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Re: problem with subset one image by another [message #59548 is a reply to message #59487] Tue, 01 April 2008 04:18 Go to previous messageGo to previous message
jameskuyper is currently offline  jameskuyper
Messages: 79
Registered: October 2007
Member
negra wrote:
...
> max_x = min(file1_xmap[1], file_xmap[1])
> min_y = max(file1_ymap[0], file_ymap[0])
> max_y = min(file1_ymap[1], file_ymap[1])

As Jean has already pointed out, you'd have to organize things
differently to use MAX() and MIN() here. However, it's far simpler to
avoid MAX() and MIN() altogether.

max_x = file1_xmap[1] < file_xmap[1]
min_y = file1_ymap[0] > file_ymap[0]
max_y = file1_ymap[1] < file_ymap[1]

It all looks pretty confusing if you're used to a language (there are
quite a few) where '<' and '>' are comparison operators, but in IDL the
corresponding comparison operators are called 'lt' and 'gt'. The '<' and
'>' operators return the minimum and maximum, respectively, of their
operands.
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