As Hannes Weisbach wrote: > I guess when you read the flash back, avrdude reads back all 8k, but > your hex file does not contain 8k of code.
Correct. Avrdude strips all trailing bytes that are 0xff (unprogrammed) when reading flash memory. (When reading EEPROM, they are kept.) > My second guess would be that the difference from 8k binary data in > the flash to the file size of 19204 bytes is caused by the Intel hex > file format. That would be a normal overhead: it must be more than twice the binary size, as each byte is encoded in two characters, and there are additional characters for the address, the record type, and the checksum. It makes most sense to read the memory contents back into a raw binary file (-U fl:r:foo.bin:r), and convert the original load file also into binary (avr-objcopy -O binary). The compare both binary files, e. g using cmp -l. -- cheers, J"org .-.-. --... ...-- -.. . DL8DTL http://www.sax.de/~joerg/ NIC: JW11-RIPE Never trust an operating system you don't have sources for. ;-) _______________________________________________ avrdude-dev mailing list avrdude-dev@nongnu.org https://lists.nongnu.org/mailman/listinfo/avrdude-dev
