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christyleomin
Joined: 08 Apr 2011
Posts: 155
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 Posted: Wed Aug 15, 2012 2:37 pm Post subject: 8 byte integer |
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I have declared as below:
| Code: | integer*8 a,b
integer c
a = b+c |
I have declared a and b as 8 bytes because at times their amx value evaluates greater than 4,294,967,296 (because a 4 byte integer can have 8 * 4 = 32 bits, so your range is from 0 to 2^32 (4,294,967,296)
Is it also necessary to declate c as 8 byte inteegr in such a case? |
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PaulLaidler
Site Admin
Joined: 21 Feb 2005
Posts: 8211
Location: Salford, UK
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| Posted: Wed Aug 15, 2012 3:36 pm Post subject: |
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| c should be OK as INTEGER*4 but you can always run a test to make sure. |
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JohnCampbell
Joined: 16 Feb 2006
Posts: 2615
Location: Sydney
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| Posted: Thu Aug 16, 2012 2:31 am Post subject: |
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| range is from 0 to 2^31 (2,147,483,648), due to sign bit. |
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davidb
Joined: 17 Jul 2009
Posts: 560
Location: UK
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| Posted: Thu Aug 16, 2012 6:45 am Post subject: |
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The maximum 32 bit integer is 0 to 2**31-1 (2,147,483,647).
The minimum 32 bit integer is -2,147,483,647 since, strictly speaking, Fortran uses a symmetric integer model, though most compilers use twos complement maths and let you go down to -2,147,483,648.
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Programmer in: Fortran 77/95/2003/2008, C, C++ (& OpenMP), java, Python, Perl |
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