Python 2, 26 bytes

from textwrap import*

fill

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Meh... built-ins are boring... instead, have a nice 87-byte solution here:

s,n=input()

x=''

for i in s.split():c=n<len(x+i);exec'print x'*c;x=x*-~-c+i+' '

print x

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Outputs trailing spaces.

PHP, 8 bytes

Admittedly not the most original solution, but PHP has a native function that matches your requirements perfectly!

wordwrap:

string wordwrap ( string $str [, int $width = 75 [, string $break = "n" [, bool $cut = FALSE ]]] )

Wraps a string to a given number of characters using a string break character.

Use like so:

$str = "Lorem ipsum dolor sit amet, consectetur adipiscing elit. Sed eget erat lectus. Morbi mi mi, fringilla sed suscipit ullamcorper, tristique at mauris. Morbi non commodo nibh. Pellentesque habitant morbi tristique senectus et netus et malesuada fames ac turpis egestas. Sed at iaculis mauris. Praesent a sem augue. Nulla lectus sapien, auctor nec pharetra eu, tincidunt ac diam. Sed ligula arcu, aliquam quis velit aliquam, dictum varius erat.";

echo wordwrap($str, 50);

Or Try it online!

JavaScript (ES6),  75 73  72 bytes

Takes input as (string)(n).

s=>n=>s.split` `.map(w=>r=(u=r?r+' '+w:w)[n]?(o+=r+`

`,w):u,o=r='')&&o+r

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Variables

The formatted output is stored in $o$ (in green below).

The updated line $u$ is defined as the concatenation of:

• the current line $r$ (in black below)


• a space if $r$ is not empty, or nothing otherwise (in orange below)


• the new word $w$ (in blue below)

We need to insert a line break whenever the $n$-th character of $u$ is set (0-indexed, in red below).

Example

$n=16$ and $s$ = "LOREM IPSUM DOLOR"

Adding "LOREM":
$$smallbegin{array}{|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|}
hline
00&01&02&03&04&05&06&07&08&09&10&11&12&13&14&15&color{red}{16}\ hline
color{blue}L&color{blue}O&color{blue}R&color{blue}E&color{blue}M&&&&&&&&&&&&\ hlineend{array}
$$

Adding "IPSUM":
$$smallbegin{array}{|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|}
hline
00&01&02&03&04&05&06&07&08&09&10&11&12&13&14&15&color{red}{16}\ hline
L&O&R&E&M&color{orange}bullet&color{blue}I&color{blue}P&color{blue}S&color{blue}U&color{blue}M&&&&&&\ hlineend{array}
$$

Adding "DOLOR":
$$smallbegin{array}{|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|}
hline
00&01&02&03&04&05&06&07&08&09&10&11&12&13&14&15&color{red}{16}\ hline
L&O&R&E&M&bullet&I&P&S&U&M&color{orange}bullet&color{blue}D&color{blue}O&color{blue}L&color{blue}O&color{blue}R\ hlineend{array}
$$

$$smallbegin{array}{|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|}
hline
00&01&02&03&04&05&06&07&08&09&10&11&12&13&14&15&color{red}{16}\ hline
color{green}L&color{green}O&color{green}R&color{green}E&color{green}M&color{green}bullet&color{green}I&color{green}P&color{green}S&color{green}U&color{green}M&color{green}hookleftarrow&&&&&\ hline
D&O&L&O&R&&&&&&&&&&&&\ hlineend{array}
$$

Perl 6, 46 29 bytes

{;*.comb(/.**{1..$_}[s|$]/)}

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Regex based solution that takes input curried, like f(n)(s) and returns a list of lines. Each line except the last has a trailing whitespace

Explanation:

{;*                         }   # Anonymous code block that returns a Whatever lambda

   .comb(/                /)    # Split the string by

          .**{1..$_}            # Up to n characters

                    [s|$]      # Terminated by a whitespace char or the end of the string

Vim, 15 bytes/keystrokes

DJ:se tw=<C-r>"

gq_

A text formatting question? I know just the tool for the job! And it even has my name in the first two keystrokes :D

<C-r> means ctrl-r.

