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Re: Bug in ISHFT? (NOT) [message #39652]

Mon, 31 May 2004 23:07

marc schellens[1]
Messages: 183
Registered: January 2000

Senior Member

Craig Markwardt wrote:
> Craig Markwardt <craigmnet@REMOVEcow.physics.wisc.edu> writes:
>
>> Would anybody call this a bug in IDL?
>>
>> IDL> print, ishft(6UL,40UL)
>> 1536
>> IDL> print, ishft(6UL,8UL)
>> 1536
>>
>> ISHFT only seems to handle integers up to 32 bits, and then wraps
>> around.
>
>
> ... or rather, I had a bug in my brain. Somehow I was mixing up
> unsigned long and unsigned long64. My mistake.

I think there was a point in bringing it up:

At least its not really consistent:

IDL> print,ishft(1b,8)
0
IDL> print,ishft(1,16)
0
IDL> print,ishft(1L,32)
1
IDL> print,ishft(1LL,64)
1

One could think that for 8 and 16 bit there is no wrap-around, but for
32 and 64 there is.
But:

IDL> print,ishft(2ULL,63)
0
IDL> print,ishft(2ULL,64)
2
So for 32 and 64 bits, ISHFT divides the second paramter modulo
32 or 64 respectively before applying it, but not for 8 and 16 bit.

marc

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		Re: Bug in ISHFT? (NOT) By: marc schellens[1] on Mon, 31 May 2004 23:07
		Re: Bug in ISHFT? (NOT) By: Craig Markwardt on Sun, 30 May 2004 21:06

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