Better way to combine number with fraction?

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 ?

Answer 1

[Stephen23]

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.

[Haisam Khaled]

Thank you sooo much

[Jon]

I'm not understanding what you would do with this string. Are you only interested in displaying the value? In that case I understand why you want to obtain a string to represent it. But, if you need to do some numerical manipulation, then you will have to at some point turn it into a double. Alternatively could keep the seconds as a double and the nanoseconds as a double and then work with those if you need to maintain the nano second precision.

[Haisam Khaled]

Hello, so these are messages from a simulator delivered from a ROS2 node, they contain coordinates of the end efffector of a robot. One of the things I need to do is to plot these actual values against the desired value. I also need to check process time of trajectory following.

[Stephen23]

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]

Yes yes I understood this part, thank you. I was just explaining why I need these time stamps in a certain format (fast).

Answer 2

[Jon]

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

[Jon]

Even if you want to use it as a string later you could always convert the double back to a string using num2str

[Stephen23]

"Even if you want to use it as a string later you could always convert the double back to a string using num2str"

With the understanding that the interim DOUBLE will have lost information that cannot be retrieved:

fprintf('%.40f\n',1706819594.685974901)

1706819594.6859748363494873046875000000000000000000

[Jon]

Good catch, I hadn't thought about the finite precision limitation, using doubles to represent large values (1e9) to nano second precision.

[Haisam Khaled]

Thank you so much, this is actually incredibly creative. I will use this method since the numbers are already "double".