Say I have a macro that does something in math mode (say, change the mathalphabet). This macro, \somecommand, takes one argument. I might want use is like this, feeding an empty argument into it: \(\somecommand{} +x\). I also want the plus sign before x to be a unary operator.

I find that whether a plus or minus sign is considered unary or binary is, while clear in theory, not always obvious upon first inspection:




\(                    +x\) \\ % unary:  "+x"
\(\zeroArgCommand     +x\) \\ % unary:  "+x"
\(\zeroArgCommand{}   +x\) \\ % binary: " + x"
\(\oneArgCommand{}    +x\) \\ % unary:  "+x"
\(\oneArgCommand{}{}  +x\) \\ % binary: " + x"


In the light of this, I am wondering about the best way of defining a unary plus/minus sign.

Intuitively I would write \(\somecommand{} {+} x\) or \(\somecommand{} {+ x}\), but I am wondering about expert opinion about a method that for sure works in all contexts.

This question was posed because in another thread, a user suggested that defining a \unaryplus macro is clearer than \somecommand {+} x. (I think what that commenter had in mind was that the reader might not be familiar with the arity of \somecommand.)

Addendum: Is there a good way to define a macro \genuineunaryplus that produces a + but that in a context <before>\genuineunaryplus<after> attaches to <after> as a unary operator but has spacing to <before> exactly as if the \genuineunaryplus were not there? That is, a macro such that if the spacing between <B> <A> is of length/type L, the expression <B>\genuineunaryplus<A> will produce <B> +<A> in the output, with the spacing between <B> and + exactly like the spacing would be between <B> and <A> if the \genuineunaryplus were not there, namely L?

Answer idea for addendum: David Carlisle's comment below outlines a possible answer for this addendum.

  • 1
    Surprisingly, the output is as expected...
    – Werner
    Feb 18, 2013 at 22:33
  • @Werner See my edit, giving a rationale for the question. Feb 18, 2013 at 22:38
  • 1
    Why not just use \unaryplus x where \newcommand{\unaryplus}{{+}}?
    – Werner
    Feb 18, 2013 at 22:49
  • @Werner See my new comments in the thread to David Carlisle's answer. Feb 18, 2013 at 23:48
  • Again, this is too generic a question. TeX has no concept of "unary operator" in your sense; just an ordinary symbol will do, so \mathord{+} would be the answer, maybe buried in a macro. Using the same symbol for two different purposes is, in general, wrong. If your + has always to behave as unary, then use a different name for it.
    – egreg
    Feb 19, 2013 at 0:16

1 Answer 1


The inspection of the math list happens after expansion (and after assignments). It is an extra stage just applicable to math mode that converts the math list into a horizontal list that is then typeset as a normal horizontal list.

So there is no real connection between the macro structure and the math spacing, It does not matter whether the thing to the left of the + has arguments or not, it just matters what it expands to (nothing in your example) so your example is equivalent to

\(                    +x\) \\ % unary:  "+x"
\(     +x\) \\ % unary:  "+x"
\({}   +x\) \\ % binary: " + x"
\(    +x\) \\ % unary:  "+x"
\({}  +x\) \\ % binary: " + x"

and as your comments show a binop like + gets binary spacing if it is between two mathord atoms such as {} or x.

Your example (from elsewhere, with \somecommand being a zero-argument macro)

\(\somecommand{} {+} x\) 


\({} {+} x\) 

Here the {+} construct makes a mathord so you get no spacing. My comment that this was probably bad markup was mainly related to the trailing {} after \somecommand it is OK to do that habitually in text mode to avoid dropping spaces but in math mode it's usually has an effect.

\(\somecommand{} {+ x}\)


\({} {+ x}\)

so here the math list has two, not three, mathord atoms: {} and +x. In this simple case it doesn't affect the spacing, but the inner expression is a single atom, so {+x} (and not x) would take any superscripts etc, and as it is an inner list linebreaking is suppressed and any white space is forced to its natural width (again not relevant here); basically in math node {...} is a box command like \hbox{....}.

There are in fact two choices for a unary math sign, a mathord or a mathop, it's easy to get an ad hoc mathord by using {+} but it is probably more consistent to declare the operators explicitly.

enter image description here

1 $a-+b$

2 $a-{+}b$

3 $a-\mathord{+}b$

4 $a-\mathop{+}b$

5 $x+y$

6 $x{+}y$

7 $x\mathord{+}y$

8 $x\mathop{+}y$

As you see from (1) if two binop atoms are adjacent the second one effectively turns into a mathord so you get the spacing as in (2) or (3) although arguably as a prefix operator giving it mathop spacing (with a small gap before its argument) is more consistent.

either way you don't want to be filling you document expressions with weird {} constructs and \mathxx primitives, just define


or whatever version you like and then use

a + \unaryplus b 

and it will all work out OK.

  • What if I would like to juxtapose x and +y, like so: x +y? Not sure whether this is useful for ordinary math, but it might be if x and y are more complicated symbols/expressions and I am using some special, domain-specific notation. Is there a macro definition that will always work? Feb 18, 2013 at 23:34
  • Just now I realize that this isn't really a matter of creating a unary plus symbol, as I'd like relational spacing on the left and no spacing on the right. As this is what also motivated my question, I'll leave this comment as is, perhaps someone has a good idea that's different from the ad-hoc \(x\;{+}y\), which feels like a hack to me. Feb 18, 2013 at 23:46
  • 1
    answer to addendum is "yes" but that's all you get at this time of night you need to rip apart the code of bm package that works out what type of atom each thing is, the instead of reconstructing a bold version of the same you'd just insert a + can be simpler as it only needs to find the first atom not iterate over a list but it's surprisingly tricky (and even trickier if you want xetex or luatex) to work out the mathclass of mathchardefed tokens. 99 times out of 100 a unary plus is going to be applied to a mathord isn't it? Feb 19, 2013 at 1:50
  • 1
    if the next token is a mathord then the space between B and + in B{+}A` is the same as the space between B and A in BA. So there is nothing to do in that case? Feb 21, 2013 at 16:24
  • 1
    @LoverofStructure you could recycle bm code which looks up the mathclass of the thing following the + but it's a lot of work for little gain, and still falls foul if you do bad markup like \zeroArgCommand{} which introduces spurious mathord atom from the {} Jun 1, 2013 at 0:02

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