Python 2, 95 bytes

s,m,n=input() print s[:n] for i in range(m-2):print s[~i]+' '*(n-2)+s[n+i] print s[1-m::-1][:n] 

Prints the first line, then the two vertical lines, then the last line.

There's got to be something shorter than writing print three times, but everything I've tried so far with saving to a variable and '\n'.join has been longer.

CJam, 31 30 bytes

At Optimizer's insistence, here is my own attempt. I'm not a fan of winning my own challenges, so I'm counting the APL family (or someone better at CJam) to beat this. ;)

l~:H;:V/(N@s{)V2-S*@(N@}H2-*W% 

Takes input in the same order as given in the question:

"Hello, World! " 5 4 

Test it here.

One byte saved thanks to Optimizer.

Explanation

Originally, I had a really nice idea for starting with the rectangle of spaces and then literally wrapping the string around it while rotating the entire grid four times. However, I couldn't seem to get that to work in case where width or height or both are 2. So I tried the naive approach (print top, loop over sides, print bottom), and surprisingly it turned out to be really short.

l~ "Read and evaluate the input."; :H; "Store the height in H and discard it."; :V/ "Store the width in V and split the input into chunks of size V."; (N "Slice off the first such chunk and push a newline."; @s "Pull up the other chunks and join them back together."; { }H2-* "Repeat this block H-2 times, printing the sides."; ) "Slice off the last character of the string."; V2-S* "Push V-2 spaces."; @( "Pull up the remaining string and slice off the first character."; N@ "Push a newline and pull up the remaining string."; W% "Reverse the remainder of the string, which is the bottom row."; 

Pyth, 47 46 45 40 37 36 bytes

This is the obvious approach implemented in Pyth. It prints the first line by indexing 0:width and then the middle, then the end.

Thanks to @Jakube for the tip with using z and Q for two inputs and using p.

AkYQ<zkV-Y2p*d-k2@zt_N@z+kN;<_<z-2Yk 

Takes input from stdin as a string and as a tuple of dimensions, newline separated:

Hello, World! 5, 4 

and writes to stdout.

Try it here.

A Double assignment kY The vars k and Y Q The dimension tuple <zk Prints out first line by doing z[:width] V-Y2 For N in height-2 p Print out everything *d Repeat " " -k2 Width-2 times @z Index z -_N1 At index -N-1 @z Index z +kN At index k+N ; Close out loop <_<z-2Yk Print last line 

J, 61 bytes

Method:

Starting from a (height-2)*(width-2) block of spaces we take the necessary amount of characters from the end of the string and add it to the current block. We repeat this 4 times. The total 5 states illustrated with the 'Hello, World! ' 5 4 example (spaces replaced with Xs for readability):

XXX !_ orld ,_W Hello XXX XX XXX! XXo _XXX, XX XXX_ XXr !XXX_ XX XXl dlroW _!d 

The code:

4 :'|:>{:((}.~{:@$);({.~{:@$)|.@|:@,])&>/^:4(|.x);'' ''$~y-2' 

Explicit function definition. The two-operand function takes a string as left argument and a list of two integers as right argument.

Example usage:

 wrap_on=.4 :'|:>{:((}.~{:@$);({.~{:@$)|.@|:@,])&>/^:4(|.x);'' ''$~y-2' 'Hello, World! ' wrap_on 5 4 Hello , ! dlroW '>v<^' wrap_on 2 2 >v ^< 

Try it online here.

Pyth, 38 37

AGHQ<zGFNC,_>z_ttH>zGj*dttGN)<>_zttHG 

I originally had a different 38 solution, but it was basically golfed solution of Maltysen's answer. So I decided to go a little bit different.

