TeX - LaTeX Stack Exchange is a question and answer site for users of TeX, LaTeX, ConTeXt, and related typesetting systems. It only takes a minute to sign up.
Sign up to join this community
I have a command with a parameter. I want to check if its value is a positive number and perform actions based on that.
\newcommand{\mycommand}[1]{
\ifnum#1>0%
%some actions
\fi
}
But I receive errors when the value of the parameter is not a number, e.g.:
\mycommand{*}
How can I check if the value of #1 is a number?
Since it appears that the \IfInteger from the xtring package perceives blank strings as integers (the empty string {} is ok, but not { }), I have defined a modified macro \IsInteger which handles that case:
\documentclass{article}
\newcommand*{\IsInteger}[3]{%
\IfStrEq{#1}{ }{%
#3% is a blank string
}{%
\IfInteger{#1}{#2}{#3}%
}%
}%
\usepackage{xstring}
\begin{document}
$2$ is \IsInteger{2}{an integer}{not an integer}\par
$2.0$ is \IsInteger{2.0}{an integer}{not an integer}
$-7$ is \IsInteger{-7}{an integer}{not an integer}\par
$-7.0$ is \IsInteger{-7.0}{an integer}{not an integer}
$2.1$ is \IsInteger{2.1}{an integer}{not an integer}\par
$-7.1$ is \IsInteger{-7.1}{an integer}{not an integer}
a is \IsInteger{a}{an integer}{not an integer}
Empty String is \IsInteger{}{an integer}{not an integer}\par
Blank String is \IsInteger{ }{an integer}{not an integer}
\end{document}
You can use IfInteger from the xstring package to test if it is an integer number:
There is also \IfDecimal which works similarly.
\documentclass{article}
\usepackage{xstring}
\begin{document}
2 is \IfInteger{2}{integer}{not an integer}
a is \IfInteger{a}{integer}{not an integer}
\end{document}
\IsInteger to compensate for that.
Mar 31, 2012 at 3:31
\ifinteger interprets negative numbers as non-integer.
It depends mostly on the expected input and also on the context where you want to use the command.
If your expected input is either a number or something that doesn't start with digits, then
\newcommand{\mycommand}[1]{%
\ifnum0<0#1\relax
#1 is a positive number%
\else
#1 is not a positive number%
\fi}
will do. For example, \mycommand{42} will do the comparison 0<042 which is true; instead, with \mycommand{*} TeX will see \ifnum0<0*\relax and it will test 0<0, which is false, so the * will be ignored as part of the "true text". Also the test from \mycommand{0} will evaluate to false.
On the other hand, with \mycommand{1x} the test will evaluate to true and give wrong results.
Another expandable way can be
\def\mycommand#1{%
\if\relax\detokenize\expandafter{\romannumeral-0#1}\relax
#1 is a number%
\else
#1 is not a number%
\fi
}
but \mycommand{0} would test true. Here \mycommand{1x} would answer that 1x is not a number.
However, the argument should not contain "dangerous" items: \mycommand{\textbf{x}} would fail miserably.
A non-expandable test can be
\makeatletter
\def\mycommand#1{%
\afterassignment\get@args\count@=0#1\hfuzz#1\hfuzz}
\def\get@args#1\hfuzz#2\hfuzz{%
\if\relax\detokenize{#1}\relax
#2 is a number%
\else
#2 is not a number%
\fi
}
\makeatother
This works also with input such as \mycommand{\textbf{1}} (the test will evaluate to false).
\mycommand{\undefined\@undefined} will fail. I am sure you know why I have used two undefined tokens.
Mar 31, 2012 at 0:27
#1 is a number by \expandafter\@firstoftwo and #1 is not a number by \expandafter\@secondoftwo. Similarly for the second solution.
Here is an expandable solution that accepts even undefined control sequences. It is admittedly acrobatic.
Some of the commands in catoptions package are redefined here for the sake of speed and experimentation.
