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Altera_Forum
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20 years ago

IORD_XDIRECT AND IOWR_XDIRECT?

Hello:

Now I have some problems about the I/O Marco IORD_XDIRECT AND IOWR_XDIRECT.For example:

I have a 16 bits slave periphery and the BEn0 ,BEn1 are both always low(means that the upper 8 bits and the lower 8 bits are both used).Make sure the front point no problem,now I use the sentense IOWR_8DIRECT (base,odd_addr,data8),the result I think is that the upper 8 bits is the data8,the lower 8 bits

is xxxxxxxx.If using even address,the result is opposit.But I am not sure it is right or wrong?

3 Replies

  • Altera_Forum's avatar
    Altera_Forum
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    20 years ago

    I look up "io.h" which define the "IORD/IOWR", and then "stb / stbio, lbu/lbuio....." in the Instruction Set Reference of NIOSII. I can't understand the specification about those instructions.

    Can someone help us to explain what they work out detailly?
  • Altera_Forum's avatar
    Altera_Forum
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    20 years ago

    Look at Table 3.5 of the Processor Reference Handbook

    <div class='quotetop'>QUOTE </div>

    --- Quote Start ---

    ldbio

    ldbuio

    stbio

    ldhio

    ldhuio

    sthio

    These operations load/store byte and half-word data from/to peripherals without caching or

    buffering.[/b]

    --- Quote End ---

  • Altera_Forum's avatar
    Altera_Forum
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    20 years ago

    Hi mountain8848,

    The stb and stbio are used to store a byte (from memory). The stbio instruction

    bypasses the data cache, the stb does not.

    The ldbu and ldbuio are used to load a byte (from memory). The lduio instruction

    bypassess the data cache, the ldbu does not. The "u" just means that the

    high-order bits in the target register are set to zero (unsigned) rather than

    being sign-extended.

    Regards,

    --Scott