Background

This is a standard textbook example to demonstrate for loops.

This is one of the first programs I learnt when I started learning programming ~10 years ago.

You are to print this exact text:

**********
**********
**********
**********
**********
**********
**********
**********
**********
**********


Specs

• You may have extra trailing newlines.
• You may have extra trailing spaces (U+0020) at the end of each line, including the extra trailing newlines.

Scoring

This is . Shortest answer in bytes wins.

• @DylanMeeus "You are to print this exact text:" Aug 4, 2016 at 12:56
• @DylanMeeus Since that is to do with the dev tools hiding repeated console outputs, and isn't native to JavaScript consoles as a whole and is not in the JavaScript spec - as well as the fact that feature can be turned off - i think it should be acceptable. Not all browsers will collapse it like that. Aug 4, 2016 at 12:58
• One of the most interesting things about this challange is that depending on your language ********** can be shorter then a loop. Makes me wonder when it's better for a given language to switch between 1 or 2 loops. Aug 5, 2016 at 9:14
• you say trailing new lines are acceptable. Are leading newlines acceptable too? Feb 10, 2017 at 2:34

51AC8, 10 bytes

10[10\*×t]


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Explanation

10[10\*×t]
10          # Push 10 (for each loop; looping var)
[        # Begin foreach loop (for 10 times)
10      # Push 10
\*    # Push '*'
×   # Multiply
t  # Pop and print
] # End foreach



K (ngn/k), 1614 13 bytes

 0:10 10#"*"


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Thanks to @Bubbler in the k tree for helping me out with this.

¯2 bytes thanks to @Bubbler

¯1 byte thanks to @Bubbler

Vyxal, jH, 654 3 bytes

×*²


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Explained

×*²
×*      # Push 100 asterisks onto the stack (as a single string) // the H flag initalises the stack with 100
²     # Split into pieces of 10 and use the j flag to join on newlines.


Alternatively:

Vyxal, 5 bytes

₁×*²⁋


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Elixir, 35 bytes

for _<-0..9,do: IO.puts"**********"


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Originally suggested to @Chester Lynn on their answer.

Commodore BASIC (C64/128, PET, VIC-20, C16/+4, THEC64/Mini, Ultimate64) ~35 tokenised BASIC bytes

