[Tex/LaTex] What does mathchar do

math-modesymbolstex-core

The package cmll defines \with (looks like '&') as

$ texdef -t latex -p cmll with

\with:
\mathchar"2026


\the\with:
8230

When I take a look at \& it is defined as

texdef -t latex \&          

\&:
\char"26


\the\&:
38

So \with and \& seem to be different, although the resuls look identical to me. What is the difference?

Best Answer

The four hexadecimal digits "kfab in a \mathchar specify

  • k is the atom type (0 = ordinary, 1 = operator, 2 = binary operation, 3 = relation, 4 = opening, 5 = closing, 6 = punctuation, 7 = variable family);

  • f the math group (font family) where the glyph should be taken from;

  • ab the slot in the font.

One can use \mathchar<15 bit number> directly or define

\mathchardef<cs>=<15 bit number>

so, for instance, after \mathchardef\with="2026 the command \with is equivalent to typing \mathchar"2026 (or \mathchar8230, for " specifies hexadecimal number).

Typing \char<8 bit number> tells TeX to use the character from the current font in the specified slot. However, when in math mode, \char"ab is equivalent to saying \mathchar"00ab, so ordinary symbol from math group 0 and the same slot. The \chardef command is the counterpart of \mathchardef and, indeed, \& is usually defined as

\chardef\&="26

Small lie: it's \chardef\&=`&, but it's not really important.

There is a big difference between $x \mathchar"2026 y$ and $x \char"26 y$, or, with the definitions above

$x \with y$

$x \& y$

enter image description here

In the first case the & symbol is spaced as it's good for a binary operation, in the second case no space is added, because we're specifying three ordinary symbols.

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