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Out of curiosity I wanted to find the limits of TeX regarding macro definitions. Assume there should be an assignment command which assigns the value of an already defined macro to a yet undefined macro, so something like

\def\newVariable{\variable}

just with the syntax usually find in other languages, i.e. using the equal sign as an assignment operator.

I guess it's not possible to write

\newVariable = \variable

as this results in an 'Undefined control sequence' when \newVariable is read. Is it possible to hook into the execution of whatever detects the undefined control sequence and look ahead with @ifnextchars for an equal sign and an macro?

What were if we'd write

newVariable = \variable

Now this should at least run without error. When \variable is expanded, is there any chance to remove the text newVariable = from the output string again (do some sort of rewinding) and use it in \variable as an option similar to \variable{newVariable}? So I guess here I would look for something like \@iflastchars. I am all open for bizarre solutions with all TeX flavours like finding the string in the output routine or in the aux or even in the PDF file produced by the last run (or any of the other output files).

I guess the answer will be "this is not possible", but as there are a lot of smart people around liking a challenge, I thought I'd give it a try. Please note, that I am fully aware of the solution with the normal TeX syntax so this is a more a theoretical exercise.

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When do you want \variable to be expanded? (It doesn't get expanded in \def\newVariable{\variable}.) –  Ian Thompson Oct 8 '12 at 18:45
    
you can use = for assigment, you just have to use \let as well: \let\newVariable = \variable. You can use = for counters and lengths: \c@section = 5, \footskip = 5mm. However, in LaTeX, you better avoid these low-level format if you don't need them. You must think of (La)TeX as of a "different programming language" that has its very strange specifics ;) –  tohecz Oct 8 '12 at 18:47
    
@IanThompson Good point. I thought it would only be possible if \variable would be expanded after reading the equal sign, as otherwise there is no chance of changing the execution. If it gets expanded it could do whatever it wants to hide the expansion to the outside if that is wanted. If you find a solution without expansion that would also be great. –  Patrick Häcker Oct 8 '12 at 18:54
    
@tohecz You are completely correct, thanks for the example. Interestingly this is possible either because a macro (ok, a primitive command) \let is left of the \newVariable (it's the same with \def) or because the thing is already defined (for counters and lengths). I guess my question could be reformulated as: Is this possible with yet undefined macros without having a macro like \let on the left? –  Patrick Häcker Oct 8 '12 at 18:58
    
The problem is that you need to use = as a "text" and "math" character with its real meaning to the "normal" people. I think that doing this is impossible because TeX reads its input lineary and invokes an error on an unknown macro in the input sequence. –  tohecz Oct 8 '12 at 19:11
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2 Answers 2

up vote 9 down vote accepted

If TeX doesn't know the meaning of a control sequence, it will complain with Undefined control sequence, unless you are assigning a meaning to the control sequence itself.

So, no, there is no way to say something like

\varB=\varA

where \varB has no previous meaning. Anyway, a syntax like this is allowed only for registers (that have previously been declared or are primitive). So, for instance,

\tolerance=\mycount

is a legitimate assignment to the primitive register \tolerance, but only when \mycount is a register of the same type (in this case an integer), be it primitive or defined by \countdef, for instance

\countdef\mycount=123

(Such an assignment, that makes \mycount equivalent to the primitive register \count123, is usually performed behind the scene by the macro \newcount.)

If one says

\varB=\varA

where \varA has a meaning and \varB hasn't there will be an error, as said before. If \varB has a meaning and it is not compatible with the assignment, there will be again an error. For instance,

\newcount\varA \varA=42
\newdimen\varB
\varB=\varA

will throw the error Illegal unit of measure (pt inserted), unless, by chance, the following tokens may be interpreted as a legal unit of measure.

Conversely,

\newdimen\varA \varA=42pt
\newcount\varB
\varB=\varA

is legal and \varB will hold the integer 2752512 (the equivalent of 42pt in scaled points: 2752512 = 42 * 65536). This is because dimensions are stored internally as integers, but it's a rather esoteric topic.

Notice that

\newcount\varA \varA=42
\newcount\varB
\varB=\varA

is very different from

\newcount\varA \varA=42
\let\varB=\varA

In the first case, changing the value stored in \varA will not change the one stored in \varB, while in the second case it will, because \varB has become an alias for \varA.

Hooking into the mechanism that discovers undefined control sequences is not possible at the user's level and requires rewriting "TeX the program".

Of course it's theoretically possible to modify the syntax rules of TeX by declaring all characters active after having created a suitable set of macros; so, for instance, each letter might store itself somewhere waiting for some operator to come along telling what to do with the saved tokens. Say

newvariable = myvar ;

where the semicolon acts as "end of instruction". The active = would be defined as "look at what comes until the semicolon, determine its type and perform the assignment to the variable whose name results from the tokens in the storage bin". There are papers by Jonathan Fine on TUGboat describing this sort of plays, see for instance "Active TeX and the DOT Input Syntax" (TUGboat, v. 20, n.3, p. 248–254). Nothing is really impossible, the problem is: is it worth the labor? Of course one must define a "programming environment" and a "typesetting environment", since, after all, TeX's job is primarily typesetting documents.

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Very informative, thanks! Any thoughts about using arguments left of the macro, i.e. some sort of postfix notation? –  Patrick Häcker Oct 8 '12 at 19:34
    
@MMM No chance: TeX works on the first token it finds. Don't forget that its primary job is to typeset text, so if it finds an A it typesets it and doesn't look ahead to see if something comes along to which this A should be passed as an argument. –  egreg Oct 8 '12 at 19:44
    
So if even you do not see a chance, this might really be impossible. Anyway, as mentioned to tohecz, I will wait until tomorrow to see if there is really no brilliant/insane solution to this artificial problem. –  Patrick Häcker Oct 8 '12 at 19:50
    
Well, there's one very theoretical way, and that is, making \ an active character. But it would be so screwed and probably kill many other things. –  tohecz Oct 8 '12 at 20:25
    
@MMM I've added some comments about some possible (devious) ways to proceed. –  egreg Oct 8 '12 at 20:27
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No, it is not possible to make TeX do some action on finding an undefined sequence, other than invoking an error. TeX core reads the input sequence lineary, so no command can really influence what was read before. Moreover, the chracter = needs to be used in other contexts as well so it cannot easily be made active.

As I point out in the comments, there are cases where = means assignment:

  • \let definition: \let\newmacro=\oldmacro
  • Lengths assignment: \topskip = 10pt
  • Counter assignment: \c@section = 5
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You have good arguments, that's why I upvoted. Nevertheless, I will wait with accepting, because I will regard this as impossible after it is either proven or if nobody found a solution in the next 24 hours. –  Patrick Häcker Oct 8 '12 at 19:26
2  
of course arguably in all your examples, = means nothing rather than means assignment, as the meaning of the expression would be the same without the = : \c@section = 5 is just syntactic sugar for \c@section5 –  David Carlisle Oct 8 '12 at 23:04
    
@DavidCarlisle unless you're wanting to be bullet-proof where you should use = with let, in the case when \oldmacro is = itself. –  tohecz Oct 10 '12 at 8:30
    
well yes but if you are paranoid then you'd also want to handle a space token as well space before or after the equals is ignored usually) –  David Carlisle Oct 10 '12 at 8:37
    
And well, D. Knuth obviously did not choose = by mistake. Imagine if it had been \let\oldmacro!\newmaco, it would have been so confusing! So he clearly used = to stress that it is an assignment. –  tohecz Oct 10 '12 at 8:42
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