To build the programs of this section, do the usual (as in the first program). To run the first program, type:
$ bin/reg1
$ echo $?
10
Again, the echo $? command shows the value returned by our program. In the
case of reg1 it should be 10.
Now, for the second program, we will use the technique from earlier. Assemble the program as usual:
$ nasm -felf64 tasks/registers/reg2.asm -o bin/reg2.o
But generate the executable from caller.c, linking with it:
$ gcc tasks/call_from_c/caller.c bin/reg2.o -o reg2
And execute:
$ bin/reg2
DEC: 65543
HEX: 10007
BIN: 0000000000000000000000000000000000000000000000010000000000000111
Notice that copying 7 (the 111 in the end of the binary number)
to AX did not clear the upper half of EAX (notice that 1 in the
17th bit). However, this behaviour does not hold for mov into
32-bit registers. If we compile and run reg3.asm (in the same way), we get:
$ bin/reg3
DEC: 3
HEX: 3
BIN: 0000000000000000000000000000000000000000000000000000000000000011
I am yet to check what happens to the 16-bit registers when we do a mov
into its 8-bit sub-registers.