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UTC since 1/1/1970, IEEE double precision floating point. Intel format

So that's a little-endian floating-point not-necessarily-integral number of seconds since the UN*X Epoch?

If so, then the answer to

So which, if any, of the encodings listed at https://gitlab.com/wireshark/wireshar... is applicable?

is "none".

You will need to fetch the floating-point value directly into a numerical value, calculate the integral part (seconds) and fractional part (as nanoseconds), construct a new NSTime from those two values, and add that to the protocol tree using subtree:add(f.time, buffer(0,8), {that value}).

UTC since 1/1/1970, IEEE double precision floating point. Intel format

So that's a little-endian floating-point not-necessarily-integral number of seconds since the UN*X Epoch?

If so, then the answer to

So which, if any, of the encodings listed at https://gitlab.com/wireshark/wireshar... is applicable?

is "none".

You will need to fetch the floating-point value directly into a numerical value, calculate the integral part (seconds) and fractional part (as nanoseconds), construct a new NSTime from those two values, and add that to the protocol tree using subtree:add(f.time, subtree:add_le(f.time, buffer(0,8), {that value}).

UTC since 1/1/1970, IEEE double precision floating point. Intel format

So that's a little-endian floating-point not-necessarily-integral number of seconds since the UN*X Epoch?

If so, then the answer to

So which, if any, of the encodings listed at https://gitlab.com/wireshark/wireshar... is applicable?

is "none".

You will need to fetch the floating-point value directly into a numerical value, calculate the integral part (seconds) and fractional part (as nanoseconds), construct a new NSTime from those two values, and add that to the protocol tree using subtree:add_le(f.time, subtree:add(f.time, buffer(0,8), {that value}).