Try it online.

 implicit: z=string from input, Q=pair of numbers from input AGHQ G=Q[0] (width), H=Q[1] (height) <zG print z[:G] _>z_ttH last H-2 chars reversed >zG all chars from the Gth position to end C, zip these 2 strings to pairs FN for each pair N: j*dttGN seperate the two chars by (G-2) spaces and print ) end for <>_zttHG print last line z[::-1][H-2:][:G] 

JavaScript (ES6), 110 115

Function with 3 parameters, returning a string

F=(s,w,h,q=h+(w-=2),t=b='')=> [for(c of s)q?t+=q<h?c+'\n'+s[w+h+w+q--]+' '.repeat(q&&w):(b+=s[w+q--],c):q] &&t+b 

Chrome version 119: no short format for functions, no default parameters. No reason to use for(of) even if it is supported

function F(s,w,h){ for(q=h+(w-=2),t=b=i=''; q; q<h?t+='\n'+s[w+h+w+q--]+' '.repeat(q&&w):b+=s[w+q--]) t+=s[i++]; return t+b } 

ES5 version 126: no for(of), no string.repeat

function F(s,w,h){ for(q=h+(w-=2),t=b=i=''; q; q<h?t+='\n'+s[w+h+w+q--]+Array(q&&-~w).join(' '):b+=s[w+q--]) t+=s[i++]; return t+b } 

Ungolfed

F=(s,w,h)=> { var q = h+(w-=2), // middle length t = '', // top and body b = ''; // bottom row for(c of s) if (q > 0) { if (q < h) { t += c+'\n'; // right side, straight t += s[w+h+w+q]; // left side, backwards if (q > 1) // body fill, except for the last line t += ' '.repeat(w) } else { t+=c, // top, straight b+=s[w+q] // bottom, backwards } --q } return t+b 

Test In Firefox/FireBug console

;[["Hello, World! ", 5, 4],["+--+|||+--+|||",4,5],[">v<^",2,2] ,["rock beats scissors beats paper beats ",11,10] ,["!\"#$%&'()*+,-./0123456789:;<=>?@ABCDEFGHIJKLMNOPQRSTUVWXYZ[\\]^_`abcdefghijklmnopqrstuvwxyz{|}~",46,3]] .forEach(test => console.log(F(...test))) 

Output

Hello , ! dlroW +--+ | | | | | | +--+ >v ^< rock beats s s c t i a s e s b o r r s epap staeb !"#$%&'()*+,-./0123456789:;<=>?@ABCDEFGHIJKLMN ~ O }|{zyxwvutsrqponmlkjihgfedcba`_^]\[ZYXWVUTSRQP 

Python 2, 97 bytes

def f(s,w,h):print s[:w];i=0;exec'print s[~i]+" "*(w-2)+s[w+i];i+=1;'*(h-2);print s[1-h:w+h-3:-1] 

Taking the direct approach.

Haskell, 164 156 bytes

import Data.List r=replicate p w h(s:t)=unlines$fst$n$n$n$n(r h$r w ' ',(w,h,s:t++[s])) n(_:b,(w,h,s))=(transpose$map reverse$(take w s):b,(h,w,drop(w-1)s)) 

The function p doesn't print the output, but returns it as a string, e.g p 4 5 "+--+|||+--+|||" -> "+--+\n| |\n| |\n| |\n+--+\n". For better display use putStr:

putStr $ p 4 5 "+--+|||+--+|||" +--+ | | | | | | +--+ 

How it works: I create a w x h block of spaces and replace the first line with the beginning of the input string. Then I rotate the block counterclockwise and repeat replacing the first line another three times.

To prevent cutting of the first character again after turn #4, I append it to the input string before starting.

"Hello World" example, 5 x 4 | Start Turn #1 Turn #2 Turn #3 Turn #4 ---------+-------------------------------------------------------------------- String | "Hello, World! H" "o, World! H" "World! H" "d! H" "" left | | Block | <empty> Hello o, W World d! H before | l l e rotating | l , r l | e olleH o l | H W ,o 

Edit: found a better way to solve the cut-off-first-character-after-turn-#4 problem.