This could ever so slightly shorter in V, but I prefer answering in vanilla vim for answers that really show off how concise vim can be for the right challenge. And the difference is so small anyway.

This could also be the following for 15 bytes as well:

:se tw=<C-r><C-w>

ddgq_

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R, 36 27 bytes

R has this as a built-in (strwrap), we return a vector of split lines.

function(s,n)strwrap(s,n+1)

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Haskell, 70 bytes

s!n|length s<=n=[s]|(t,_:d)<-splitAt(until((<'!').(s!!))pred n)s=t:d!n

Python 2, 74 bytes

s,n=input()

while s:i=n;exec"i-=' '<(s+' '*n)[i];"*n;print s[:i];s=s[i+1:]

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Java (JDK), 46 44 bytes

Basically a pure regex solution in Java, almost certainly the shortest I've written.

Cheers to Kevin for helping to cut down the bytes in the regex even further!

n->s->s.replaceAll(".{1,"+n+"}( |$)","$0n")

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Using a curried lamdba, it creates a regex to greedily match up to n characters followed by either a space or end of string. It then replaces those characters with themselves followed by a newline.

Mathematica, 16 bytes

InsertLinebreaks

Built-in function. Takes a string and an integer as input and returns a string as output.

InsertLinebreaks["string", n]

 inserts newline characters to make no line longer than n characters.

C (gcc), 68 bytes

i;b(s,n,l)char*s,*l;{for(i=n;*++s;i--||(i=l-s+n,*l=10))l=*s-32?l:s;}

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Thanks to ceilingcat, save 2 bytes by moving global char*l to parameter.

Retina 0.8.2, 37 bytes

.+$

$*

!`(?=S.*¶(1)+)(?<-1>.)+(?=s)

Try it online! Takes s and n on separate lines. Explanation:

.+$

$*

Convert n to unary.

(?=S.*¶(1)+)(?<-1>.)+(?=s)

Match non-whitespace, then look ahead to n and count it as $#1. Then go back and use a balancing group to match up to n characters followed by whitespace.

!`

Output the matches as a list of lines.

Charcoal, 19 bytes

Ｎθ←Ｆ⪪Ｓ «¿‹⁺ＬιⅈθＭ→⸿ι

Try it online! Link is to verbose version of code. Takes input of n and s on separate lines. Explanation:

Ｎθ

Input n.

←

Move the cursor left one square to balance the right movement from the first iteration of the loop.

Ｆ⪪Ｓ «

Split the string on spaces and loop over the words.

¿‹⁺Ｌιⅈθ

Calculate whether the next word will reach the right edge.

Ｍ→

If it will not then move one square right.

⸿

If it will then start a new line.

ι

Output the word.

Red, 125, 117, 114 112 bytes

func[s n][d: 0 parse s[any[to" "p:" "opt[[to" "| to end]q:(if(-1 - d + index? q)> n[p/1: #"^/"d: index? p])]]]s]

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05AB1E, 18 bytes

õs#vDy«g²›i,}yðJ}?

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Explanation:

õ                   # Push an empty string "" to the stack

 s                  # Swap to take the (implicit) string input

  #                 # Split it by spaces

   v            }   # For-each `y` over the words:

    D               #  Duplicate the top of the stack

                    #  (which is the empty string in the very first iteration)

     y«             #  Append the current word `y`

       g            #  Get its length

        ²›i }       #  If its lengthy is larger than the second input:

           ,        #   Pop and output the current duplicated value with trailing newline

             yð     #  Push the word `y` and a space " "

               J    #  Join the entire stack together

                 ?  # After the loop, output the last part as well (without newline)

Powershell, 40 83 bytes

Test case with n=80 added.

param($s,$n)$s-split' '|%{if(($o+$_|% le*)-lt$n){$o+=' '*!!$o+$_}else{$o;$o=$_}}

$o

Test script:

