提交 b65490a3 编写于 作者: C Cole Robinson

qemu: Properly report a startup timeout error

The timeout errors were unconditionally being overwritten by the less
helpful 'unable to start guest' error.
上级 80de0a8c
......@@ -1351,15 +1351,15 @@ qemudReadLogOutput(virDomainObjPtr vm,
buf[got] = '\0';
if (got == buflen-1) {
qemuReportError(VIR_ERR_INTERNAL_ERROR,
_("Out of space while reading %s log output"),
what);
_("Out of space while reading %s log output: %s"),
what, buf);
return -1;
}
if (isdead) {
qemuReportError(VIR_ERR_INTERNAL_ERROR,
_("Process exited while reading %s log output"),
what);
_("Process exited while reading %s log output: %s"),
what, buf);
return -1;
}
......@@ -1371,7 +1371,8 @@ qemudReadLogOutput(virDomainObjPtr vm,
}
qemuReportError(VIR_ERR_INTERNAL_ERROR,
_("Timed out while reading %s log output"), what);
_("Timed out while reading %s log output: %s"),
what, buf);
return -1;
}
......@@ -1547,12 +1548,8 @@ qemudWaitForMonitor(struct qemud_driver* driver,
virStrerror(errno, ebuf, sizeof ebuf));
}
if (ret < 0) {
/* Unexpected end of file - inform user of QEMU log data */
qemuReportError(VIR_ERR_INTERNAL_ERROR,
_("unable to start guest: %s"), buf);
if (ret < 0)
return -1;
}
VIR_DEBUG("Connect monitor to %p '%s'", vm, vm->def->name);
if (qemuConnectMonitor(vm) < 0)
......
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