CSSE2310 2021 SEM2 exam answers UQAttic.net
CSSE2310: 2021 SEM2 exam answers
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It’s the day of the CSSE2310 2022 Sem 1 exam so time to try the one from last sem lmao
CSSE2310 2021 SEM2 exam answers UQAttic.net
a) ls -a ~/etc
Etc is a directory in root so wouldn’t it just be /etc [+1]
Mv /tmp/a4data/*.spec ~/data [+1]
Mkdir exam
Gcc -lpthread -lm threadcalc.c server.c -o netserver (unsure)
Gcc -pthread -lm threadcalc.c server.c -o netserver (to link pthread it’s just -pthread, no “l”)
svn commit
Grep tcp /etc/services
Grep -v nameserver /etc/resolv.conf
Grep r=255 ~/colours | grep -v g=255 >> ./tmp/red
Grep -c “Mount Cootha” addresses
CSSE2310 2021 SEM2 exam answers UQAttic.net
Grep -c set /etc/vimrc >> ./vim/set.count
Ln -s /local/courses/csse2310/bin/testa3.sh a3test
Uint8_t foo[5]; (-1)
int8_t foo[5]; -> asks for signed integers [+1]
void (*foo)(int);
Pthread_t foo; [+1]
Float foo = 3.14;
float foo = 3.14f;
Int foo[2]; [+1]
Struct {
Char c;
Int i;
} foo;
CSSE2310 2021 SEM2 exam answers UQAttic.net
13,456 - SEGFAULT [+1]
20,000 - 183,840 [+1]
60,000 - UNKNOWN [+1]
206,495 - 23,107,231 [+1]
CSSE2310 2021 SEM2 exam answers UQAttic.net
2^43 Bytes = 2^13 GiB
Entries per page = 8KiB/8 = 1KiB = 1024 Bytes
3rd level size = 1024 * 8KiB = 8192KiB = 8MiB
2nd level size = 1024 * 8MiB = 8192MiB = 8GiB
Num 3rd level page = 1000MiB/8MiB = 125 pages
Total mem = (1 + 1 + 125) * 8KiB = 1016KiB
Num 3rd level page = 10,000MiB/8MiB = 1250 pages
Now need at least 2 2nd level pages to store references
Total mem = (1 + 2 + 1250) * 8KiB = 10024KiB
Page table size = (1 + 1 + 2 + 125 + 1125) * 8KiB = 10 032 KiB
Top Level Table:
2^20 entries * 8 bytes per entry = 2^23 bytes = 8 MiB = 8192 KiB
8192KiB + 125 * 8KiB
= 9192 KiB (+1)
2^30 entries * 8 bytes each
= 2^20 * 2^10 * 2^3 bytes
= 8192 MiB (+1)
Explanation please?
Exam consultation 2 covers a similar example but;
CSSE2310 2021 SEM2 exam answers UQAttic.net
- For 9f
- Consider the visual address, and how there are a total of 43 bits, with 13 being
used for offset. If only a single level exists, that means that there exists only one
block with a total of 30 bits assigned by the visual address to that one block.
- So, the one first level block contains 2^30 entries
- Each entry is 8 bytes, so 2^30 * 8 = 2^33 bytes = 2^13 MiB = 8192MiB
- For 9e
- From previous questions, we know that each level contains 10 bits for a 3-level
visual address in this example, so it would have been
| 10 bits | 10 bits | 10 bits | 13 bits offset |
With the change from 3 levels to two, the remaining bits after the 2nd level
become part of the first, so it becomes
| 20 bits | 10 bits | 13 bits offset |
- The size of the first level page is equal to the number of entries * entry size
= 2^20 * 8
And the size of the 2nd level entries are 8KiB each
- We also know the number of second level entries from before, which was 125
(just do address range (1,000 MiB) / page size (8 KiB)). Find size of second level
with num entries * page size = 125 * 8KiB
CSSE2310 2021 SEM2 exam answers UQAttic.net
6 (+1)
- I got 5 (nevermind, got 6 now…below diagram is incorrect at point G)
5 (+1)
CSSE2310 2021 SEM2 exam answers UQAttic.net
4 (+1)
17 (+1)
31 (+1)
x
CSSE2310 2021 SEM2 exam answers UQAttic.net
Network A:
Lowest address - 34.15.160.1 = 34.15.160.0000 0001
Highest address - 34.15.160.128 = 34.15.160.1000 0000
CIDR =
Netmask =
Broadcast =
Network B:
Lowest address - 34.15.160.1 = 34.15.160.0000 0001
Highest address - 34.15.160.128 = 34.15.160.1000 0000
2^4 - 3 = 13 (CIDR is /28?, so 32 - 28 = 4 bits for ports. 2^4 = 16. 3 already in use by the
network, so 16 - 3 = 13 total remaining addresses. Would be good to get some confirmation on
this)
/28 -> 16 addresses, 4 in use, 2 reserved, therefore 10 unused addresses.
