Error in foreach loop when number range defined with a macro

Question

I am having problems to implement a loop that runs over the months of the year and the days of each month. A simple representative example of my current approach is shown below.

\documentclass{article} 

\usepackage{pgffor} 
\usepackage{ifthen}

\newcommand\Ndays[1]{\ifthenelse{\equal{#1}{February}}{28}%
                 {\ifthenelse{\equal{#1}{March}}{31}{30}}}   

\begin{document}

\foreach \m in {February,March,April}
  {
  month:\m \par
  \newcommand\Nd{\Ndays{\m}}
  ndays: \Nd \par
  \foreach \n in {1,...,31}
    {
    day:\n \par
    }
  }
\end{document}

Evidently, each month has a different number of days, so I decided to define a dictionary associating each month with the number of days through the "Ndays" macro. Now, I need to include the correct number of days for each month in the example above. If I naively substitute "31" by "\Nd" in the inner loop, an error is found when I compile this code with pdflatex:

! Undefined control sequence.
<argument> \equal 
                  {\m }{February}
l.23   }

I played around with different macro definitions and \expandafter-s, without success. Moreover, if I define the macro as:

\newcommand\Nd{28}

the naive substitution does not raise errors. Is there any simple solution and explanation to this?

Answer 1

You can try this. Exploiting (for once) the fact that \foreach works in a group hence LaTeX does not complain with the \newcommand. HenceElse one would use \renewcommand and an initial \newcommand\foo{}.

\documentclass{article} 

\usepackage{pgffor} 
\usepackage{ifthen}

\newcommand\Ndays[1]{\ifthenelse{\equal{#1}{February}}
                 {\newcommand\Nd{28}}%
                 {\ifthenelse{\equal{#1}{March}}
                     {\newcommand\Nd{31}}%
                     {\newcommand\Nd{30}}}}   

\begin{document}

\foreach \m in {February,March,April}
  {
  month:\m \par
  \Ndays{\m}%
  ndays: \Nd \par
  \foreach \n in {1,...,\Nd}
    {
    day:\n \par
    }
  }
\end{document}

Answer 2

The obvious glitch (at least for who has some experience in macro expansion) is that \Nd does not expand to the number of days, but to the set of instruction for printing the number of days.

The following macros associate days to months (also taking care of leap years) in an expandable way.

\documentclass{article}
\usepackage{multicol}
\usepackage{xparse}

\ExplSyntaxOn
\NewDocumentCommand{\printmonths}{O{\c_sys_year_int}m}
 {% #1 is an optional year
  % #2 is a comma separated list of months
  \clist_map_inline:nn { #2 }
   {
    \manual_print_month:nn { #1 } { ##1 }
   }
 }
\cs_new_protected:Nn \manual_print_month:nn
 {
  \par
  Month:~#2 \par
  Ndays:~\manual_print_days:nn { #1 } { #2 } \par
  \int_step_inline:nnnn { 1 } { 1 } { \manual_print_days:nn { #1 } { #2 } }
   {
    day:~##1 \par
   }
 }

\cs_new:Nn \manual_print_days:nn
 {
  \str_case:nn { #2 }
   {
    {January}{31}
    {February}{\manual_february:n { #1 }}
    {March}{31}
    {April}{30}
    {May}{31}
    {June}{30}
    {July}{31}
    {August}{31}
    {September}{30}
    {October}{31}
    {November}{30}
    {December}{31}
   }
 }

\cs_new:Nn \manual_february:n
 {
  \bool_if:nTF
   {
    \int_compare_p:n { \int_mod:nn { #1 } { 4 } != 0 }
    ||
    (
     \int_compare_p:n { \int_mod:nn { #1 } { 100 } = 0 }
     &&
     \int_compare_p:n { \int_mod:nn { #1 } { 400 } != 0 }
    )
   }
   { 28 } { 29 }
 }

\ExplSyntaxOff

\begin{document}

\begin{multicols}{3}

\printmonths[2100]{February,March,April}

\end{multicols}

\end{document}

Apart from the complication for February, the macros are straightforward: we map the argument as a comma separated list. For each item the function \manual_print_month:nn is executed, which steps from 1 to the number of days and does the printing.

You can check that the output for

\printmonths[2100]{February,March,April}

is as follows (no leap year)

A possibly more efficient version of \manual_february:n with lazy evaluation of conditionals:

\cs_new:Nn \manual_february:n
 {
  \bool_lazy_or:nnTF
   {
    \int_compare_p:n { \int_mod:nn { #1 } { 4 } != 0 }
   }
   {
    \bool_lazy_and_p:nn
     {
      \int_compare_p:n { \int_mod:nn { #1 } { 100 } = 0 }
     }
     {
      \int_compare_p:n { \int_mod:nn { #1 } { 400 } != 0 }
     }
   }
   { 28 } { 29 }
 }