Re: list mutability (very basic question)

*To*: mathgroup at smc.vnet.net*Subject*: [mg131704] Re: list mutability (very basic question)*From*: Bob Hanlon <hanlonr357 at gmail.com>*Date*: Sun, 22 Sep 2013 02:45:21 -0400 (EDT)*Delivered-to*: l-mathgroup@mail-archive0.wolfram.com*Delivered-to*: l-mathgroup@wolfram.com*Delivered-to*: mathgroup-outx@smc.vnet.net*Delivered-to*: mathgroup-newsendx@smc.vnet.net*References*: <20130921083830.751226A65@smc.vnet.net>

You are missing part of the quote. It reads: "It is important to understand that ReplacePart always creates a new list.It does not modify a list that has already been assigned to a symbol the way m[[...]] = val does." Set has attribute HoldFirst which prevents x[[1]] in your second example from turning into {0}[[1]] (your first example) and causing the error Set::setps: {0} in the part assignment is not a symbol. >> Attributes[Set] {HoldFirst, Protected, SequenceHold} Bob Hanlon On Sat, Sep 21, 2013 at 4:38 AM, Alan <alan.isaac at gmail.com> wrote: > I just realized that I am unclear on a very basic point: in what ways are > lists mutable? > > Example: > {0}[[1]]=1; (* a. this is an error *) > x={0} > x[[1]]=1 (* b. this is not an error *) > ReplacePart[x, 1 -> 2] (* c. this is differently not an error *) > > > http://reference.wolfram.com/mathematica/tutorial/ManipulatingListsByTheirIndices.html > It is important to understand that ReplacePart always creates a new list. > It does not modify a list that has already been assigned to a symbol the > way does. > > I think this means that after b. (i.e., Set[Part[x,1],1]) that x still > references the same location in memory. If so, why is a. an error? (I > realize that someone will think that saying that {0} is not a L-value will > be a helpful answer, and perhaps it should be, but I'm looking for a > different angle on the answer.) > > Thanks, > Alan Isaac > >

**References**:**list mutability (very basic question)***From:*Alan <alan.isaac@gmail.com>