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That wasn't the part that confused me. What confused me was what you expected to be encoded into the instruction. Your math indicated that you multiple n by 3 and added 1 to it to get your 13 bytes. So that means you intend to use 4 bytes to store an address in 32 bits which implied to me that you intended to have a fixed address encoded. But what if the address in in a register as would often

Whoops - sorry for the confusion. n would be set in stone beforehand. I basically meant to indicate that we'd either be looking at a 32 bit or 64 bit system, ie 4 byte or 8 byte addresses. On Tue, Mar 20, 2018, 1:07 PM Craig Topper <craig.topper at gmail.com> wrote: > Hi Gus, > > When you say "n byte destination" you mean you want to encode an n byte > address as