Converting ros::time to double value without uint_t

asked 2016-02-18 05:39:07 -0500

user23fj239
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I had a look at the ros::durationBase and it has functions toSec toNsec to convert the following into one int64_t which would exceed a double in c++;

int32_t     nsec
int32_t     sec

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answered 2016-02-18 06:24:41 -0500

mgruhler

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Both, ros::durationBase (API Doc) as well as ros::timeBase (API Doc) contain the respective toSec() functions which return a double. What you did is not required!

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So this is also ok? ( I could not get it to run with the function first)

double h_time = h.stamp.toSec();

user23fj239 ( 2016-02-18 08:35:37 -0500 )edit

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answered 2016-02-18 05:39:21 -0500

user23fj239
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So I did this:

    double h_time = double(h.stamp.sec) + double(h.stamp.nsec)*1e-9;

Basically the same what the function does: 00093 double toSec() const { return (double)sec + 1e-9*(double)nsec; };

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So what is exactly your question here?

gvdhoorn ( 2016-02-18 05:45:46 -0500 )edit

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Converting ros::time to double value without uint_t

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