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[CS61C FA20] Lecture 13.6 - Compilation, Assembly, Linking, Loading: Example — Transcript

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0:00and welcome back we're at our final
0:02lecture
0:03c-a-l-l we saw the whole process let's
0:06actually take an example
0:07let's take the most complicated program
0:09ever not let's just do
0:11hello world so this hello.c
0:14is not your normal hello world um i've
0:16got a main i'm ignoring my rx c arc v
0:19for now
0:20i've got a printf but my printf isn't
0:22just hello world my printf has
0:24two arguments it is hello string
0:28and then i have a percent s and then the
0:29world is going to be fed into that
0:31percent s
0:32so i really have two arguments to that
0:33print f
0:35two strings and one string has a percent
0:37s
0:38formatting directive and the other is
0:40the other string world and that's going
0:42to fit into that hello world to make
0:43hello world and i return 0. that's it
0:46so i say i'm done i save this i include
0:49my
0:49my library standard io so i know about
0:51printf and now i compile this
0:54what do i get this beautiful piece of
0:57code
0:58you could have hand authored you
1:01understand this but let's actually take
1:02a deep dive into
1:03what every single line does let's do
1:04this together
1:06so first i have a text segment a text
1:09segment says that i'm going to have
1:10the fo the following is going to be my
1:13code so
1:14text says that the directive for that
1:16dotted line 2
1:17says i need to align this code to 2 to
1:20the whatever that number is bytes so i'm
1:22saying this is going to be word aligned
1:23aligned to every 32 bits
1:27this is saying i've got a the following
1:29is a
1:30global symbol that main label that
1:33you're seeing here
1:34i'm declaring that to be a global symbol
1:35other people can know about
1:37okay i've got my main label
1:40as i have with most functions if you're
1:42going to make any function call from
1:44there and then return
1:45i've got to be able to make room for ra
1:48and maybe other arguments
1:49so here i am i'm going to reduce the
1:51stack pointer by
1:52four bytes to allocate some stuff there
1:56now first thing i always do is i store
1:58ra
1:59into sp12 the top guy the top of that
2:02frame that's my
2:03frame i'm going to store ra there so if
2:05i make a function call i'd be able to
2:06restore that and get back to where i was
2:08that's pretty cool now i've got
2:12two strings i'm going to reference if
2:13you remember this line let's go back
2:15i've got two strings here's two strings
2:16i need to be set up
2:18these two strings need to be set up as
2:20arguments to
2:21printf so how do i set those up well
2:24let's compute the address of string one
2:26here's string one down here
2:27here's string one down here here's
2:29string one dot here's string two here
2:32notice the the formatting string is in
2:34is in string one okay
2:36so i say louis add-i
2:39i get the high 20 bits put them in the
2:42upper
2:42immediate area and then i add i the low
2:4512 bits and i put that in there okay
2:47then i do the same thing here so this is
2:50in some sense
2:51this is going to be my load address
2:54of string 1 and string 2 into a0 and a1
2:56because remember
2:57i want to call printf printf has two
2:59arguments a0 and a1
3:02so here it is call printf i've set up a0
3:05and a1 i call printf
3:08now i now i just i'm done so now i
3:11restore this restore ra fix the stack
3:14pointer
3:16load into a0 my return value i put that
3:18zero in here and i say return
3:21it's pretty clean dot section r0 data
3:24says i'm going to be in a read-only data
3:26area from here on
3:27in okay so r0data says read are not zero
3:30o r o means read only data i'm saying
3:33this is going to be
3:35i'm gonna i'm aligning this to four
3:36bytes this is a little bit different
3:38than this align this says
3:39how many bytes i want to align the
3:40following and i align these based on
3:43a word aligned and now here below here
3:45is my data okay
3:46that's it now you're gonna say am i done
3:49well that's the
3:50that s file is there more work to do
3:52with the next stage well let's see what
3:53we've got
3:54let's see if the next stage is we
3:56assemble this down
3:58here's my main code okay starts at zero
4:01every every single guy starts at zero
4:03this is a file by itself so i start this
4:05at zero and then eventually
4:06the final resting place has a different