$f = {



param($s,$n)$s-split' '|%{if(($o+$_|% le*)-lt$n){$o+=' '*!!$o+$_}else{$o;$o=$_}}

$o



}



@(

,(50, "Lorem ipsum dolor sit amet, consectetur adipiscing elit. Sed eget erat lectus. Morbi mi mi, fringilla sed suscipit ullamcorper, tristique at mauris. Morbi non commodo nibh. Pellentesque habitant morbi tristique senectus et netus et malesuada fames ac turpis egestas. Sed at iaculis mauris. Praesent a sem augue. Nulla lectus sapien, auctor nec pharetra eu, tincidunt ac diam. Sed ligula arcu, aliquam quis velit aliquam, dictum varius erat.",

"Lorem ipsum dolor sit amet, consectetur adipiscing",

"elit. Sed eget erat lectus. Morbi mi mi, fringilla",

"sed suscipit ullamcorper, tristique at mauris.",

"Morbi non commodo nibh. Pellentesque habitant",

"morbi tristique senectus et netus et malesuada",

"fames ac turpis egestas. Sed at iaculis mauris.",

"Praesent a sem augue. Nulla lectus sapien, auctor",

"nec pharetra eu, tincidunt ac diam. Sed ligula",

"arcu, aliquam quis velit aliquam, dictum varius",

"erat.")

,(80, "Lorem ipsum dolor sit amet, consectetur adipiscing elit. Sed eget erat lectus. Morbi mi mi, fringilla sed suscipit ullamcorper, tristique at mauris. Morbi non commodo nibh. Pellentesque habitant morbi tristique senectus et netus et malesuada fames ac turpis egestas. Sed at iaculis mauris. Praesent a sem augue. Nulla lectus sapien, auctor nec pharetra eu, tincidunt ac diam. Sed ligula arcu, aliquam quis velit aliquam, dictum varius erat.",

"Lorem ipsum dolor sit amet, consectetur adipiscing elit. Sed eget erat lectus.",

"Morbi mi mi, fringilla sed suscipit ullamcorper, tristique at mauris. Morbi non",

"commodo nibh. Pellentesque habitant morbi tristique senectus et netus et",

"malesuada fames ac turpis egestas. Sed at iaculis mauris. Praesent a sem augue.",

"Nulla lectus sapien, auctor nec pharetra eu, tincidunt ac diam. Sed ligula arcu,",

"aliquam quis velit aliquam, dictum varius erat.")

) | %{

    $n,$s,$expected = $_

    $result = &$f $s $n

    "$result"-eq"$expected"

    # $result   # uncomment this line to dispaly a result

}

Output:

True

True

Java 8, 135 bytes

n->s->{String r="",S=s.split(" "),t=r;for(int i=0;i<S.length;)if((t+S[i]).length()>n){r+=t+"n";t="";}else t+=S[i++]+" ";return r+t;}

Try it online.

Explanation:

n->s->{                      // Method with integer & String parameters and String return

  String r="",               //  Result-String, starting empty

         S=s.split(" "),   //  Input-String split by spaces

         t=r;                //  Temp-String, starting empty as well

  for(int i=0;i<S.length;)   //  Loop `i` in the range [0, amount_of_words):

    if((t+S[i]).length()>n){ //   If `t` and the word are larger than the integer input:

      r+=t+"n";             //    Add `t` and a newline to the result

      t="";}                 //    And reset `t` to an empty String

     else                    //   Else:

       t+=S[i++]+" ";        //    Append the word and a space to `t`

                             //    (and then increase `i` by 1 with `i++` for the next word

                             //     of the next iteration)

  return r+t;}               //  Return the result-String appended with `t` as result

Japt, 24 bytes

¸r@[XY]q(X·Ì+S+Y Ê>V?R:S

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Thanks to Bubbler for removing some duplication.

Explanation:

¸                           :Split into words

 r@                         :Reduce back into a string by:

         X                  : Start with the already reduced portion

          ·Ì                : Get the last line

            +S+Y            : Add a space and the next word

                 Ê>V?       : If the length of that line is greater than n:

                     R      :  Choose newline

                      :     : Otherwise:

                       S    :  Choose space

   [XY]q(                   : Join the next word to the result using the chosen character

JavaScript, 40 bytes

s=>n=>eval(`s.match(/.{1,${n}}( |$)/g)`)

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APL (Dyalog Unicode), 14 bytes^SBCS

Infix function; left argument is n, right argument is n.

