Re: linux-user - time64 question

[Laurent Vivier]

Le 05/05/2020 à 23:38, Sid Manning a écrit :
> I’m looking at a testcase failure when my target uses 64bit time in
> msg.h (struct msqid_ds).  I’ve been able to get around this but changing
> target_msqid_ds like so:
> 
> 
> @@ -3900,18 +3901,9 @@ static inline abi_long do_semop(int semid,
> abi_long ptr,
> unsigned nsops)
>  struct target_msqid_ds
>  {
>      struct target_ipc_perm msg_perm;
> -    abi_ulong msg_stime;
> -#if TARGET_ABI_BITS == 32
> -    abi_ulong __unused1;
> -#endif
> -    abi_ulong msg_rtime;
> -#if TARGET_ABI_BITS == 32
> -    abi_ulong __unused2;
> -#endif
> -    abi_ulong msg_ctime;
> -#if TARGET_ABI_BITS == 32
> -    abi_ulong __unused3;
> -#endif
> +    abi_ullong msg_stime;
> +    abi_ullong msg_rtime;
> +    abi_ullong msg_ctime;
>      abi_ulong __msg_cbytes;
>      abi_ulong msg_qnum;
>      abi_ulong msg_qbytes;
> 
> It seems like either should have worked but I get garbage back in some
> of the elements below msg_time fields without the change.
> 
> If time_t is 64bits then it seems like stime/rtime/ctime should be
> abi_ullong.
> 
> My target is Hexagon and the TARGET_ABI_BITS is 32.

The structure has been changed into the kernel for the y2038 and the
change has not been reflected into qemu (and it needs).

See kernel commit:

c2ab975c30f0 ("y2038: ipc: Report long times to user space")

Thanks,
Laurent