[Tex/LaTex] the difference between \relax and {}

macrostex-core

As I understand it, the point of \relax is to stop a macro from swallowing up subsequent material (such as whitespace). But {} (\bgroup\egroup) serves the same function. Why do we need both?

Best Answer

Try the following couple examples:

\documentclass{article}

\begin{document}

\[
\begin{array}{rl}
+9mm   & +9m^2 \\
[+9mm] & [+3m]^2 \\{}
+9mm   & +9m^2 \\{}
[+9mm] & [+3m]^2 \\\relax
+9mm   & +9m^2 \\\relax
[+9mm] & [+3m]^2
\end{array}
\]

\end{document}

enter image description here

If you leave there \\ alone, the [+9mm] is interpreted as an optional argument to \\ and transformed into a vertical space.
If you use {}, then it inserts an "empty thing" into the formula and the + is improperly spaced.
Only the variant with \relax works correctly here, because it really does nothing.

Very similar examples can be used to show that \relax is the correct terminator of \dimexpr and \numexpr expansion etc.

Related Solutions

[Tex/LaTex] the difference between \let and \def

The difference is in the time at which the ‘right hand side’ is evaluated.

Thus \let\foo\bar defines \foo to have the value that \bar had at the point of definition. On the other hand, \def\foo{\bar} in effect defines \foo to have the value that \bar has at the point of use.

Consider:

\def\bar{hello}
\let\fooi\bar
\def\fooii{\bar}
\fooi +\fooii

\def\bar{goodbye}
\fooi +\fooii

This produces

hello+hello
hello+goodbye

This is a simple process.

However it's also a subtle one, so it might be worth highlighting a few key points:

When TeX encounters control sequences such as \fooi, it evaluates them; if these are macros (that is, they have been defined by \def, or \let equal to something which was defined by \def), then the result is that they will expand to other tokens, which TeX will then examine in turn, and so on, recursively, until what's left is either ‘primitive’ control sequences or letters (I'm simplifying a little bit).
\fooi expands directly to the characters hello (because \bar initially did, and \fooi was defined to have the same value).
\fooii, in contrast, expands to \bar, which is then immediately reexamined and reexpanded. In the first case, \bar expands to hello and in the second case to goodbye. The definition of \fooii hasn't changed, but \bar has been redefined in between.
Getting a clear idea of the process of this recursive expansion is very helpful when learning how to develop and debug TeX macros.

[Tex/LaTex] Difference between \let\foo\relax and \def\foo{} for disabling

When you

\def\foo{}

you are giving \foo an empty definition, but \foo still 'exists' using any of the tests used for this (such as \ifdefined or LaTeX's \@ifundefined). The same effect can be achieved by doing

\let\foo\empty

which is very slightly more efficient as it points to an already-used memory location (not a worry nowadays). When \foo has such a definition, it is expandable, and so

\edef\baz{\foo}

will result in \baz being empty, as \foo expands to nothing at all.

On the other hand

\let\foo\relax

makes \foo equal to the \relax primitive. That is a 'do nothing' operation, but importantly is not expandable. So in this case

\edef\baz{\foo}

leaves \baz with definition '\foo'. That can be useful: it's a way of temporarily preventing a macro from doing anything while retaining it's appearance in other code. On the other hand, sometimes you don't want that: it depends on the context. When \foo is equal to \relax, whether it is regarded as 'existing' by the various tests is more variable. TeX automatically creates control sequences equal to \relax in various cases, and so some tests will regard anything equal to \relax as 'not defined'.

So which is better depends on your use case.

Best Answer

Related Solutions

[Tex/LaTex] the difference between \let and \def

[Tex/LaTex] Difference between \let\foo\relax and \def\foo{} for disabling

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