sizeof in fortran

Question

It's quite common in C-code to see stuff like:

malloc(sizeof(int)*100);

which will return a pointer to a block of memory big enough to hold 100 ints. Is there any equivalent in fortran?

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Use case:

I have a binary file which is opened as:

open(unit=10,file='foo.dat',access='stream',form='unformatted',status='old')

I know that the file contains "records" which consist of a header with 20 integers, 20 real numbers and 80 characters, then another N real numbers. Each file can have hundreds of records. Basically, I'd like to read or write to a particular record in this file (assuming N is a fixed constant for simplicity).

I can easily calculate the position in the file I want to write if I know the size of each data-type:

header_size = SIZEOF_INT*20 + SIZEOF_FLOAT*20 + SIZEOF_CHAR*80
data_size = N*SIZEOF_FLOAT
position = (record_num-1)*(header_size+data_size)+1

Currently I have

!Hardcoded :-(
SIZEOF_INT = 4
SIZEOF_FLOAT = 4
SIZEOF_DOUBLE = 8
SIZEOF_CHAR = 1

Is there any way to do better?

constraints:

• The code is meant to be run on a variety of platforms with a variety of compilers. A standard compliant solution is definitely preferred.

Answer 1

In your use case I think you could use

inquire(iolength=...) io-list

That will give you how many "file storage units" are required for the io-list. A caveat with calculating offsets in files with Fortran is that "file storage unit" need not be in bytes, and indeed I recall one quite popular compiler by default using a word (4 bytes) as the file storage unit. However, by using the iolength thing you don't need to worry about this issue.