Someone made mention recently that you can't assign to a variable using
the deref operator on a numeric expando, but you can using a param:
# This works
alias testalias1 (varname) {
@ *varname = 'booya'
}
# This does not work
alias testalias2 {
@ *0 = 'booya'
}
The reason for this has to do with how tokens are handled in the math
parser. There are three types of tokens:
1) A bare variable name varname
2) A string 'booya'
3) A number 0
When you use the deref operator on a variable name, it converts the rvalue of
that variable into an lvalue.
assign varname testing
# this assigns 'this is a test' to $testing
*varname = 'this is a test'
# So *varname is the same as $testing...
*varname == 'this is a test'
# So this is always true
*varname == [$($varname)]
When you use the deref operator on a *number*, it expands to the rvalue of
the numeric expando of that number, and does *NOT* convert that rvalue into
an lvalue. This is what allows you to do:
@ foo = *0
But the thing that allows you to do that is the same thing that doesn't
allow you to do:
@ *0 = 'testing'
Because *0 is $0 and not $($0):
# This is false!
*0 == [$($0)]
# This is true
*0 == [$0]
# This is true!
*varname = [$($varname)]
# This is false!
*varname = [$varname]
Derefing variable names and numbers are at odds with each other. So it is
not possible for both
@ *0 = 'testing'
and @ foo = *0
to both work at the same time. So for backwards compatability, we will
maintain the current behavior, even though it's not consistent.
All is not lost -- you *can* assign to a variable name passed as $0, you
just have to deref it a second time, because *0 yields an rvalue, and you
can just deref it again to turn it into an lvalue:
# This assigns to the variable passed as $0
@ *(*0) = 'testing
Consequently, I recommend we make no change to how this works.
Comments, thoughts, questions, improvements, suggestions?
Jeremy
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