The two use different algorithms to accumulate the sum (I have added comments to the relevant portion of the V6 assembler one, to help understand it): V6: mov $buf,r2 / Pointer to buffer in R2 2: movb (r2)+,r4 / Get new byte into R4 (sign extends!) add r4,r5 / Add to running sum adc r5 / If overflow, add carry into low end of sum sob r0,2b / If any bytes left, go around again Read the description of MOVB in the PDP-11 Processor manual. V7: while ((c = getc(f)) != EOF) { nbytes++; if (sum&01) sum = (sum>>1) + 0x8000; else sum >>= 1; sum += c; sum &= 0xFFFF; } I'm not clear on some of that, so I'll leave its exact workings as an exercise, but I'm pretty sure it's not a equivalent algorithm (as in, something that would produce the same results); it's certainly not identical. (The right shift is basically a rotate, so it's not a straight sum, it's more like the Fletcher checksum used by XNS, if anyone remembers that.) Among the parts I don't get, for instance, sum is declared as 'unsigned', presumably 16 bits, so the last line _should_ be a NOP!? Also, with 'c' being implicitly declared as an 'int', does the assignment sign extend? I have this vague memory that it does. And does the right shift if the low bit is one really do what the code seems to indicate it does? I have this bit that ASR on the PDP-11 copies the high bit, not shifts in a 0 (check the processor manual). That is, of course, assuming that the compiler implements the '>>' with an ASR, not a ROR followed by a clear of the high bit, or something. Noel