4:08value but i'm starting here from zero
4:10and now this is the this is assuming
4:12that this is like
4:13the first instruction this is four that
4:15means that's going to be this
4:16instruction okay so it's basically
4:18saying these are the addresses of what
4:19these guys are
4:20and this is just the file so in this
4:21file is just this
4:23okay just make sure you know this is all
4:24that contains in that file
4:26nothing else this is just for you to see
4:29that this is just for you to see what's
4:30happening
4:31and this is just for you to remember
4:32where things are but only this stuff
4:34here this is all that's stored in a
4:36dotto file okay does that make sense
4:38all right so remember we had to make
4:40room on the stack
4:41we to make room for four bytes on the
4:43stack i store
4:44our a now i've got my luau louis
4:48addie that's for my first string that's
4:50for my a0
4:52that's for string one here's my louis at
4:54i for string two i still got zeros here
4:56i don't know where those values are
4:58going to be so i'm just putting zeros
4:59i'm letting you know that this is what's
5:00going to be here okay
5:01and what's this one do you remember
5:02where this is right here what's this
5:04line do you guys remember what that next
5:05thing after i
5:05set up a0 and a1 what's my what's the
5:08next thing i do you're right
5:10i've got a jowl i gotta call printf
5:13this is the call to printf right here
5:15okay gotta remember what that is
5:17then i restore my stack sorry i
5:20load back ra i restore my stack i load
5:23in a0
5:24as my return value and i return okay so
5:26that's what main does
5:27and all i've got to do is figure out how
5:29to then those guys in red
5:30that's the job of the linker well done
5:35so what do i do i fill those in
5:39this is the value of string one boop
5:42this is the value not really the value
5:43but this is what has to be there in
5:45those values to be able to handle string
5:46two
5:47now a0 has some combination of this as
5:50the upper 20 and lower 12
5:51upper 20 lower 12 for string one and
5:54string two
5:54ie a0 and a1 this is where
5:58printf lives okay love that
6:02and well that's at least the uh that's
6:04at least the uh that 20 bits there for
6:06that
6:07and then that's all i need to do now
6:09you're going to say well that was great
6:11except that how did you calculate those
6:14values
6:15for this okay so the target's address
6:17for string
6:18one is 20 in hex
6:21in the upper 20 bits and a 10 in the
6:24lower 12 bits
6:25sounds good that doesn't seem crazy so
6:28here's my instruction sequence louie
6:31here's 20 here and add i the lower 12.
6:35seems reasonable there except it doesn't
6:38work
6:40why doesn't it work immediates
6:43in risk five are always sign extended
6:46we knew that but maybe we forgot about
6:48it a
6:50if you remember a is 10 10 has a 1
6:53in the upper bit so the highest the 12th
6:56bit
6:56the 11th bit by index but the 12th bit
6:59by counting from 1
7:00that's a one well when i sign extend
7:03it's going to be all ones
7:05so it looks like this i signed it to be
7:06all like this all
7:08ones see that all ones
7:12when i add this by the way what's all
7:13ones if i were to say like think about
7:14number two's complement what's all ones
7:17negative one so if i just look at the
7:19upper 20 bits
7:21and just add this value to this value
7:24just forget again i had only though
7:26those guys i don't this this is all
7:27zeros right so that's not going to
7:28affect the upper 20 bits okay a10
7:30doesn't mean the lower 20
7:31lower 12 bits so i'm fine with that but
7:33if i if i just consider those two
7:34numbers even though
7:35they're the upper 20 if i just consider
7:37them by themselves all f's
7:38which is really minus one and 20 what do
7:41you get
7:43you get one less than 20 which is one f
7:46so i'm off i'm off by exactly that one
7:49this is
7:50actually because this counts for a
7:51negative one i'm actually off by that
7:54one
7:54if that makes any sense so to fix
7:58it i take my 20 i add 1 to it
8:02to compensate for the fact that when i
8:03sign extended i'm going to lose the one
8:05and then when i add the a10 on the lower
8:0812 sign extend it remember you can only
8:10add 32 bits to 32 bits in the systems
8:12it's not like
8:12oh i only add the 12. now you gotta add
8:1432 to 32 to get 32. that's what our alus
8:16do
8:17our arithmetic logic units they only
8:18know how to add 32 to 32 to get 32.
8:20so then i'll get the right answer 28 10.