Postscript, 62 bytes

This of course uses binary tokens, but it is equivalent to:

/Courier findfont setfont 0 h moveto s [ w {1 0} repeat pop pop h {0 -1} repeat pop pop w {-1 0} repeat pop pop h {0 1} repeat ] xyshow 

Here is a hexdump of the file (xxd round.ps):

0000000: 91c7 9243 9295 3020 6892 6b73 5b77 7b31 ...C..0 h.ks[w{1 0000010: 2030 7d92 8392 7592 7568 7b30 202d 317d 0}...u.uh{0 -1} 0000020: 9283 9275 9275 777b 2d31 2030 7d92 8392 ...u.uw{-1 0}... 0000030: 7592 7568 7b30 2031 7d92 835d 92c3 u.uh{0 1}..].. 

Run as:

gs -dw=11 -dh=10 -ss="rock beats scissors beats paper beats " round.ps 

The output is really small (as a result of not scaling the font at all), so you need to zoom in a fair bit to see it.

This takes advantage of the xyshow operator to write out the string using custom character spacings. In this case, I use the negative vertical spacing to write down, then negative horizontal space to write backwards, then positive vertical space to write upward. Because of this, I don't need to use any sort of string manipulation.

><>, 82 80 + 3 = 83 bytes

:2-&\ v!?:<oi-1 /?(0:i \~ao{2-{~} \{:?!v1-}o&:& >:?v!~{{o}ao4. ^ >" "o1- o;!?l< 

Esolang page for ><> (Fish)

This turned out to be shorter than I expected. It uses the straightforward approach of printing the first line, then the columns padded with the central spaces, then the last line.

Input the string via STDIN and the height and width via command line with the -v flag, like so:

py -3 fish.py round.fish -v <height> <width> 

Explanation

:2-& Put W-2 in the register :?!v1-io Directly print the first W characters of the input i:0(?/ Read the rest of the input ~ao{2-{~} Pop a few leftovers 0s from above, decrement H by 2 and print a newline Stack now consists of H = H-2 at the bottom and the rest of the input reversed [loop] {:?!v If H is 0... ~ Pop the 0 l?!;o Print the rest of the (reversed) input Otherwise... 1-} Decrement H o Output the top of stack &:& Copy I = W-2 from the register :? If I is nonzero... " "o1- Print a space and decrement I, then repeat from the previous line {{o}ao Print the bottom input character and output a newline 4. Jump to the start of the loop (note that I = 0 is leftover from above) 

Bash+coreutils, 124

A shell script to get you started:

echo "${3:0:$1}" fold -1<<<"${3:$1*2+$2-2}"|tac|paste - <(fold -1<<<"${3:$1:$2-2}")|expand -t$[$1-1] rev<<<"${3:$1+$2-2:$1}" 

Pass input as command-line args:

$ ./roundnround.sh 5 4 "Hello, World! " Hello , ! dlroW $ 

JavaScript, 161 160 158 bytes

The method I've come up with turned out way too long, but oh well, it's been practice. (Also, I got it to spell out r+o[u]+'\n':d.)

function f(o,w,n){s=o.slice(0,w)+'\n';o=o.slice(w);n-=2;r='';for(u=w-2;u--;)r+=' ';for(u=d=0;d=o[2*n+w+~u],u<w+n;u++)s+=(u<n)?(d||' ')+r+o[u]+'\n':d;return s} 

For input that doesn't make sense the output is undefined (literally, and a bunch of times), but it works for all the test cases.

Groovy, 140

f={a,x,y->println a.substring(0,x);(1..y-2).each{println a[a.length()-it]+' '*(x-2)+a[it+x-1]}println a.substring(x+y-2,2*x+y-2).reverse()} 

call:

f('rock beats scissors beats paper beats ',11,10) 

output:

rock beats s s c t i a s e s b o r r s epap staeb 

K, 55 54 bytes

Using the same approach as randomra's J implementation; Start with a block of spaces and add from the tail of the string to the edge while rotating four times:

f:{`0:*4{((,r#*|x),|:'+*x;(r:-#*x)_*|x)}/((y-2)#" ";x)} 

And some examples:

 f["Hello,_World!_";4 5] Hello _ , ! _ dlroW f[">v<^";2 2] >v ^< 

Breaking it down a little for readability,

Generate an NxM block:

 t:2 3#!6 (0 1 2 3 4 5) 