0a$="**********":fori=.to9:?a$:next


Very simply, a string is created called a$; this is of 10 asterisks. This is printed to the screen in a for/next loop, which is from zero to nine inclusive. Each string is printed onto a new row on the screen. A few tokenised BASIC bytes could be saved with: 0fori=.to9:?"**********":next  Though I think this would be less performant (not that performance will matter with such a trivial BASIC listing). • A minor tweak and it's also valid Applesoft Basic 0a$="**********":fori=0to9:?a$:next Aug 26, 2021 at 2:55 • You can also do 0?"**********":a=a+1:ifa<10thengoto in Commodore BASIC; without a line number following the goto command, Commodore BASIC will goto 0 as long as there is a line zero Aug 26, 2021 at 12:58 V (vim), -v 11 bytes i*<esc>yl9pyy9p  Try it online! Insert a * and copy it 9 times towards left and copy the line and paste it nine times. Also I am a sock of someone :P -4 as I am an idot. • Ooh nice, a fellow sockpuppet Oct 1, 2021 at 15:32 • 9i*<esc>yy9p? Oct 15, 2021 at 11:50 • Also, this is 11 bytes, not 15 bytes, because <esc> is one byte Oct 15, 2021 at 11:56 • dom!!!e @pxeger – flax Oct 16, 2021 at 7:52 Python 3, 25 24 bytes -1 byte thanks to @Jo King *map(print,['*'*10]*10),  Try it online! Unpacks the map object so it actually prints the output, instead of optimising it away. Cascade, 32 27 bytes " * * * *@ *} *| */ * *. */  Try it online! Aha! It turns out just printing ten asterisks ten times does beat out my more complex answer. Unfortunately though, writing this program exposed not one, but two bugs in my interpreter. Below is my longer, but higher effort solution: Cascade, 32 bytes _9'2 ^*| \ !] (.n n?\ \%!| n)+\  Try it online! One of my more compact Cascade programs. I'm still not sure if it would be shorter to just print 10 asterisks 10 times instead of using a counter, but I'm satisfied with this, especially how the bottom intersects with the top. Explanation The easiest way to understand the code is know the basic structure. Each instruction can take up to three inputs, each one below it.  + lcr  These are the left, center and right arguments. In this case, the + instruction is dyadic, meaning it takes the left and right arguments. Most instructions are either dyadic or monadic (taking one argument, the center one). The /\| instructions take only the argument they are pointing to, and the ! instructions skips over the center and takes the argument two below it. Each of these arguments can themselves be code instructions, meaning they chain together in a prefix like notation. For example, this code could be represented by the recursive Lisp-like code: code = (if (0==(n=dec n)) doboth (print ( if (% n (+9 2)) ('*') else (inc 9) )) (code) )  If that doesn't make sense, here's an expanded look at the code, which is still a little confusing, but it at least has an idea of which parts are connected to each other through the |/\Xs. Note that the top and bottom rows are the overlap between the two (and same with the left and right). \ |/|\ \ _ 9 ' 2 / \ | / ^ * | / \ | \ ! ] \ \ \/ \ ( . n/ | | | n ? \ /|\ \ \ % ! | X \\ | n ) +/\ \ |/|\ \  Starting from the top left corner, we have the first check (_) which executes the right branch only if the left is successful. The left goes down to the ], which sets the n variable (initially 110) to the decrement (() of n, i.e. n=n-1. The check then takes the result of this (the new value of n) and continues if it is positive. This moves onto the branch instruction on the right (^), which executes both the left and right branches. The left prints (.) the value given by the choice instruction (?). This branches depending on the center value, which is the modulo (%) of n and the addition (+) of 9 and 2. Note that this wraps around to the top again for those digits. If n%(9+2) is 0, then we branch left, which navigates around the % and returns the increment ()) of 9, printing a newline. If it is not divisible by 11, then we go right, skip over the + with a !, and return the character (') of * to print. Now the right branch of the ^ skips over the n, then goes right, down, and right again, wrapping around both the right edge and the bottom edge to loop back to the _ in the top left. This now loops over the exact same code until n has reached 0, printing ten asterisks and then a newline. Minim, 42 37 35 Bytes New solution halts by checking if [0] > 99. $<42._^++[0]%10.$<10._^[0]>99.C=-1.  With whitespace and comments: $< 42.         ; Print 42 as unicode '*'
_^ ++[0] % 10. ; Increment index 0 and skip next stmt if index 0 mod 10 is nonzero
$< 10. ; Print 10 as unicode '\n' _^ [0] > 99. ; Skip next stmt if index 0 is greater than 99 C = -1. ; Set program counter to -1 (advances to 0 afterwards)  Old solutions halted by checking [0] == 100... $<42._^++[0]%10.$<10._^[0]==100.C=-1.  ... or used labels and gotos. _>1.$<42._^++[0]%10.$<10._<?([0]-100)._>0.  GitHub Repository • Nice answer! Did you make this language? (Also, welcome to Code Golf!) Aug 13, 2021 at 3:27 • @RedwolfPrograms Thank ya kindly~ Yes indeed! This is an old esolang (2017) that I resurrected last month, and have been in the process of finishing. I didn't do too bad for my first entry, I suppose! And thank you~ Aug 13, 2021 at 3:37 • Also just realized that my program's size is the ASCII value of the asterisk! How serendipitous~ Aug 13, 2021 at 4:11 • When I went to the esolangs article, I instantly recognised this as being from Truttle server. Welcome to the wonderful world of Code Golf! Aug 13, 2021 at 11:17 Python 3, 29 23 bytes print(('*'*10+'\n')*10)  Thanks to orlp for shaving off 6 bytes. • We require either a full program or a function. An expression is not sufficient. – orlp Aug 4, 2016 at 9:35 • Yes, I'm referring to your second paragraph. – orlp Aug 4, 2016 at 9:36 • print((10*"*"+"\n")*10) is shorter. – orlp Aug 4, 2016 at 9:37 • darnit we had the same answer :( Aug 4, 2016 at 9:50 • Changing print to exit will save 1 byte and still output the result, there wasn't specified if it should be printed on standard output. Aug 4, 2016 at 13:50 Golisp, 34 bytes for[range@10{(_)writeln@*["**"5]}]  Due to a "bug", I can't concatenate strings... • Congratulations! Aug 4, 2016 at 10:29 • @LeakyNun For what? Aug 4, 2016 at 10:30 • For developing a language. Aug 4, 2016 at 10:30 • @LeakyNun Just look at my GitHub repos... Aug 4, 2016 at 10:35 • I didn't say "for developing your first language" Aug 4, 2016 at 10:35 SpecBAS - 18 bytes ?(("*"*10)+#13)*10  ? is shorthand for PRINT, #13 is the equivalent of \n in other languages. C#, 79 bytes class P{void Main(){for(int i=0;i++<10;)System.Console.Write("**********\n");}}  • This won't run without static void right? – pay Aug 4, 2016 at 15:49 • @pay has void already not sure if it needs static will check later when I have chance Aug 4, 2016 at 15:59 golflua, 22 characters ~@i=0,9 w"**********"$


Sample run:

bash-4.3$golflua -e '~@i=0,9 w"**********"$'
**********
**********
**********
**********
**********
**********
**********
**********
**********
**********


