Re: Problem with my split routine I am not understanding



On Thu, 28 Dec 2006 10:30:01 -0800, Beliavsky wrote:

integer :: bpos = 1, pos = 1, idx = 1

I have not read your message carefully, but do you realize that the
line above is equivalent to

integer, save :: bpos = 1, pos = 1, idx = 1


I did not realize that. Your solution of not initializing bpos, pos and
idx in that manner fixed the problem. Thank you for the pointer.

Jeremy
.



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