Rotate 90 degrees by using transpose(+) and reverse-each (|:'):

 |:'+t (3 0 4 1 5 2) 

So if we have a block of spaces t and a string s, we can prepend a slice of the tail of s to t:

 s: 12 18 17 8 9 12 18 17 8 9 (,(-#t)#s),|:'+t (8 9 3 0 4 1 5 2) 

We use the form 4 {[x] ... }/( ... ) to repeatedly apply a function to a tuple consisting of the string and the matrix we're building. Each time we do this rotate-and-concatenate step we also chop down the string.

edit:

Another idea is to try splitting the input string into the fragments we want at each rotation, which simplifies the main body of the program. Unfortunately this works out to be slightly longer at 56 bytes:

f:{`0:((y-2)#" "){|:'(,y),+x}/(+\(0,y[0 1 0]-2 1 1))_|x} 

If there's a better way to calculate those split points I'm open to suggestions.

edit2:

Rearranging slightly lets me remove a pair of parentheses. 54 bytes!

f:{`0:((y-2)#" "){|:'(,y),+x}/(0,+\y[0 1 0]-2 1 1)_|x} 

K, 80 68 bytes

f:{[s;y;n]`0:(,n#s),({s[(#s)-x+2],((n-2)#" "),s@n+x}'!y-2),,n#|-4!s} 

Shortened from 80 thanks to @JohnE.

Original:

f:{s:x;n:z;`0:(,s@!n),({s[(#s)+-2-x],({" "}'!n-2),s@n+x}'!y-2),,(|s@!-4+#s)@!n} 

I barely even know how this thing works.

Example usage:

f["Hello, world! ";5;4] 

There are some possible optimizations, but I keep making Kona segfault...

R, 178

This is an unnamed function taking s, w, h as parameters. I wish there was a nicer way to split the string.

function(s,w,h){W=w+h-1;H=w+W-1;S=strsplit(s,'')[[1]];O=array(" ",c(h,w+1));O[,w+1]="\n";O[1:h,c(w,1)]=c(S[w:W],S[(W+W-1):H]);O=t(O);O[1:w,c(1,h)]=c(S[1:w],S[H:W]);cat(O,sep='')} 

Ungolfed

W=w+h-1; # additional index points H=w+W-1; # additional index points S=strsplit(s,'')[[1]]; # vectorize the string O=array(" ",c(h,w+1)); # create an array of spaces O[,w+1]="\n"; # set newlines O[1:h,c(w,1)]=c(S[w:W],S[(W+W-1):H]); # build middles lines O=t(O); # transpose array O[1:w,c(1,h)]=c(S[1:w],S[H:W]); # build top and bottom lines cat(O,sep='') # cat out results 