⎕CY'dfns'⋄wrap

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⎕CY copy in the dfns library

⋄ then

wrap^[c] use the wrap^[n] function

^[c] code of that function
^[n] notes for that function

Golfed version of wrap, 59 bytes^SBCS

{⍺≥≢⍵:⍵⋄(t↑⍵),2↓⎕TC,⍺∇⍵↓⍨t+b⊃⍨t←⊃⌽⍺,g/⍨⍺≥g←⍸(⍺+1)↑b←' '=⍵}

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{…} dfn; ⍺ is left argument (width), ⍵ is right argument (string)

 ≢⍵ tally (number of characters) of string

 ⍺≥…: if width is greater than or equal to that, then:

  ⍵ return the string

 ⋄ otherwise:

  ' '=⍵ Boolean mask where blanks are equal to the string

  b← store in b (for blanks)

  (…)↑ take the following number of elements from that:

   ⍺+1 one more than the width

  ⍸ indices where true

  g← store in g (for gaps)

  ⍺≥ Boolean mask where the width is greater than or equal to that

  g/⍨ filter the gap indices by that

  ⍺, append that to the width

  ⊃⌽ pick the last element of that (lit. pick the first of the reversed)

  t← store in t (for take)

  b⊃⍨ use that to pick an element from the mask of blanks

  t+ add that to t

  ⍵↓⍨ drop that many characters from the string

  ⍺∇ recurse on that with the same left left argument

  ⎕TC, append that to the list of terminal control characters (8:HT, 10:NL, 13:CR)

  2↓ drop the first two character from that (leaving just a leading 13:CR)

  (…), append that to the following:

   t↑⍵ the first t characters of the string

Thanks to @Erik the Outgolfer, a golfed version :

Python 3, 94 bytes

def f(t,n):

 while t:i=n+(t[min(len(t)-1,n)]==" "or-t[n-1::-1].find(' '));print(t[:i]);t=t[i:]

Try it online!

# Python 3, 130 bytes

def f(t,n):

 l=

 while len(t):

  i=(n-t[:n][::-1].find(' '),n+1)[t[min(len(t)-1,n)]==" "]

  l.append(t[:i])

  t=t[i::]

 return l

Try it online!

Not so golfed version...

JavaScript + HTML + CSS, 117 64 bytes

-53 bytes courtesy of @Neil

n=50

s="Lorem ipsum dolor sit amet, consectetur adipiscing elit. Sed eget erat lectus. Morbi mi mi, fringilla sed suscipit ullamcorper, tristique at mauris. Morbi non commodo nibh. Pellentesque habitant morbi tristique senectus et netus et malesuada fames ac turpis egestas. Sed at iaculis mauris. Praesent a sem augue. Nulla lectus sapien, auctor nec pharetra eu, tincidunt ac diam. Sed ligula arcu, aliquam quis velit aliquam, dictum varius erat."

f=(n,s)=>document.body.innerHTML+=`<tt><p style=width:${n}ch>${s}`

f(n,s)

Jelly, 12 bytes

ḲŒṖK€€ḣ€ƑƇṪY

Try it online!

Unfortunately, this is too slow to work for the provided test case in under a minute over TIO.

C# (.NET Core), 162 bytes

stringt(string n,int a){var g="";for(int i=0;i++<Math.Floor((double)n.Length/a);)g+=$"^.{{{i*a-1}}}|";return Regex.Split(n,$@"(?n)(?<=({g.Trim('|')})S*)s");}}

This function uses a regex which matches the closest whitespace that is near the nth or multiple of nth character and splits the string based on it.

Try it online!

The TIO link is a full program, and the function has a static keyword so the function can be called from main.

Test Regex

C# (Visual C# Interactive Compiler), 78 bytes

s=>n=>System.Text.RegularExpressions.Regex.Replace(s,".{1,"+n+"}( |$)","$0n")

Try it online!

