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Home/ Questions/Q 6060587
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Editorial Team
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Editorial Team
Asked: May 23, 20262026-05-23T08:50:34+00:00 2026-05-23T08:50:34+00:00

I have the following c++ method: typedef unsigned long p3tOffsetType; p3tOffsetType buildString(std::string string) {

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I have the following c++ method:

typedef unsigned long p3tOffsetType;
p3tOffsetType buildString(std::string string)
{
    for (stringMap::const_iterator string_iterator = strings.begin(); string_iterator != strings.end(); ++string_iterator)
    {
        if (string_iterator->second == string)
            return string_iterator->first;
    }
    p3tOffsetType new_string_offset = string_offset;
    strings[string_offset] = string;
    string_offset += string.size() + 1;

    return new_string_offset;
}

What does the function do? I can give more of the code if needed.
The code is a snippet from a P3T file packer found in the source P3TBuilder (version 2.7).

I need to know this because I am trying to

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  1. Editorial Team
    Editorial Team
    2026-05-23T08:50:35+00:00Added an answer on May 23, 2026 at 8:50 am

    Assuming that strings is a map<p3tOffsetType, std::string> and that string_offset is initialized to zero, it seems to do the following: Imagine that you call the method a few times with, say, "Hello", "Hi", and "Hey", and that you would treat all these strings as C-strings and store them in the same char array. The array elements would then be {'H', 'e', 'l', 'l', 'o', '\0', 'H', 'i', '\0', 'H', 'e', 'y', '\0'}. The starting indices of the three strings are 0, 6, and 9, respectively. What the method does is to create a map that maps these starting indices to the strings, so strings[0] == "Hello", strings[6] == "Hi", and strings[9] == "Hey". Also, it eliminates duplicates, so calling the method again with "Hello" would leave the map unchanged.

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