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    I'm reading the implementation of the buddy algorithm.
    I was wondering what happens if someone tries to free a block of order k
    while one
    of the buddies of order k-1 which make this block is already free and
    the other isn't.
    From the implementation, it seems to me that this will lead to an
    incorrect behaviour but I'm not sure of my interpretation.

    So I would like to know :
    Is this allowed ? Or should one make sure not to free already free
    blocks ?

    Thanks in advance.

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