Defining macros within macros

The way to think of this is that ## gets replaced by # in just the same way that #1 gets replaced by “whatever is the first argument”.

So if you define a macro:

\newcommand\a[1]{+#1+#1+#1+}

or (using the TeX primitive \def):

\def\a#1{+#1+#1+#1+}

and use it as \a{b}, the macro expansion produces “+b+b+b+”, as most people would expect.

However, if we now replace part of the macro:

\newcommand\a[1]{+#1+\newcommand\x[1]{xxx#1}}

then \a{b} will give us the rather odd

+b+''\newcommand{x}[1]{xxxb}''

so that the new \x ignores its argument.

If we use the TeX primitive:

\def\a#1{+#1+\def\x #1{xxx#1}}

\a{b} will expand to “+b+\def\x b{xxxb}”. This defines \x to be a macro delimited by b, and taking no arguments, which is surely not what was intended!

Actually, to define \x to take an argument, we need

\newcommand\a[1]{+#1+\newcommand\x[1]{xxx##1}}

or, using the TeX primitive definition:

\def\a#1{+#1+\def\x ##1{xxx##1}}

and \a{b} will expand to

+b+''\def\x #1{xxx#1}''

because #1 gets replaced by “b” and ## gets replaced by #.

To nest a definition inside a definition inside a definition then you need ####1, doubling the number of # signs; and at the next level you need 8 #s each time, and so on.


Source: Defining macros within macros

composition/macros/definir_une_macro_a_l_interieur_d_une_autre_macro.txt · Dernière modification: 2018/12/04 00:25 par jejust
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