Test run

> (function(s,w,h){W=w+h-1;H=w+W-1;S=strsplit(s,'')[[1]];O=array(" ",c(h,w+1));O[,w+1]="\n";O[1:h,c(w,1)]=c(S[w:W],S[(W+W-1):H]);O=t(O);O[1:w,c(1,h)]=c(S[1:w],S[H:W]);cat(O,sep='')})("Hello, World! ",5,4) Hello , ! dlroW > (function(s,w,h){W=w+h-1;H=w+W-1;S=strsplit(s,'')[[1]];O=array(" ",c(h,w+1));O[,w+1]="\n";O[1:h,c(w,1)]=c(S[w:W],S[(W+W-1):H]);O=t(O);O[1:w,c(1,h)]=c(S[1:w],S[H:W]);cat(O,sep='')})("+--+|||+--+|||",4,5) +--+ | | | | | | +--+ > (function(s,w,h){W=w+h-1;H=w+W-1;S=strsplit(s,'')[[1]];O=array(" ",c(h,w+1));O[,w+1]="\n";O[1:h,c(w,1)]=c(S[w:W],S[(W+W-1):H]);O=t(O);O[1:w,c(1,h)]=c(S[1:w],S[H:W]);cat(O,sep='')})(">v<^",2,2) >v ^< > (function(s,w,h){W=w+h-1;H=w+W-1;S=strsplit(s,'')[[1]];O=array(" ",c(h,w+1));O[,w+1]="\n";O[1:h,c(w,1)]=c(S[w:W],S[(W+W-1):H]);O=t(O);O[1:w,c(1,h)]=c(S[1:w],S[H:W]);cat(O,sep='')})("rock beats scissors beats paper beats ",11,10) rock beats s s c t i a s e s b o r r s epap staeb > # Escaped the \ as well > (function(s,w,h){W=w+h-1;H=w+W-1;S=strsplit(s,'')[[1]];O=array(" ",c(h,w+1));O[,w+1]="\n";O[1:h,c(w,1)]=c(S[w:W],S[(W+W-1):H]);O=t(O);O[1:w,c(1,h)]=c(S[1:w],S[H:W]);cat(O,sep='')})("!\"#$%&'()*+,-./0123456789:;<=>?@ABCDEFGHIJKLMNOPQRSTUVWXYZ[\\]^_`abcdefghijklmnopqrstuvwxyz{|}~",46,3) !"#$%&'()*+,-./0123456789:;<=>?@ABCDEFGHIJKLMN ~ O }|{zyxwvutsrqponmlkjihgfedcba`_^]\[ZYXWVUTSRQP > 

T-SQL, 307

While still horrendously long, this turned out to be quite a bit easier (and shorter) than I thought in a query. Implemented as an inline table valued function for T-SQL.

CREATE FUNCTION f(@S VARCHAR(MAX),@ INT,@H INT)RETURNS TABLE RETURN WITH R AS(SELECT 2i,LEFT(@S,@)S,STUFF(@S,1,@,'')+'|'R UNION ALL SELECT i+1,CASE WHEN i<@H THEN LEFT(RIGHT(R,2),1)+REPLICATE(' ',@-2)+LEFT(R,1)ELSE REVERSE(LEFT(R,@))END,STUFF(STUFF(R,LEN(R)-1,1,''),1,1,'')FROM R WHERE i<=@H)SELECT S FROM R 

This recurses through the string @h times. First recursion clips @W characters from the string. The middle recursions take the last and first from remaining string with string padding between. The last recursion reverses what is left. There is a few lost characters dealing with the way SQL Server treats trailing spaces on VARCHARS.

Test Run

WITH TestSet AS ( SELECT * FROM (VALUES ('Hello, World! ',5,4), ('+--+|||+--+|||',4,5), ('>v<^',2,2), ('rock beats scissors beats paper beats ',11,10), ('!"#$%&''()*+,-./0123456789:;<=>?@ABCDEFGHIJKLMNOPQRSTUVWXYZ[\]^_`abcdefghijklmnopqrstuvwxyz{|}~andfoure',50,3) ) Test(S,W,H) ) SELECT x.S FROM TestSet CROSS APPLY ( SELECT S FROM dbo.F(S,W,H) )x S ---------------------------- Hello , ! dlroW +--+ | | | | | | +--+ >v ^< rock beats s s c t i a s e s b o r r s epap staeb !"#$%&'()*+,-./0123456789:;<=>?@ABCDEFGHIJKLMNOPQR e S ruofdna~}|{zyxwvutsrqponmlkjihgfedcba`_^]\[ZYXWVUT (24 row(s) affected) 

Pyth, 33 bytes

J-vw2<zQFkC,_z<>zQJj*-Q2dk;<>_zJQ 

Demonstration.

