By calling any kernel API you are invoking code that is GPL'd. It is not like EXPORT_SYMBOL functions are BSD-licensed (and even if they were, they themselves c

By calling any kernel API you are invoking code that is GPL'd. It is not like EXPORT_SYMBOL functions are BSD-licensed (and even if they were, they themselves call GPL-licensed code). In fact even making a normal system call from userspace invokes GPL'd code. So I think there are only two interpretations. One is that EXPORT_SYMBOL_GPL attempts to claim a copyright on the interface, and not the code. The other is that redistributing a GPL-incompatible object that invokes GPL'd code is always infringement, and (possibly) the special exception for userspace in Linux's copyright statement, https://github.com/torvalds/linux/blob/master/LICENSES/exceptions/Linux-syscall-note https://github.com/torvalds/linux/blob/master/LICENSES/excep... , is the only thing that makes normal userspace processes non-infringing, and that regular EXPORT_SYMBOL is basically also an explicit exception (which may or may not even be legally meaningful - during the dma-buf discussion, various copyright holders expressed that they did not interpret EXPORT_SYMBOL as an exception).