x86-urgent-2022-05-08

 - Prevent FPU state corruption. The condition in irq_fpu_usable() grants
   FPU usage when the FPU is not used in the kernel. That's just wrong as
   it does not take the fpregs_lock()'ed regions into account. If FPU usage
   happens within such a region from interrupt context, then the FPU state
   gets corrupted. That's a long standing bug, which got unearthed by the
   recent changes to the random code.

 - Josh wants to use his kernel.org email address