e92868536d
* Clarify recognition of "command form" by MATLAB parser. Expand a bit the explanation; remove an unnecessary non-capturing group, replace \s+ by \s at the end (we just need to check for the presence of a space). * MATLAB: Correctly parse 1./x as (1)(./)(x).
10 lines
178 B
Text
10 lines
178 B
Text
# 1./x is (1)(./)(x), not (1.)(/)(x)
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---input---
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1./x
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---tokens---
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'1' Literal.Number.Integer
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'./' Operator
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'x' Name
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'\n' Text.Whitespace
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