8:24so i've got to have this like
8:25compensating factor of this extra
8:28uh it's actually 2 to the 12. i got this
8:30extra 2 to the 12. that's up there
8:32to do that so what is this
8:35so by the way how do i figure out what
8:37this value is what do i figure out what
8:38this
8:38total value was this is going to be the
8:41immediate right the immediate that i put
8:42in there is going to be the a10
8:44well what is if i sign extended to all
8:46fs what does that become
8:47well if i have a number and it looks
8:49like it's negative right i have this
8:50number what
8:51is it let's figure out what it is all fs
8:53means the upper guy is a one if it's all
8:55upper guys and one
8:56two's complement is it means it's a
8:57negative number how do i figure what
8:58that value is
8:59invert the bits and add one and that's
9:01what the negative number was okay
9:03so let's take a look at this let's
9:04complement all fs and a10 well if i
9:07flip all those it becomes a zero flip a
9:10a
9:10is 10 10 remember 10 10 okay
9:1410 10 for a well the opposite of that is
9:17101. so there's a 5 there okay
9:21so that 101 becomes this 5.
9:24this one is one off of well flip that
9:27it's all ones except for the lowest guy
9:29that's an e
9:30so this guy is an e if it's a zero flip
9:33those bits it's an
9:34f 5ef is my value let's add that one now
9:37again
9:385ef remember to figure out what the
9:41two's complement is
9:42i flip it and add one so 5ef
9:45plus 1 is
9:495 5f0 0 is 15 20
9:52okay so that means this number
9:56was negative 15 20. okay
9:59so minus 15 20.
10:03so here we go instruction sequence is
10:06minus 15 20. add i and by the way
10:10let's do this here we go loop
10:14this calculates 28 10.
10:17so therefore i'll go back here so
10:19therefore
10:21what i have is this number
10:25is minus 15 20 which is what i want
10:28i want that number and i want this one
10:32to be 21. remember so the plus 1 goes up
10:36there i need to add 2 to the 12. that's
10:3821.
10:39the lower side is minus 15 20. and that
10:41calculates the address of string 1. so
10:43remember
10:44louis 21 in hex at i
10:47minus 15 20. and that's why whoops let
10:51me show you here
10:52that's why look look this is
10:55louis 21 at i
10:591520 huh okay
11:02pretty cool right in conclusion
11:07what does a compiler do it takes a
11:09high-level language
11:10here c converts it down to the assembler
11:13language
11:14which you can have different assembler
11:15languages if you changed a different isa
11:17it'd be a different assembly language
11:19risk five is different than mips okay so
11:21those are different assembly languages
11:23we're living in the risk five world now
11:25assembler removes the pseudo
11:27instructions sets up two tables
11:29a symbol table of all the symbols it
11:30knows about in that file and a
11:32relocation table
11:33of all the things that need to be fixed
11:35in the linker stage
11:36the linker goes to the symbols takes all
11:39of them in aggregate figures out are
11:40there any duplicate symbols error
11:42or any missing symbols error okay if i'm
11:45calling somebody but i have no symbols
11:46there error also once i have them
11:49once that's all covered i have to now
11:51place them all together concatenate all
11:52the text first
11:54all the text first all the data next and
11:57fixing those links i've got it now
11:59working it out out
12:00and sets whatever and makes use of
12:02whatever debugging that persists that
12:03continue there okay so whatever other
12:05header information in the edit out file
12:07done the loader takes that as we said
12:10before
12:11sets up the stack takes it and takes a
12:13disk disk file
12:14makes it make some memory for the system
12:16copies the text and data into that
12:18memory this is i'm summarizing this
12:19whole lecture in two in one minute
12:21copies all the all that text and data
12:23from edit out
12:24into that memory sets the stack pointer
12:26with any arguments on the command line
12:28sets up a0 and a1 to be the value of arc
12:31c and arc v
12:33and sets the stack pointer clears the
12:35registers and says
12:37go and watch his back happily as a as a
12:40clam as this thing continues to run and
12:42when it's all done
12:42whatever the return value was it passes
12:45it back through through the os
12:46to the top level and we're all done
12:50amazing that's the goal that's the high
12:52level picture
12:53and you now understand how compiling
12:56assembling linking and loading works in
12:58a risk five system
12:59congratulations give yourselves a hand
13:01high five
13:03well done good stuff we'll see you folks
13:05at the next module
13:06take care

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