A command which concatenates a string an arbitrary number of times

Question

I'm trying to define a command \repeatstring{#1}{#2} which repeats the string #2 #1 times. Here is what I tried: (it is based on what I found in How to concatenate strings into a single command?)

\documentclass[11pt, a4paper]{article}
\usepackage{etoolbox}
\newcounter{countdown}
\newcommand\concathere{}
\newcommand\repeatstring[2]{
    \setcounter{countdown}{#1}
    \renewcommand\concathere{}
    \whileboolexpr
        {test {\ifnumcomp{\thecountdown}{>}{0}{true}{false}}}
        {
        \addtocounter{countdown}{-1}
        \appto\concathere{#2}          % here the concatenation happens
        }
    \concathere
    }
\begin{document}
\repeatstring{5}{abc}
\end{document}

\concathere is meant to contain the string the desired number of times. When I run \repeatstring{5}{abc} the output is simply "true". Nothing more, nothing less. Interestingly, when I replace the final line of my command by aaa\concathere bbb the result is "true aaabbb". This should indicate where things go wrong, but I can't figure it out. Probably it simply prints "true" because 5>0, and then stops for some reason.

Answer 1

You are misunderstanding how the logical <expression> should be given. The result should be 'logically true', not the text true:

\documentclass{article}

\usepackage{etoolbox}
\newcounter{countdown}
\newcommand\concathere{}
\newcommand\repeatstring[2]{%
    \setcounter{countdown}{#1}%
    \renewcommand\concathere{}%
    \whileboolexpr
        {test {\ifnumcomp{\thecountdown}{>}{0}}}% 
        {%
        \addtocounter{countdown}{-1}%
        \appto\concathere{#2}% here the concatenation happens
        }%
    \concathere
    }
\begin{document}
\repeatstring{5}{abc}

\end{document}

Notice that in the expression part I've just got something that will give a logical result. There are a few examples in the etoolbox manual: see page 21 for example.

(I've prevented spurious spaces appearing in the result by adding appropriate % at the end of lines: not relevant to the issue but important for real use.)

Answer 2

Repetition via \romannumeral trick

The number is multiplied with 1000 to convert it to a roman number. Then TeX produces a long string consisting of letter m, whose length is the original number.

Then \repeatstringX looks at the next token, if it is an m, then the string unit is output. Otherwise the next token is the end marker F and the loop stops.

\documentclass[11pt, a4paper]{article}

\newcommand\repeatstring[2]{%
  \def\tempParam{#2}%
  \expandafter\repeatstringX\romannumeral\the\numexpr(#1)\relax000 F%
}
\newcommand*{\repeatstringX}[1]{%
  \csname repeatstring#1\endcsname
}
\newcommand*{\repeatstringm}{%
  \tempParam
  \repeatstringX
}
\newcommand*{\repeatstringF}{}

\begin{document}
  [\repeatstring{5}{abc}]
\end{document}

Fixing the MWE

• If \ifnumcomp is used inside the expression for test, then the arguments for true and false are omitted, see the description of test in the documentation of etoolbox.

• There are many unwanted white spaces, caused by the end of lines in the definition of \repeatstring.

Fixed MWE:

\documentclass[11pt, a4paper]{article}
\usepackage{etoolbox}
\newcounter{countdown}
\newcommand\concathere{}
\newcommand\repeatstring[2]{%
  \setcounter{countdown}{#1}%
  \renewcommand\concathere{}%
  \whileboolexpr
     {test {\ifnumcomp{\thecountdown}{>}{0}}}
     {%
       \addtocounter{countdown}{-1}%
       \appto\concathere{#2}% here the concatenation happens
     }%
  \concathere
}
\begin{document}
  [\repeatstring{5}{abc}]
\end{document}

Answer 3

There are various ways to accomplish your needs.

\documentclass[11pt]{article}
\usepackage{xparse} % for D

% for A, B, C
\newcounter{mycount}

\makeatletter
\newcommand\repeatstringA[2]{%
  \setcounter{mycount}{#1}%
  \ifnum\themycount>0
    #2%
    \addtocounter{mycount}{-1}%
    \expandafter\@firstofone
  \else
    \expandafter\@gobble
  \fi
  {\repeatstringA{\themycount}{#2}}%
}

\newcommand\repeatstringB[2]{%
  \setcounter{mycount}{#1}%
  \@whilenum{\value{mycount}>0}\do{#2\addtocounter{mycount}{-1}}%
}

\newcommand\repeatstringC[2]{%
  \ifnum#1>0
    \expandafter\@firstofone
  \else
    \expandafter\@gobble
  \fi
  {#2\expandafter\repeatstringC\expandafter{\the\numexpr#1-1\relax}{#2}}%
}
\makeatother

\ExplSyntaxOn
\DeclareExpandableDocumentCommand{\repeatstringD}{mm}
 {
  \prg_replicate:nn { #1 } { #2 }
 }
\ExplSyntaxOff

\begin{document}
\repeatstringA{5}{abc}

\repeatstringB{5}{abc}

\repeatstringC{5}{abc}

\repeatstringD{3*2-1}{abc}
\end{document}

The last one is particularly appealing. The first two are essentially equivalent.