\documentclass[a4paper]{article}
\usepackage{catoptions}
\makeatletter
\new@def\cptifdef#1{%
\cptifblank{#1}{%
\@secondoftwo
}{%
\csname @\ifx#1\@undefined second\else
\ifx#1\relax second\else first\fi\fi oftwo\endcsname
}%
}
\new@def\cptifundef#1{\cptifdef{#1}\@secondoftwo\@firstoftwo}
\new@def\cptifleftbraced#1{%
\cptifblank{#1}{%
\@secondoftwo
}{%
\csname @\if\expandafter\cpt@car\detokenize{#1}\car@nil
\expandafter\cpt@car\string{\car@nil\ifnum0=`}\fi
first\else second\fi oftwo\endcsname
}%
}
\new@def\cptifxpandable#1{%
\cptifleftbraced{#1}{%
\@secondoftwo
}{%
\expandafter\ifx\noexpand#1#1%
\expandafter\@secondoftwo
\else
\expandafter\@firstoftwo
\fi
}%
}
\newcommand\cpttwooftwo[2]{#1#2}
\newcommand\cptifblank[1]{%
\expandafter\ifx\expandafter\noboundary\@gobble#1.\noboundary
\expandafter\@firstoftwo
\else
\expandafter\@secondoftwo
\fi
}
\newcommand\cptifcmdeq[2]{%
\ifx#1#2\cpt@quark
\expandafter\@firstoftwo
\else
\expandafter\@secondoftwo
\fi
}
\long\def\cptifsolo#1{%
\if0\pdfstrcmp{\detokenize\expandafter
{\cpttwooftwo#1{}{}}}{\detokenize{#1{}}}%
\expandafter\@firstoftwo
\else
\expandafter\@secondoftwo
\fi
}
\new@def\cptxpand#1{%
\romannumeral-`\q\cptifblank{#1}{\space}{\cpt@xpand#1\xpand}%
}
\def\cpt@xpand#1#2\xpand{%
\cptifsolo{#1}{%
\cptifundef{#1}{%
\cpt@@xpand{#2}{\noexpand#1}%
}{%
\cptifxpandable{#1}{%
\cpt@@xpand{#2}{\expandafter\cpt@xpand#1\xpand}%
}{%
\cpt@@xpand{#2}{\noexpand#1}%
}%
}%
}{%
\cpt@@xpand{#2}{\cpt@xpand#1\xpand}%
}%
}
\def\cpt@@xpand#1{%
\expandafter\cptswap\expandafter{\romannumeral-`\q
\cptifblank{#1}{\space}{\cpt@xpand#1\xpand}}%
}
\def\ifinteger#1{%
\if\cptifblank{#1}{1}{\expandafter\ifinteger@a\romannumeral-`\q
\cpt@xpand#1\xpand\cpt@nnil}00%
\expandafter\@firstoftwo
\else
\expandafter\@secondoftwo
\fi
}
\def\ifinteger@a#1{%
\cptifcmdeq#1\cpt@nnil{}{%
\ifinteger@b#1\cpt@nil0123456789\cpt@nnil
}%
}
\def\ifinteger@b#1\cpt@nil#2#3\cpt@nnil{%
\expandafter\cptifcmdeq\cpt@car#1\car@nil#2{%
\ifinteger@a
}{%
\cptifblank{#3}{%
1\cpt@removetonnil
}{%
\ifinteger@b#1\cpt@nil#3\cpt@nnil
}%
}%
}
% Tests
\def\cmda{01}
\def\cmdb{xy}
\let\cmdc\undefined
\edef\cmdd{\ifinteger{\cmda}{True}{False}}
\edef\cmde{\ifinteger{\cmdc}{True}{False}}
\edef\cmdf{\ifinteger{\cmda\cmdb\cmdc}{True}{False}}
\show\cmdf
I know, i'm "a bit late" to the party, but this seems to work fine:
\makeatletter
\def\mycommand#1{%
\sbox\z@{\@tempcnta=0#1\relax}%
\expandafter\ifdim\wd\z@>\z@\relax
#1 is NOT an integer
\else
#1 is an integer
\fi}
\makeatother
This works as long as you try to find positive integers (i.e., counters), not floats or negative numbers.
\box0 is already occupied in your context. Try replacing \sbox\z@ with \sbox\@tempboxa and \wd\z@ with \wd\@tempboxa or maybe declare your own box register (\newbox\mybox) and use that for measuring, instead.
\ifnum, (La)TeX does form numbers by expanding expandable tokens. Thus your request implies also checking whether the tokens forming the "given parameter" at the stage of expansion form an algorithm which terminates at all and which does not trigger errors. This is the halting problem.