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8/12/2019 De Thi LapTrinhMang Full Khong Loi Giai
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Nhm Lp trnh mng Khoa CNTT trng i hc Kinh doanh v Cng nghH Ni
cng n tp
Mn hc: Lp trnh mng
Cu 1: Trong cc pht biu vmng my tnh sau, pht biu no ng?a) Mng my tnh l cc my tnh c kt ni vi nhau.
b) Mng my tnh bao gm: cc my tnh, thit bmng m bo bic kt ni, phn
mm cho php thc hin vic giao tip gia cc my
c) Mng my tnh gm: cc my tnh, dy mng, vmng
d) Mng my tnh gm: cc my tnh, dy mng, vmng, hub
Cu 2: Di gc a l, mng my tnh c phn thnh nhng mng no?
Cu 3: Tiu ch no dng phn loi mng?
Cu 4: Nhng yu tno sau y lin quan ti vic lc chn kin trc mng?
a) Slng my tnh trong mng
b) Tc truyn thng trong mng
c) a im lp t v khnng ti chnh
d) CA, B, C
Cu 5: Cc c trng kthut ca mng my tnh?
A. ng truyn/Tc /Kin trc/Hiu hnh.
B. ng truyn/Kthut chuyn mch/Tc /Hiu hnh.
C. ng truyn/Kthut chuyn mch/Kin trc/Tc .D. ng truyn/Kthut chuyn mch/Kin trc/Hiu hnh.
Cu 6: Trong M hnh OSI c my lp
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Nhm Lp trnh mng Khoa CNTT trng i hc Kinh doanh v Cng nghH Ni
Cu 7: lp no cung cp cc phng tin ngi sdng c thtruy nhp c
vo mi trng OSI v cung cp cc dch vthng tin phn tn
Cu 8. Tng no trong m hnh OSI lm vic vi cc tn hiu in:
CU 9: Lp no (Layer) trong m hnh OSI chu trch nhim m ho (encryption)
dliu?
Cu 10: Tng no trong m hnh OSI c chc nng dng to ra nhng gi tin ?
Cu 11. Sphn on dng dliu xy ra ti tng no ca m hnh OSI ?
Cu 12. Tng no trong m hnh OSI c chc nng chuyn !i dliu ?
Cu 13 : Lp no trong m hnh OSI thc hin vic chn ng v chuyn tip
thng tin; thc hin kim sot lung dliu v ct/hp dliu?
Cu 14. Giao thc TCP/IP no c sdng ti tng Application ca m hnh OSI?
Cu 15. cp pht ng a chIP, ta c thsdng dch vc giao thc no:
Cu 16. Protocol l:
Cu 17. Lp no sau y chc sdng trong m hnh TCP/IP ?a. Application
b. Network
c. Transport
d. Internet
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Nhm Lp trnh mng Khoa CNTT trng i hc Kinh doanh v Cng nghH Ni
Cu 24. Loi mng no m tt ccc ti nguyn c tp trung ti cc Server v cc
ti nguyn c qun l chia s#mt cch c t!chc ?
Cu 25: Cc c trng kthut ca mng my tnh?A. ng truyn/Tc /Kin trc/Hiu hnh.
B. ng truyn/Kthut chuyn mch/Tc /Hiu hnh.
C. ng truyn/Kthut chuyn mch/Kin trc/Tc .
D. ng truyn/Kthut chuyn mch/Kin trc/Hiu hnh.
Rem cu 1A
HA(1) = Socket l g?
Rem cu 2A
HA(2) = TCP service l g?
Rem cu 3A
HA(3) = Socket API dng cchg?
Rem cu 4A
HA(4) = Trong cc bc gii thut thit kcho client, bc no sau y dnh ring
cho gii thut dng TCP?
TA(4,1) = Xc nh a chserver.
TA(4,2) = To socket.
TA(4,3) = Kt ni n serverTA(4,4) = Gi nhn dliu theo giao thc lp ng dng thit k.
Rem cu 5A
HA(5) = Lp no sau y sdng to server trong thvin java.net?
