topic Using realloc-lhs with size 0 array in Intel® Fortran Compiler

Using realloc-lhs with size 0 array

OP1 — Mon, 31 Oct 2011 21:51:19 GMT

When A is an allocatable array, provided that the option /assume:realloc_lhs is used, the following syntax is correct:

A= [1,2,3]

But how do I allocate A to size 0 (short of a proper ALLOCATE statement)? A = [] does not work (with IVF 11.1). It's a bit of an academic questoin, but for debugging purposes it's quite helpful.

Thanks,

Olivier

Using realloc-lhs with size 0 array

Arjen_Markus — Tue, 01 Nov 2011 07:18:45 GMT

You might try:

A = [ (i, i=1,0) ]

I do not know why your original statement does not work, but the above creates a zero-length

array explicitly, so I expect it to do the trick.

Regards,

Arjen

Using realloc-lhs with size 0 array

IanH — Tue, 01 Nov 2011 07:52:56 GMT

Put a type-spec in the array-constructor.

A = [ INTEGER :: ]

Using realloc-lhs with size 0 array

OP1 — Tue, 01 Nov 2011 14:48:37 GMT

Thanks Ian - it works! It does look a bit weird, certainly.

Olivier

Using realloc-lhs with size 0 array

jimdempseyatthecove — Wed, 02 Nov 2011 15:47:01 GMT

>>it works! It does look a bit weird

What is the data type on the rhs with

A = []

?

Fortran does not have a "void" type, but it does have an empty array of (specified) type.

Using realloc-lhs with size 0 array

OP1 — Wed, 02 Nov 2011 16:24:03 GMT

Jim,

It's an INTEGER allocatable array. I was looking for the convenience of A = []. Ian suggestion works.

Olivier