MBO L2 data : can I truncate keys?

duncan_kerr
duncan_kerr
in Elektron

Im looking at MBO L2 data and I can see that the "action" keys can be printed, and look like:

'3339 6563*******************************39ec'

'3330*************************30'

'1****************************1'

(where * are spaces and pad key to 50 chars wide)

however, when I look at the keys in Refinitiv app OMMViewer, you truncate the keys and only seem to use the right hand end of the keys.

My question is - can you explain how the keys work, and what the significance of the lft and right hand bit of the keys is? Can I make any assumptions with spaces in the gap in the middle of the key?

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Best Answer

  • Jirapongse
    Jirapongse
    Answer ✓

    @duncan_kerr

    Refer the RDM usage guide, the data type of the MBO key is buffer.

    image

    The data in the buffer is a byte array. When printing to a screen, the byte array can be represented as a hexadecimal string, such as '3339 6563'. Moreover, this data can also be represented as an ASCII string.

    For example, from the ASCII table, '3339 6563' represents '39ec' in an ASCII string.

    image

    Therefore, the left-hand side should be a hexadecimal string while the right-hand side is an ASCII string.

    Hexadecimal string => ASCII string

    '3339 6563' => '39ec'
    '3330'      => '30'

    It depends on what methods or libraries that the application uses to print the buffer. Some methods can display the buffer in this format.

    '3339 6563*******************************39ec'
    '3330*************************30'

    Then, OMM Viewer uses ASCII strings for display. This is my assumption.

    However, I am not sure about the last example.

    '1****************************1'

    What RICs are you using to retrieve that data?

Answers

  • duncan_kerr
    duncan_kerr

    OMM VIewer seems to truncate away some of the key:

    image

  • duncan_kerr
    duncan_kerr

    I see, the last example was from memory, and wrong. What I was thinking of, was:

    KEY ADD "31 1"

    If my understanding is correct I can extract the key from the ascii representation of the buffer by disgarding everything up to and including the last space in the string, so:


    31****1 => 1

    '3339 6563*******************************39ec' => 39ec

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