(6 + (2^10 * 4) + (2^20 * 2)) * 32KiB = 67,240,128 KiB
Thought it would be
32KiB/8 bytes = 4096 pointers per block
(6 + (4096 * 4) + ((4096^2) * 2)) * 32KiB = 1,074,266,304 KiB[+4]
(6 + 4096 *4) * 32 = 524480 KiB [+1]
CSSE2310 2021 SEM2 exam answers UQAttic.net
4 -> 3 for block reads at file, 1 for single indirect at block 6
4 - don’t get the above explanation, but the memory starts in direct pointer 4, uses 5 and 6, and
then uses the first pointer in the first single indirect, for a total of 4 blocks.
Inode
Permissions
Owner
Timestamp
Block pointers
(12 total for this
question)
Meta Data
Size
Can store data in block pointer section of inode such that, 12 pointers * 8 bytes per pointer = 96
bytes total storage [+1]
Assumption: that there is a way to mark an iNode as containing binary data and not a pointer to
a block, as otherwise the kernel may attempt to use binary data as a block address/index.
^^ This was a guess unsure if it’s right
CSSE2310 2021 SEM2 exam answers UQAttic.net
Chmod g-r exam (unsure)
Chmod g-rx (I think get access to file inside a dic means interacting with the dic, which needs x
(I tried on moss: without x, I cannot cd into the dic [+1]
Chmod o+x resources (unsure)
Chmod o+r resource chmod g-r resource(tested on moss, [ls] is related to r
4? 6 minus link between itself and “.”? [+1]
0, link still exists with file results[+1]
Alice, bob ->permissions are given to all with r, x for user, group and other, so i’d assume
anyone can use this transaction
CSSE2310 2021 SEM2 exam answers UQAttic.net
alice, bob, dave, eve, fred ? anyone not part of course as far as i can tell as well as alice (the
user)
Alice, dave, eve fred
Bob, dave, fred, eve
Bob, carol, dave, eve
CSSE2310 2021 SEM2 exam answers UQAttic.net
Q10)
CSSE2310 2021 SEM2 exam answers UQAttic.net
1 #include
2 #include
3 #include
4 #include
5 #include
6 #include
7
8 int main(int argc, char* argv[]) {
9 int infile = open(argv[1], O_RDONLY);
10 int fd[2];
11 pipe(fd);
12 dup2(fd[1], STDOUT_FILENO);
13 dup2(infile, STDIN_FILENO);
14 pid_t pid;
15
16 if (pid = fork()) {
17 close(fd[1]);
18 } else {
19 // child
20 execlp(argv[3], argv[3], NULL);
21 }
22
23 int outfile = open(argv[2], O_WRONLY);n
24 dup2(fd[0], STDIN_FILENO);
25 dup2(outfile, STDOUT_FILENO);
26 execvp(argv[4], argv + 4);
27 return 0;
28 }
Seems to work but maybe not the best way…. I wonder if its ok to just exec /bin/bash lmao
MakeFile
CC=gcc
CFLAGS=-Wall -pedantic -std=gnu99
.PHONY:=clean
all: execpipe
#Generate executable by linking object files
CSSE2310 2021 SEM2 exam answers UQAttic.net
execpipe: execpipe.c
$(CC) $(CFLAGS) execpipe.c -o execpipe
clean:
rm -f *.o
rm -f execpipe