copy on write

[88888 Dihedral]

在 2012å¹´1月14日星期六UTC+8上åˆ6æ—¶48分29秒，Evan Driscoll写é“：
Well, in any associative operation with an identity implemented in a computer language personally I believe the operator overloading part in C++ is another teasing trick.


Do we have to work out the algebra here?

 

[88888 Dihedral]

在 2012å¹´1月14日星期六UTC+8上åˆ6æ—¶48分29秒，Evan Driscoll写é“：
Well, in any associative operation with an identity implemented in a computer language personally I believe the operator overloading part in C++ is another teasing trick.


Do we have to work out the algebra here?

 

[Wolfram Hinderer]

But isn't it equally true if we say that z = t[1], then t[1] += x is syntactic sugar for z = z.__iadd__(x)? Why should that fail, if z can handle it?

It's more like syntactic sugar for
y = t; z = y.__getitem__(1); z.__iadd__(x); y.__setitem__(1, z)

It's clear that only the last expression fails, after the mutation has
taken place.

Just in case you wonder about the y: you need it for more complicated
cases.
t[1][1] += [4] is syntactic sugar for
y = t.__getitem__(1); z = y.__getitem__(1); z.__iadd__([4]);
y.__setitem__(1, z)

That makes clear why there's no exception in this code:

t = (0, [1, [2, 3]])
t[1][1] += [4]
t

(0, [1, [2, 3, 4]])