Better way to combine number with fraction?

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Haisam Khaled
Haisam Khaled 2024-2-1
评论： Haisam Khaled 2024-2-3
I have this struct output (from ROS2 message):
MessageType: 'builtin_interfaces/Time'
sec: 1706819594
nanosec: 685974901
I want to combine sec and nanosec in one value like this:
1706819594.685974901
I used the following line of code (and it worked):
zBase.header.stamp.sec+ sprintf(".%d",zBase.header.stamp.nanosec)
Is there a better way ?

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采纳的回答

Stephen23
Stephen23 2024-2-1
编辑：Stephen23 2024-2-1
Ran in:
Note that you will need to use SPRINTF (or COMPOSE etc) to get the right output when NANOSEC has fewer than the full nine digits:
s = 1706819594;
ns = 685974901;
sprintf("%d.%09d",s,ns)
ans = "1706819594.685974901"
ns = 123;
sprintf("%d.%09d",s,ns)
ans = "1706819594.000000123"
Whereas your format string will given an incorrect output:
s + sprintf(".%d",ns)
ans = "1706819594.123"
If the input number has more than nine digits then the output will be wrong anyway.
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Stephen23
Stephen23 2024-2-3
Ran in:
If the two values are numeric and you want a single numeric output then skip the intermediate text:
format long G
s = 1706819594;
ns = 685974901;
out = s + ns/1e9
out =
1706819594.68597
As mentioned elsewhere in this thread, converting to numeric will irretrievably lose information.
Haisam Khaled
Haisam Khaled 2024-2-3
Yes yes I understood this part, thank you. I was just explaining why I need these time stamps in a certain format (fast).

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更多回答（1 个）

Jon
Jon 2024-2-1
If you want to use it as a numerical value I would do this
val = str2double(zBase.header.stamp.sec) + str2double(zBase.header.stamp.nanosec)/1e9
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Jon
Jon 2024-2-1
编辑：Jon 2024-2-1
Good catch, I hadn't thought about the finite precision limitation, using doubles to represent large values (1e9) to nano second precision.
Haisam Khaled
Haisam Khaled 2024-2-2
Thank you so much, this is actually incredibly creative. I will use this method since the numbers are already "double".

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