Lines Matching defs:byte
300 * Now, the NAND chip with 2K page(data chunk is 512byte) shows below:
445 * contains one byte for every ECC chunk, and is also padded to the
458 * We need to compute the byte and bit offsets of
1183 * Get the byte from the data area that overlays the block mark. Since
1185 * physical block mark won't (in general) appear on a byte boundary in
1190 /* Get the byte from the OOB. */
1239 * ECC data are not byte aligned and we may have
1240 * in-band data in the first and last byte of
1351 * rely on the first byte of the auxiliary buffer to contain
1479 * 2) In read operations, the first byte of the OOB we return must reflect the
1514 * | into the first byte of | |
1550 /* Read the block mark into the first byte of the OOB buffer. */
1581 * byte boundaries.
1621 * guaranteed to be aligned on a byte boundary).
1636 /* Align last ECC block to align a byte boundary */
1666 * byte boundaries.
1711 /* Align last ECC block to align a byte boundary */
1731 * If required, swap the bad block marker and the first byte of the
1838 * and starts in the 12th byte of the page.
1938 loff_t byte;
1966 * Compute the chip, page and byte addresses for this block's
1971 byte = block << chip->phys_erase_shift;
1989 ret = chip->legacy.block_markbad(chip, byte);
2060 * (2) the size of the ECC parity is byte aligned.