Avoiding referencing (was: Python sequences reference - problem?)

[Duncan Booth]

Thorsten Kampe <thorsten at thorstenkampe.de> wrote in news:amejvb$55qsi$1 at ID-
77524.news.dfncis.de:

> Obviously I did not pass 'foo' "by value" to function 'part' but by 
> reference and obviously not only local variable 'seq' was modified in 
> the function but "outer" variable 'foo', too.
> 
> So how to deal with that?
> 
You missed out option number 4:

Don't write code that modifies its arguments unless you always want the 
arguments modified.

In general, if you want to modify an argument, but don't want the change 
reflected back to the caller, make a copy of the argument. If you want the 
change reflected back to the caller always, then go ahead and modify the 
argument, but make it clear from the function name that it does that. If 
you want the function to sometimes modify its argument, then write it so 
that it never modifies the argument but returns a modified value as a 
result. Then the caller can decide whether to use that result or EXPLICITLY 
ignore it.

For the specific example you gave, there was no reason to modify the 
argument, the code is just as simple without either copying or modifying:

>>> def part(seq, indices):
	partition = []
	start = 0
	for slice in indices:
		partition.append(seq[start:slice])
		start = slice
	partition.append(seq[start:])
	return partition

>>> foo = [11, 22, 33, 44, 55, 66, 77]
>>> part(foo, [2, 3])
[[11, 22], [33], [44, 55, 66, 77]]
>>> foo
[11, 22, 33, 44, 55, 66, 77]
>>> 

-- 
Duncan Booth                                             duncan at rcp.co.uk
int month(char *p){return(124864/((p[0]+p[1]-p[2]&0x1f)+1)%12)["\5\x8\3"
"\6\7\xb\1\x9\xa\2\0\4"];} // Who said my code was obscure?