TA(5,1) = Socket
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Nhm Lp trnh mng Khoa CNTT trng i hc Kinh doanh v Cng nghH Ni
TA(5,2) = ServerSocket
TA(5,3) = SocketServer
TA(5,4) = DatagramSocket
Rem cu 1BHB(1) = Phng thc no sau y dng ly a chip ca my ch?
TB(1,1) = getHostAddress()
TB(1,2) = getHostName()
TB(1,3) = getLocalHost()
TB(1,4) = getLocalHostName()
Rem cu 2B
HB(2) = Phng thc no sau y dng ly tn ca my ch?
TB(2,1) = getHostAddress()
TB(2,2) = getHostName()
TB(2,3) = getLocalHost()
TB(2,4) = getLocalHostName()
Rem cu 3B
HB(3) = on chng trnh sau y lm cng vic g? & vbCrLf &_
try { & vbCrLf &_
InetAddress myHost = InetAddress.getLocalHost(); & vbCrLf &_
System.out.println(myHost.getHostAddress()); & vbCrLf &_
} catch (UnknownHostException e) { & vbCrLf &_
System.out.println(Khong tim thay dia chi host); & vbCrLf &_}
TB(3,1) = In ra a chmy ch.
TB(3,2) = In ra tn my ch.
TB(3,3) = Ly a chmy ch
TB(3,4) = Chng trnh thng bo l$i
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Nhm Lp trnh mng Khoa CNTT trng i hc Kinh doanh v Cng nghH Ni
Rem cu 4B
HB(4) = on chng trnh sau y lm cng vic g? & vbCrLf &_
try { & vbCrLf &_
System.out.println(myHost.getHostName()); & vbCrLf &_} catch (UnknownHostException e) { & vbCrLf &_
System.out.println(Khong tim thay dia chi host); & vbCrLf &_
}
TB(4,1) = In ra a chmy ch.
TB(4,2) = In ra tn my ch.
TB(4,3) = Ly a chmy ch
TB(4,4) = Chng trnh thng bo l$i
Rem cu 5B
HB(5) = Phng thc no trva chm socket kt ni n?
TB(5,1) = getInetAddress()
TB(5,2) = getInetLocalAddress()
TB(5,3) = getLocalAddress()
TB(5,4) = getSocketName()
Rem cu 6B
HB(6) = Phng thc no trva chsocket?
TB(6,1) = getInetAddress()
TB(6,2) = getInetLocalAddress()
TB(6,3) = getLocalAddress()TB(6,4) = getSocketName()
Rem cu 7B
HB(7) = Phng thc no trva c!ng socket kt ni n?
TB(7,1) = getPort()
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Nhm Lp trnh mng Khoa CNTT trng i hc Kinh doanh v Cng nghH Ni
3. Socket socket = new Socket(args[i],80); & vbCrLf &_
4. System.out.print("Connected to " + socket.getInetAddress()); & vbCrLf &_
5. System.out.print(" on port " + socket.getPort()); & vbCrLf &_
6. System.out.print(" from port " + socket.getLocalPort()); & vbCrLf &_
7. System.out.println(" of " + socket.getLocalAddress()); & vbCrLf &_8. } catch (UnknownHostException e) { & vbCrLf &_
9. e.printStackTrace(); & vbCrLf &_
10. } catch (SocketException e) { & vbCrLf &_
11. e.printStackTrace(); & vbCrLf &_
12. } catch (IOException e) { & vbCrLf &_
13. e.printStackTrace(); & vbCrLf &_
14. } & vbCrLf &_
15. }
TB(9,1) = 1,3
TB(9,2) = 2,8
TB(9,3) = 6,12
TB(9,4) = Khng sai dng no
Rem cu 10B
HB(10) = Chng trnh sau sai %dng no? & vbCrLf &_
1. for (int i = 0; i < args; i++) { & vbCrLf &_
2. try { & vbCrLf &_
3. Socket socket = new socket(args[i],80); & vbCrLf &_
4. System.out.print("Connected to " + socket.getInetAddress()); & vbCrLf &_
5. System.out.print(" on port " + socket.getPort()); & vbCrLf &_6. System.out.print(" from port " + socket.getLocalPort()); & vbCrLf &_
7. System.out.println(" of " + socket.getLocalAddress()); & vbCrLf &_
8. } catch (UnknownHostException e) { & vbCrLf &_
9. e.printStackTrace(); & vbCrLf &_
10. } catch (SocketException e) { & vbCrLf &_
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12. } & vbCrLf &_
13. } catch (IOException e) { & vbCrLf &_
14. e.printStackTrace(); & vbCrLf &_
}
TB(12,1) = Khi client kt ni n server trn c!ng 5000, server trvngy thnghin thi ca hthng.