MATLAB, 101

function f(H,W,S) w=1:W;h=(0:H-3).';n=W+H-2;S(3*n)=' ';S([w;[2*n-h,3*n*ones(H-2,W-2),h+W+1];n-w+W+1]) 

C++, 398 bytes

Compiler used - GCC 4.9.2 with -std=c++14 flag

#include<bits/stdc++.h> using namespace std;string s;vector<vector<char>> M;int w,h,p,i,j;void F(int x,int y){if(p<s.size()&&(((!y||y==h-1)&&x>=0&&x<w)||((!x||x==w-1)&&y>=0&&y<h))&&!M[y][x])M[y][x]=s[p++],F(x+1,y),F(x,y+1),F(x-1,y),F(x,y-1);}int main(){getline(cin,s);cin>>w>>h;M.resize(h,vector<char>(w,0));F(0,0);while(i<h){j=0;while(j<w){if(!M[i][j])M[i][j]=32;cout<<M[i][j++];}i++;cout<<endl;}} 

Test it here.

Explanation

#include<bits/stdc++.h> using namespace std; string s; // input string vector<vector<char>> M; // output matrix int w, h, p, i, j; // w = width // h = height // p = iterator over s // i, j = iterators used later for printing answer void F( int x, int y ) { // If the coordinates (x, y) are either on the first row/column or the last row/column and are not already populated with the input characters, populate them if ( p < s.size() && ( ( ( y == 0 || y == h - 1 ) && x >= 0 && x < w ) || ( ( x == 0 || x == w - 1 ) && y >= 0 && y < h ) ) && !M[y][x] ) { M[y][x] = s[p++]; F( x + 1, y ); F( x, y + 1 ); F( x - 1, y ); F( x, y - 1 ); } } int main() { getline( cin, s ); cin >> w >> h; // Input taken !! M.resize( h, vector<char>( w, 0 ) ); // Fill the matrix with null characters initially F( 0, 0 ); // This function does all the work // Now printing the matrix while ( i < h ) { j = 0; while ( j < w ) { if ( !M[i][j] ) { M[i][j] = ' '; // Replace '\0' with ' ' } cout << M[i][j++]; } i++; cout << endl; } } 

Perl, 193 195 bytes

($s,$w,$h,$i,$y)=(@ARGV,0,2); $o.=substr$s,$i,$w; $i+=$w; $o.=sprintf"\n%s%*s",substr($s,2*($w+$h)-$y++-3,1)||' ',$w-1,substr($s,$i++,1)while$y<$h; print$o."\n".reverse(substr($s,$i,$w))."\n"; 

I'm sure this can be greatly improved. I'm a noob. >,<

Java 11, 180 bytes

(s,w,h)->{var r=s.substring(0,w)+"\n";int i=w;for(var S=s.split("");i<w+h-2;)r+=S[3*w+2*h-i-5]+" ".repeat(w-2)+S[i++]+"\n";return r+new StringBuffer(s.substring(i,i+w)).reverse();} 

Try it online (NOTE: String.repeat(int) is emulated as repeat(String,int) for the same byte-count, because Java 11 isn't on TIO yet.)

Explanation:

(s,w,h)->{ // Method with String & 2 int parameters and String return-type var r=s.substring(0,w)+"\n"; // Result-String, starting at the the first row of output, // which is a substring in the range [0, `w`) int i=w; // Index-integer, starting at `w` for(var S=s.split(""); // Split the input-String into a String-array of characters i<w+h-2;) // Loop `i` in the range [`w`, `w+h-2`) r+= // Append the result-String with: S[3*w+2*h-i-5] // The character at index `2*w+2*h-4 - i+w-1` +" ".repeat(w-2) // Then append `w-2` amount of spaces +S[i++] // Then append the character at index `i` +"\n"; // And a trailing new-line return r // After the loop, return `r` as result +new StringBuffer(s.substring(i,i+w)).reverse(); // Appended with the last row of output, // which is a substring in the range [`i`, `i+w`) reversed 

Charcoal, 4 bytes

ＢＮＮＳ 

Try it online (verbose) or try it online (pure).

Explanation:

Basically a builtin for this challenge. Take the first two integer-inputs as width and height, and prints a box with the third input-string as border:

Box(InputNumber(), InputNumber(), InputString()) ＢＮＮＳ