Machine code: Difference between revisions

This example perform these: At Datamem(1) put 500, eax=5100, eax add Datamem(1), eax add 5, store eax to Datamem(0). We have an option to clear eax, or use it to return value as error code.

We have to leave all other registers, and stack as we found it.

 

 

EmbByte(0x31, 0xC0) ' xor eax, eax

Ret() ' Return

Return DataMem, 0:=a, 1:=b

Try ok {

}

}

Sub Ret()

Return ExecMem, Address:=0xC3