Credit goes to @LukeStevens for coming up with the Java version... Apparently .NET makes you import the RegularExpressions namespace in order to do a replace :(

Here is my original version that splits on the space character and uses LINQ to join them back together:

C# (Visual C# Interactive Compiler), 91 bytes

s=>n=>s.Split(' ').Aggregate((a,w)=>a+(a.Length-a.LastIndexOf('n')+w.Length>n?'n':' ')+w)

Try it online!

Dart, 112 bytes

f(s,n){var l=[''];s.split(' ').forEach((w){if((l.last+w).length<=n)l.last+=w+' ';else l.add(w+' ');});return l;}

Try it online!

APL(NARS), 48 chars, 96 bytes

{⊃⍵{⍺≥≢⍵:⊂⍵⋄k←1+⍺-' '⍳⍨⌽r←⍺↑⍵⋄(⊂k↑r),⍺∇k↓⍵}⍨⍺+1}

test:

  f←{⊃⍵{⍺≥≢⍵:⊂⍵⋄k←1+⍺-' '⍳⍨⌽r←⍺↑⍵⋄(⊂k↑r),⍺∇k↓⍵}⍨⍺+1}

  s←"Lorem ipsum dolor sit amet, consectetur adipiscing elit. Sed eget erat lectus. Morbi mi mi, fringilla sed suscipit ullamcorper, tristique at mauris. Morbi non commodo nibh. Pellentesque habitant morbi tristique senectus et netus et malesuada fames ac turpis egestas. Sed at iaculis mauris. Praesent a sem augue. Nulla lectus sapien, auctor nec pharetra eu, tincidunt ac diam. Sed ligula arcu, aliquam quis velit aliquam, dictum varius erat."

  50 f s

Lorem ipsum dolor sit amet, consectetur adipiscing 

elit. Sed eget erat lectus. Morbi mi mi, fringilla 

sed suscipit ullamcorper, tristique at mauris.     

Morbi non commodo nibh. Pellentesque habitant      

morbi tristique senectus et netus et malesuada     

fames ac turpis egestas. Sed at iaculis mauris.    

Praesent a sem augue. Nulla lectus sapien, auctor  

nec pharetra eu, tincidunt ac diam. Sed ligula     

arcu, aliquam quis velit aliquam, dictum varius    

erat.                                              

C, 63 bytes

b(a,n)char*a;{while(strlen(a)>n){for(a+=n;*a-32;--a);*a++=10;}}

The function of this exercise b(a,n) would break the line "a" as exercise said,
in the way not change its length (if we see the result as one string) because
change some spaces in n or new line in place. The input string "a" should have no n character too in it for b() function (it could have n in input string for bs())

b(a,n) function would be ok only because the restriction of this exercise,
that impose each word of "a" string has length < n if this is not true, that function can go

to one infinite loop...(very wrong in my way of see so i copy too the function more good because
in that case would return -1 and not would go to one infinite loop; it is bs(a,n) below)I not exclude both functions are bugged...

#define R(x,y) if(x)return y

#define U unsigned

U bs(char*a,U n)

{U c,q,r=1,i,j;

 R(!a||n<1||n++>0xFFFF,-1);

 for(j=c=i=0;;++i,++c)

    {R(i==-1,-1);q=a[i];

     if(q==10)goto l;

     if(c>=n){R(i-j>n,-1);a[i=j]=10;l:c=-1;++r;}

     R(!q,r);

     if(q==32)j=i;

    }

}

result of b() passed on one function that add line lenght each line

Lorem ipsum dolor sit amet, consectetur adipiscing [50]

elit. Sed eget erat lectus. Morbi mi mi, fringilla [50]

sed suscipit ullamcorper, tristique at mauris. [46]

Morbi non commodo nibh. Pellentesque habitant [45]

morbi tristique senectus et netus et malesuada [46]

fames ac turpis egestas. Sed at iaculis mauris. [47]

Praesent a sem augue. Nulla lectus sapien, auctor [49]

nec pharetra eu, tincidunt ac diam. Sed ligula [46]

arcu, aliquam quis velit aliquam, dictum varius [47]

erat. [5]