TB(12,2) = Khi client kt ni n server trn c!ng 5000, server in ra thng tin ca
socket kt ni n.
TB(12,3) = Khi client kt ni n server trn c!ng 5000, server trvthng tin v
ngy vit chng trnh.
TB(12,4) = Khi client kt ni n server trn c!ng 5000, server yu cu username
v password.
Rem cu 13B
HB(12) = on chng trnh sau thc hin cng vic g? & vbCrLf &_
1. ServerSocket server; & vbCrLf &_
2. Socket connection; & vbCrLf &_
3. PrintStream p; & vbCrLf &_
4. try { & vbCrLf &_
5. server = new ServerSocket(5000); & vbCrLf &_
6. while (true) { & vbCrLf &_
7. connection = server.accept(); & vbCrLf &_
8. p = new PrintStream(connection.getOutputStream()); & vbCrLf &_
9. p.println(The program was written on 22th, Dec, 2012); & vbCrLf &_
10. connection.close(); & vbCrLf &_11. server.close(); & vbCrLf &_
12. } & vbCrLf &_
13. } catch (IOException e) { & vbCrLf &_
14. e.printStackTrace(); & vbCrLf &_
}
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Nhm Lp trnh mng Khoa CNTT trng i hc Kinh doanh v Cng nghH Ni
TB(13,1) = Khi client kt ni n server trn c!ng 5000, server trvngy thng
hin thi ca hthng.
TB(13,2) = Khi client kt ni n server trn c!ng 5000, server in ra thng tin ca
socket kt ni n.
TB(13,3) = Khi client kt ni n server trn c!ng 5000, server trvthng tin vngy vit chng trnh.
TB(13,4) = Khi client kt ni n server trn c!ng 5000, server yu cu username
v password.
Rem cu 1C
HC(1) = Cho on chng trnh sau, khi client kt ni vo c!ng 5000, nu g lnh
Hello, server c phn ng g? & vbCrLf &_
1. try { & vbCrLf &_
2. ServerSocket server = new ServerSocket(5000); & vbCrLf &_
3. Socket client = server.accept(); & vbCrLf &_
4. InputStream inStream = client.getInputStream(); & vbCrLf &_
5. OutputStream outStream = client.getOutputStream(); & vbCrLf &_
6. Scanner in = new Scanner(inStream); & vbCrLf &_
7. PrintWriter out = new PrintWriter(outStream,true); & vbCrLf &_
8. out.println("Hello! Enter \"exit\" to logout"); & vbCrLf &_
9. boolean done = false; & vbCrLf &_
10. while (!done && in.hasNextLine()) { & vbCrLf &_
11. String line = in.nextLine(); & vbCrLf &_
12. if (line.equals("browse web")) { & vbCrLf &_
13. out.println("Mo chuong trinh duyet web"); & vbCrLf &_14. } else if (line.equals("calculator")) & vbCrLf &_
15. out.println("Mo chuong trinh may tinh dien tu"); & vbCrLf &_
16. else { out.println("Echo " + line); } & vbCrLf &_
17. if (line.trim().equals("exit")) { done = true;} & vbCrLf &_
18. }client.close(); } catch (IOException e) {}
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TC(1,1) = Echo Hello
TC(1,2) = browse net
TC(1,3) = calculator
TC(1,4) = Hello
Rem cu 2C
HC(2) = Cho on chng trnh sau, khi client kt ni vo c!ng 5000, nu g lnh
browse net, server c phn ng g? & vbCrLf &_
1. try { & vbCrLf &_
2. ServerSocket server = new ServerSocket(5000); & vbCrLf &_
3. Socket client = server.accept(); & vbCrLf &_
4. InputStream inStream = client.getInputStream(); & vbCrLf &_
5. OutputStream outStream = client.getOutputStream(); & vbCrLf &_
6. Scanner in = new Scanner(inStream); & vbCrLf &_
7. PrintWriter out = new PrintWriter(outStream,true); & vbCrLf &_