If you want to produce a control sequence containing the repetitions, then the changes are easy:

\documentclass[11pt]{article}
\usepackage{xparse} % for D

% for A, B, C
\newcounter{mycount}

% a container
\newcommand{\concathere}{}

\makeatletter
\newcommand\repeatstringA[2]{%
  \renewcommand{\concathere}{}%
  \setcounter{mycount}{#1}%
  \ifnum\themycount>0
    \expandafter\def\expandafter\concathere\expandafter{\concathere #2}%
    \addtocounter{mycount}{-1}%
    \expandafter\@firstofone
  \else
    \expandafter\@gobble
  \fi
  {\repeatstringA{\themycount}{#2}}%
}

\newcommand\repeatstringB[2]{%
  \setcounter{mycount}{#1}%
  \@whilenum{\value{mycount}>0}\do{%
     \expandafter\def\expandafter\concathere\expandafter{\concathere #2}%
     \addtocounter{mycount}{-1}%
  }%
}

\newcommand\repeatstringC[2]{%
  \ifnum#1>0
    \expandafter\@firstofone
  \else
    \expandafter\@gobble
  \fi
  {%
   \expandafter\def\expandafter\concathere\expandafter{\concathere #2}%
   \expandafter\repeatstringC\expandafter{\the\numexpr#1-1\relax}{#2}%
  }%
}
\makeatother

\ExplSyntaxOn
\NewDocumentCommand{\repeatstringD}{mm}
 {
  %\tl_clear:N \concathere
  %\prg_replicate:nn { #1 } { \tl_put_right:Nn \concathere { #2 } }
  % a faster method suggested by Bruno Le Floch
  \tl_set:Nx \concathere { \prg_replicate:nn { #1 } { \exp_not:n { #2 } } }
 }
\ExplSyntaxOff

\begin{document}
\repeatstringA{5}{abc}\concathere

\repeatstringB{5}{abc}\concathere

\repeatstringC{5}{abc}\concathere

\repeatstringD{3*2-1}{abc}\concathere
\end{document}

At the end of the execution of \repeatstringX{5}{abc} the macro \concathere will contain abcabcabcabcabc.

Instead of

\expandafter\def\expandafter\concathere\expandafter{\concathere #2}

(where \def is used for efficiency), you can of course use

\appto\concathere{#2}

provided you have loaded etoolbox.

None of these methods will overflow the input stack size; however, big numbers may overflow other parts of the memory.

Answer 4

Here, I set up a recursive loop. Works with macros, too.

\documentclass[11pt]{article}
\newcounter{mycount}
\def\repeatstring#1#2{%
  \setcounter{mycount}{#1}%
  \ifnum\value{mycount}>0\relax#2%
    \addtocounter{mycount}{-1}%
    \repeatstring{\value{mycount}}{#2}%
  \fi%
}
\begin{document}
\repeatstring{5}{abc}

\repeatstring{3}{\today}
\end{document}

As egreg points out in my comments, the above method will fail if the stack size (5000) is exceeded by the repeat count. He also provides the remedy, which in the context of my MWE, can be achieved in the following way (by expanding the \fi before proceeding to the next recursion):

\documentclass[11pt]{article}
\newcounter{mycount}
\def\repeatstring#1#2{%
  \setcounter{mycount}{#1}%
  \ifnum\value{mycount}>0\relax#2%
    \addtocounter{mycount}{-1}%
    \def\tmp{\repeatstring{\value{mycount}}{#2}}%
    \expandafter\tmp%
  \fi%
}
\begin{document}
\repeatstring{5}{abc}

\repeatstring{3}{\today}

\repeatstring{5555}{i$\!\!$ }
\end{document}

Answer EDITED to replace \themycount with \value{mycount}, per egreg's recommendation.

Answer 5

ConTeXt solution

\def\repeatstring#1#2{\edef\concathere{\dorecurse{#1}{#2}}}
\starttext
\repeatstring{5}{abc}\concathere
\stoptext