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redo time.monotonic() to avoid double precision #343

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Merged
merged 2 commits into from
Oct 17, 2017
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redo time.monotonic() to avoid double precision #343

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6afe5e1
Set DRVSTR on output pins to strong (more current capability).
dhalbert Oct 16, 2017
49acf09
redo time.monotonic() to avoid double precision
dhalbert Oct 17, 2017
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3 changes: 3 additions & 0 deletions atmel-samd/common-hal/digitalio/DigitalInOut.c
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Original file line number Diff line number Diff line change
Expand Up @@ -80,6 +80,9 @@ void common_hal_digitalio_digitalinout_switch_to_output(
pin_conf.input_pull = PORT_PIN_PULL_NONE;
port_pin_set_config(self->pin->pin, &pin_conf);

// Turn on "strong" pin driving (more current available). See DRVSTR doc in datasheet.
system_pinmux_pin_set_output_strength(self->pin->pin, SYSTEM_PINMUX_PIN_STRENGTH_HIGH);

self->output = true;
self->open_drain = drive_mode == DRIVE_MODE_OPEN_DRAIN;
common_hal_digitalio_digitalinout_set_value(self, value);
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4 changes: 3 additions & 1 deletion shared-bindings/time/__init__.c
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Expand Up @@ -52,7 +52,9 @@
//| :rtype: float
//|
STATIC mp_obj_t time_monotonic(void) {
return mp_obj_new_float(common_hal_time_monotonic() / 1000.0);
uint64_t time64 = common_hal_time_monotonic();
// 4294967296 = 2^32
return mp_obj_new_float(((uint32_t) (time64 >> 32) * 4294967296.0f + (uint32_t) (time64 & 0xffffffff)) / 1000.0f);
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How does precision change with this? Would we get more precision by dividing by 1000.0f earlier?

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I could distribute the division among the two halves, but the precision is still lost due to single-precision floating point. So I don't believe it will make a difference:

>>> (2.0**22+1)/1000 - (2.0**22+0)/1000 
0.0
>>> (2.0**22+2)/1000 - (2.0**22+1)/1000 
0.00195313
>>> (2.0**22+3)/1000 - (2.0**22+2)/1000 
0.0
>>> (2.0**22+4)/1000 - (2.0**22+3)/1000 
0.00195313
>>>

}
MP_DEFINE_CONST_FUN_OBJ_0(time_monotonic_obj, time_monotonic);

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