8. out.println("Hello! Enter \"exit\" to logout"); & vbCrLf &_
9. boolean done = false; & vbCrLf &_
10. while (!done && in.hasNextLine()) { & vbCrLf &_
11. String line = in.nextLine(); & vbCrLf &_
12. if (line.equals("browse web")) { & vbCrLf &_
13. out.println("Mo chuong trinh duyet web"); & vbCrLf &_
14. } else if (line.equals("calculator")) & vbCrLf &_
15. out.println("Mo chuong trinh may tinh dien tu"); & vbCrLf &_
16. else { out.println("Echo " + line); } & vbCrLf &_
17. if (line.trim().equals("exit")) { done = true;} & vbCrLf &_18. }client.close(); } catch (IOException e) {}
TC(2,1) = Echo browse net
TC(2,2) = Mo chuong trinh duyet web
TC(2,3) = Mo chuong trinh may tinh dien tu
TC(2,4) = Thoat chuong trinh
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Rem cu 3C
HC(3) = Khi client kt ni vo c!ng 5000, chng trnh s&hin thci g? &
vbCrLf &_
1. try { & vbCrLf &_
2. ServerSocket server = new ServerSocket(5000); & vbCrLf &_3. Socket client = server.accept(); & vbCrLf &_
4. InputStream inStream = client.getInputStream(); & vbCrLf &_
5. OutputStream outStream = client.getOutputStream(); & vbCrLf &_
6. Scanner in = new Scanner(inStream); & vbCrLf &_
7. PrintWriter out = new PrintWriter(outStream,true); & vbCrLf &_
8. out.println("Hello! Enter \"exit\" to logout"); & vbCrLf &_
9. boolean done = false; & vbCrLf &_
10. while (!done && in.hasNextLine()) { & vbCrLf &_
11. String line = in.nextLine(); & vbCrLf &_
12. if (line.equals("browse web")) { & vbCrLf &_
13. out.println("Mo chuong trinh duyet web"); & vbCrLf &_
14. } else if (line.equals("calculator")) & vbCrLf &_
15. out.println("Mo chuong trinh may tinh dien tu"); & vbCrLf &_
16. else { out.println("Echo " + line); } & vbCrLf &_
17. if (line.trim().equals("exit")) { done = true;} & vbCrLf &_
18. }client.close(); } catch (IOException e) {}
TC(3,1) = Hello! Enter exit to logout
TC(3,2) = Mo chuong trinh duyet web
TC(3,3) = Mo chuong trinh may tinh dien tu
TC(3,4) = Ngt kt ni
Rem cu 4C
HC(4) = Khi client kt ni vo c!ng 5000, khi client gi chu$i calculator, server
phn ng nhthno? & vbCrLf &_
1. try { & vbCrLf &_
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7. PrintWriter out = new PrintWriter(outStream,true); & vbCrLf &_
8. out.println("Hello! Enter \"exit\" to logout"); & vbCrLf &_
9. boolean done = false; & vbCrLf &_
10. while (!done && in.hasNextLine()) { & vbCrLf &_
11. String line = in.nextLine(); & vbCrLf &_12. if (line.equals("browse web")) { & vbCrLf &_
13. out.println("Mo chuong trinh duyet web"); & vbCrLf &_
14. } else if (line.equals("calculator")) & vbCrLf &_
15. out.println("Mo chuong trinh may tinh dien tu"); & vbCrLf &_
16. else { out.println("Echo " + line); } & vbCrLf &_
17. if (line.trim().equals("exit")) { done = true;} & vbCrLf &_
18. }client.close(); } catch (IOException e) {}
TC(5,1) = 1, 4, 6
TC(5,2) = 2, 5, 10
TC(5,3) = 11, 14, 18
TC(5,4) = Khng sai dng no
1
Rem cu 6C
HC(6) = on chng trnh sau sai %dng no? & vbCrLf &_
1. try & vbCrLf &_
2. ServerSocket server = new ServerSocket(5000); & vbCrLf &_
3. Socket client = server.accept(); & vbCrLf &_
