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Jim B 3532601532 _WINDLL being defined doesn't mean libssh2 was built as a DLL (#1578) (#1793)
https://github.com/libssh2/libssh2/issues/1578
WINDLL means that the current project using this library is a DLL; not
that this library was built as a DLL. This check for _WINDLL is
incorrect, and the LIBSSH2_EXPORTS flag should be used instead.

Description of problem - build libssh2 as a static library to be
included in a dynamic library; if libssh2 uses _WINDLL then the names
generated are `__imp_*` which do not exist in the static build of the
library.

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Co-authored-by: d3x0r <d3x0r@no-reply.github.com>
2026-04-06 08:26:08 -07:00
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2023-06-07 08:18:55 +02:00