Lines Matching refs:byte
349 * Now, the NAND chip with 2K page(data chunk is 512byte) shows below:
524 * contains one byte for every ECC chunk, and is also padded to the
624 * contains one byte for every ECC chunk, and is also padded to the
637 * We need to compute the byte and bit offsets of
1387 * Get the byte from the data area that overlays the block mark. Since
1389 * physical block mark won't (in general) appear on a byte boundary in
1394 /* Get the byte from the OOB. */
1443 * ECC data are not byte aligned and we may have
1444 * in-band data in the first and last byte of
1556 * rely on the first byte of the auxiliary buffer to contain
1701 * 2) In read operations, the first byte of the OOB we return must reflect the
1736 * | into the first byte of | |
1772 /* Read the block mark into the first byte of the OOB buffer. */
1803 * byte boundaries.
1843 * guaranteed to be aligned on a byte boundary).
1858 /* Align last ECC block to align a byte boundary */
1888 * byte boundaries.
1933 /* Align last ECC block to align a byte boundary */
1953 * If required, swap the bad block marker and the first byte of the
2060 * and starts in the 12th byte of the page.
2160 loff_t byte;
2188 * Compute the chip, page and byte addresses for this block's
2193 byte = block << chip->phys_erase_shift;
2211 ret = chip->legacy.block_markbad(chip, byte);
2282 * (2) the size of the ECC parity is byte aligned.