Open a #resource using FileOpen function

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Hello. Is that possible to do this considering the FileOpen function works with specific directories only?

[Alain Verleyen]

Yashar Seyyedin:
Hello. Is that possible to do this considering the FileOpen function works with specific directories only?

No, a resource is not a file. So you don't need to open it.

It's directly available as a variable you have to process. For example :

#resource "\\your_data_file.bin" as SCustomStruct data[]

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Alain Verleyen #:

No, a resource is not a file. So you don't need to open it.

Thanks