Public bug reported:
Hello,
If you check the qemu/util/qemu-timer.c you will find the following
function:
597: int64_t qemu_clock_get_ns(QEMUClockType type)
598: {
....
602: switch (type) {
....
606: case QEMU_CLOCK_VIRTUAL:
607: if (use_icount) {
608: return cpu_get_icount();
Now on line 606, in case we requested QEMU_CLOCK_VIRTUAL, and we are using
icount, the value of cpu_get_icount(); will be returned.
However if I understand correctly, in order to convert icount to ns, you
must use take into account the icount shift -- as defined in the
documentation: "The virtual cpu will execute one instruction every 2^N
ns of virtual time.".
Therefor, the correct value to return would be
cpu_icount_to_ns(cpu_get_icount()), where cpu_icount_to_ns is defined in
cpus.c:
296: int64_t cpu_icount_to_ns(int64_t icount)
297: {
298: return icount << icount_time_shift;
299: }
Best Regards,
Humberto "SilverOne" Carvalho
** Affects: qemu
Importance: Undecided
Status: New
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https://bugs.launchpad.net/bugs/1783422
Title:
qemu_clock_get_ns does not take into account icount_time_shift
Status in QEMU:
New
Bug description:
Hello,
If you check the qemu/util/qemu-timer.c you will find the following
function:
597: int64_t qemu_clock_get_ns(QEMUClockType type)
598: {
....
602: switch (type) {
....
606: case QEMU_CLOCK_VIRTUAL:
607: if (use_icount) {
608: return cpu_get_icount();
Now on line 606, in case we requested QEMU_CLOCK_VIRTUAL, and we are using
icount, the value of cpu_get_icount(); will be returned.
However if I understand correctly, in order to convert icount to ns,
you must use take into account the icount shift -- as defined in the
documentation: "The virtual cpu will execute one instruction every 2^N
ns of virtual time.".
Therefor, the correct value to return would be
cpu_icount_to_ns(cpu_get_icount()), where cpu_icount_to_ns is defined
in cpus.c:
296: int64_t cpu_icount_to_ns(int64_t icount)
297: {
298: return icount << icount_time_shift;
299: }
Best Regards,
Humberto "SilverOne" Carvalho
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