyth, 19 bytes

VQ+*dN\\)V_Q+*dhN\/


Try it online!

TIL multiplying a string by a negative value multiplies it by its absolute value.

# Batch File, 117 + 9 bytes

Executed using cmd/v/q/c.

for /l %%i in (1,1,%1)do (for /l %%j in (2,1,%%i)do set %%i= !%%i!)&echo !%%i!\
for /l %%i in (%1,-1,1)do echo !%%i!/


# Pyth, 18 bytes

js.e+R?k\/\\b_B*L;


Try it online!

# Python 3.8 (pre-release), 80 bytes

lambda n:'\n'.join([' '*i+'\\'for i in(r:=range(n))]+[' '*(n+~i)+'/'for i in r])


Try it online!

Same approach as the one I suggested to @Artemis Fowl, but takes advantage of the walrus operator.

# Swift 5/Xcode 10.2.1, 91 bytes

(0..<n*2).map{String(repeating:" ",count:$0<n ?$0:n*2-$0-1)+($0<n ?"\\":"/")+"\n"}.joined()


Try it online!

# Pepe, 91 bytes

rErEeEeEEEeeREeEREErereeereeERrEeeEeeeeeRrEeeeReereRerEEEererEeeEeEEEErEEEEREErereeereeERee


Try it online!

# Perl 5, 59 bytes

$_--;say$"x$_."\\"for(0..$_);say$"x$_."/"for(reverse 0..$_)  Example execution: perl -nE '$_--;say$"x$_."\\"for(0..$_);say$"x$_."/"for(reverse 0..$_)'


Waits for an integer to be provided on STDIN when executed. Once provided, it will display the arrow.