Tomoichi Takahashi wrote:
> 2. Table 2. in a case of multi properties, encoding format seems
> followings :) In entrance property in building obejct KobeMap,
> "1" is set in the 3rd field. Am I right?
>
> \begin{table}[p]
> \center
> \begin{tabular}{|l|l|l|}
> \hline
> Field & Data type & Meaning\\
> \hline
> Header & int & Property type\\
> Length & int & Size of property data in bytes\\
> Value & int & Property value\\
> \hline
> \multicolumn{3}{l}{in a case of multi properties}\\
> \hline
> Field & Data type & Meaning\\
> \hline
> Header & int & Property type\\
> Length & int & Size of property data in bytes\\
> Value & int & Number of property values\\
> Value & int & Property value$_{1}$\\
> & : & \\
> Value & int & Property value$_{n}$\\
> \hline
> \end{tabular}
>
> \caption{Property encoding format}
> \label{table_property_format}
> \end{table}
>
Hi.
Yes, that is correct. For the entrance property the number of values is
*usually* 1, but it might be something else. Your code should be able to
handle both cases.
Pseudocode for reading both kinds of property type is as follows:
if (property is int) {
return readIntValue();
}
if (property is int array) {
int size = readIntValue();
int[] data = new int[size];
for (int i=0;i<size;++i) data[i] = readIntValue(i);
return data;
}
I hope that helps.
Cheers,
Cameron.
-- Cameron Skinner Artificial Intelligence Group Department of Computer Science The University of Auckland email: cam@cs.auckland.ac.nz phone: +64 9 3737599 x82924 fax: +64 9 3737453 Post: Department of Computer Science The University of Auckland Private Bag 92019 Auckland New Zealand _______________________________________________ robocup-rescue-s mailing list robocup-rescue-s@mailman.cc.gatech.edu https://mailman.cc.gatech.edu/mailman/listinfo/robocup-rescue-sReceived on Tue Apr 18 04:48:04 2006
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