drop() by default?

[RobinHankin]

Most of the time I want the drop()ed value:

> e(1)^2
Element of a Clifford algebra, equal to
scalar ( 1 )
> drop(e(1)^2)
[1] 1
> 

Maybe drop() should be called by default to eliminate this needless repetition. Maybe it should be an option()?

[RobinHankin]

Also, equality testing is infelicitous:

> e(1)^2 == 1
Error in Ops.clifford(e(1)^2, 1) : 
  Generic '==' only compares two clifford objects with one another
> drop(e(1)^2)  == 1
[1] TRUE
> 

[RobinHankin]

Simply changing the last line of clifford() from return(out) to return(drop(out)) is not a good idea: it requires a Kafkaesque sequence of code changes because quite often you actually want a scalar to be a Clifford object

[RobinHankin]

The problem with the earlier suggestion of adding return(drop(out)) to the end of clifford() is this: I want idiom 4 + e(1) to work. Currently this is done by coercing the 4 to a clifford, then dispatching to clifford_plus_clifford(). But the suggestion makes as.clifford(4) return numeric 4, and so the clifford addition fails:

> clifford_plus_clifford(4,e(1))
Error in x[[2]] : subscript out of bounds
> 

[RobinHankin]

There are a few functions in the package that should have a drop argument added; see also issue #59.