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Efficient translate from 0x12345678 to 0x87654321

liaoo hi

Dear all,
I met a problem due to big-endian of 8051. The phenomenon is as follows:

32-bit value A[4] is in XDATA space
Ex. offset 0 = 0x12; offset 1 = 0x34; offset 2 = 0x56; offset 3 = 0x78

I used the following to get this value:
*((UINT32 *) 0);

For 8051 we got "0x12345678" because of big-endian...

Is there any "efficient"(favor size) method to transform it to 0x87654321 ?

I tried to use: A[0]+A[1]<<8+A[2]<<16+A[3]<<24 but the "machine code" gets bigger...

Thanks !

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  • 0 in reply to liaoo hi

    As Per said, the format of the data sent to and received from the flash drive should be defined in the interface specification for the drive.

    it is then up to you to ensure that both your Windows driver writes according to that specification, and your 8051 "driver" reads according to that specification.

    clearly, one of them is broken!

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  • 0 in reply to liaoo hi

    As Per said, the format of the data sent to and received from the flash drive should be defined in the interface specification for the drive.

    it is then up to you to ensure that both your Windows driver writes according to that specification, and your 8051 "driver" reads according to that specification.

    clearly, one of them is broken!

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