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Re: List insert at index that is well out of range - behaves like append that too SILENTLY

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  Re: List insert at index that is well out of range - behaves like append that too SILENTLY MRAB <python@mrabarnett.plus.com> - 2014-09-16 13:04 +0100

#77928 — Re: List insert at index that is well out of range - behaves like append that too SILENTLY

On 2014-09-16 06:58, dieter wrote:
> Harish Tech <technews.full@gmail.com> writes:
>
>> Let me demonstrate the problem I encountered :
>>
>> I had a list
>>
>>  a = [1, 2, 3]
>>
>> when I did
>>
>> a.insert(100, 100)
>>
>> [1, 2, 3, 100]
>>
>> as list was originally of size 4 and I was trying to insert value at index
>> 100 , it behaved like append instead of throwing any errors as I was trying
>> to insert in an index that did not even existed .
>>
>>
>> Should it not throw
>>
>> IndexError: list assignment index out of range
>
> At least the documentation states that what you observe is the intended
> behaviour.
>
> According to the documentation, "a.insert(i, x)" is
> equivalent to "a[i:i] = x" (i.e. a slice assignment) and
> if in a slice "a[i:j]" "i" or "j" are larger then "len(a)",
> then it is replaced by "len(a)".
>
> If this is not what you want, derive your own list type
> and override its "insert" method.
>
It's nice to be able to say a.insert(p, x) even when p == len(a)
instead of having a special case:

if p == len(a):
     a.append(x)
else:
     a.insert(p, x)

Of course, what to do when p > len(a) is another matter.

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