C++, 75 bytes

#include<cstdio>
int main(){for(int i=0;i<10;++i)std::puts("**********");}

• Why the downvote? Aug 4, 2016 at 13:00
• I believe this was an automatic downvote from the community user. Your post was auto-flagged as low quality (since it only contained code) and Leaky Nun's edit caused an automatic unowned downvote. See meta.stackexchange.com/q/236883 Aug 4, 2016 at 13:03
• That's actually not true. I don't think it's fair to assume. Aug 4, 2016 at 13:55
• You can save 1 byte by changing the for-loop to: for(int i=0;++i<11;) Aug 4, 2016 at 14:07
• Use int main(i) and remove the int i=0. Then, replace i<10;++i with i++<10;. -7 Aug 4, 2016 at 14:58

T-SQL, 50 bytes

print replicate(replicate('*',10)+char(10),10)

• not an T-SQL expert but wouldn't 'print replicate('**********'+char(10),10)' be shorter? Aug 5, 2016 at 8:25

C, 47 bytes

main(i){while(i<111)putchar(10|!!(i++%11)<<5);}


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Not as compact as the other C answer (putchar is such a long name!), but I don't use the asterisk character in my program. It treats the output as a 11 by 10 grid, where the 11th character is the newline. It then computes the ASCII for '*' (10 + 32 = 42) or '\n' (10) for each position.

I could save one byte with this approach if I were to change the character expression to: 42-!(i++%11)*32, but that would require an asterisk.

I have no idea how to score this, but I decided to give it a shot since there's an answer on Notepad.

Here's the sequence:

* * * * * CTRL (hold) A D D D D D D D D D D

• If codegolf.stackexchange.com/a/88747/14732 is valid (and has upvotes), this is more than perfectly valid. Aug 4, 2016 at 20:27
• @Mego As long as that one exists, this one will exist as well Aug 5, 2016 at 7:47

Bash (pure), 30 bytes

printf '%.s**********
' {0..9}


Test it on Ideone.

How it works

Before calling the shell built-in printf, Bash expands the glob {0..9} to 0 1 2 3 4 5 6 7 8 9.

The format string

%.s**********


specifies a string whose first 0 characters are included in the output (%.s), followed by ten asterisks and a linefeed. printf repeats the format string as many times as needed to consume all arguments. Since .%s is an empty string, this results in the desired output.

MSX BASIC, 40 30 bytes

1FORR=1TO10:FORC=1TO10:?"*";:NEXT:?:NEXT


Update: D'oh... much shorter if I just print whole rows:

1FORR=1TO10:?"**********":NEXT

• -1 byte: 1FORR=0TO9:?"**********":NEXT Jul 2, 2018 at 9:55

Clojure, 38 bytes

(apply print(repeat 10"**********\n"))


Simple enough

• I'm afraid, that should be 38 bytes. Currently it prints only 8 x 8. Aug 5, 2016 at 13:17
• You are completely right. I keep posting wrong cases this whole day. Aug 5, 2016 at 13:52

(*****):*(
)*:*::*:**S


Scheme, 51 Bytes

(map (lambda (n)(display "**********\n"))(iota 10))


Ruby, 26 20 bytes

10.times{puts'*'*10}


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Explanation

Runs 10 times; each time, it prints * (10 times), with an automatic line break at the end.

Kudos to OrangeFlash81 for saving me 6 bytes.

• I think 10.times{puts'*'*10} works as well and is shorter. Aug 7, 2016 at 8:26

Xtend, 33 bytes

[for(_:0..9)print('**********
')]


... a Java dialect; [] is a lambda definition

{{iiii}ii{c}{ddd}ddc{d}}


how it works:

{foo} does foo exactly ten times. first, it starts a ten loop, in this loop, it sets the counter to 42, prints it ten times (as char (asterisk)), decrements it to 10, prints it (newline), subtracts 10, then loops again, doing this 10 times.

• 23 bytes by setting the counter to 10 initially
– ovs
Sep 22, 2021 at 7:50

Code:

In cell A1 (12 bytes):

=REPT(B1,10)


In cell B1 (11 bytes):

**********        # Contains New Line
# <--


Result:

Previous Attempt #1 (31 bytes):

=REPT(REPT("*",10)&CHAR(10),10)


Previous Attempt #2 (also 31 bytes):

=REPT("**********"&CHAR(10),10)


S.I.L.O.S 45 bytes

All whitespace necessary due to my poor interpreter design.

a = 10
lbla
printLine **********
a - 1
if a a


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• 19 bytes off Aug 27, 2016 at 20:46

Lua, 34 bytes

for i=1,10 do print"**********"end


Befunge-93 (PyFunge), 28 bytes

#<90v0-1_@#:,
#^_ >1+"*",:9


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