mov eax,[HexStr] What is placed in EAX here is the address given to the variable HexStr when the program is loaded into memory Hi, after trying to give a sense to this statement, I realized that this is wrong. What is moved in eax are the first 4 letters of HexStr . mov ebx,HexStr gives in ebx the address of HexStr , where the value now in eax is located. As I checked this also on Insight There many of this errors in Duntemann book, and it is the 3rd edition. This is a major problem, I think, when one is starting on a subject of this kind and has to confront with doubts created by such kind of misspelling. Thanks |