4. InputStream inStream = server.getInputStream(); & vbCrLf &_
5. OutputStream outStream = client.getOutputStream(); & vbCrLf &_6. Scanner in = new Scanner(outStream); & vbCrLf &_
7. PrintWriter out = new PrintWriter(outStream,true); & vbCrLf &_
8. out.println("Hello! Enter \"exit\" to logout"); & vbCrLf &_
9. boolean done = false; & vbCrLf &_
10. while (!done && in.hasNextLine()) { & vbCrLf &_
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11. String line = in.nextLine(); & vbCrLf &_
12. if (line.equals("browse web")) { & vbCrLf &_
13. out.println("Mo chuong trinh duyet web"); & vbCrLf &_
14. } else if (line.equals("calculator")) & vbCrLf &_
15. out.println("Mo chuong trinh may tinh dien tu"); & vbCrLf &_16. else { out.println("Echo " + line); } & vbCrLf &_
17. if (line.trim().equals("exit")) { done = true;} & vbCrLf &_
18. }client.close(); } catch (IOException e) {}
TC(6,1) = 1, 4, 6
TC(6,2) = 2, 5, 10
TC(6,3) = 11, 14, 18
TC(6,4) = Khng sai dng no
Rem cu 7C
HC(7) = on chng trnh sau sai %dng no? & vbCrLf &_
1. try { & vbCrLf &_
2. ServerSocket server = new ServerSocket(5000); & vbCrLf &_
3. Socket client = server.accept(); & vbCrLf &_
4. InputStream inStream = client.getInputStream(); & vbCrLf &_
5. OutputStream outStream = client.getOutputStream(); & vbCrLf &_
6. Scanner in = new Scanner(inStream); & vbCrLf &_
7. PrintWriter out = new PrintWriter(outStream,true); & vbCrLf &_
8. out.println("Hello! Enter \"exit\" to logout"); & vbCrLf &_
9. boolean done = false; & vbCrLf &_
10. while (!done && in.hasNextLine()) { & vbCrLf &_11. String line = out.nextLine(); & vbCrLf &_
12. if (line.equals("browse web")) { & vbCrLf &_
13. out.println("Mo chuong trinh duyet web"); & vbCrLf &_
14. } else if (line.equal("calculator")) & vbCrLf &_
15. out.println("Mo chuong trinh may tinh dien tu"); & vbCrLf &_
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16. else { out.println("Echo " + line); } & vbCrLf &_
17. if (line.trim().equals("exit")) { done = true;} & vbCrLf &_
18. }client.close() } catch (IOException e) {}
TC(7,1) = 1, 4, 6
TC(7,2) = 2, 5, 10TC(7,3) = 11, 14, 18
TC(7,4) = Khng sai dng no
Cu 1: Truyn dliu theo phng thc UDP l dng truyn dliu?
a. Khng chnh xcb. Chnh xcc. m bod. Tt cu ng
Cu 2: Truyn dliu theo phng thc UDP ?
e. Khng i h'i nhiu ti nguyn ca hthnga. i h'i nhiu ti nguyn ca hthngb. Chim dng nhiu ti nguyn !cngc. Khng p n no ng
Cu 3: Truyn dliu theo phng thc UDP c sdng trong ng dng?
f. Mang tnh cht thng boa. Mang tnh cht cnh bob. Mang tnh cht tm kim thng tinc. Mang tnh cht trao !i thng tin
Cu 4: UDP l vit tt ca?
Cu 5: UDP sdng cc lp l?
Cu 6: UDP sdng m hnh mng no?
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Nhm Lp trnh mng Khoa CNTT trng i hc Kinh doanh v Cng nghH Ni
Cu 7: DatagramSocket c dng truyn v nhn cc?
a. Socketb. ServerSocketc. DatagramPacketd. DatagramSocket
Cu 8: Chiu di dliu ti a c tha vo DatagramPacket l khong?
Cu 9: Trn cng mt my tnh, c!ng trong giao thc TCP v UDP c thtrng
nhau hay khng?
Cu 10: Khi nhn c mt DatagramPacket t"mt qu trnh khc gi n, ta c
thly thng tin a chmy g%i trn DatagramPacket ny b(ng phng thc?
a. Public InternetAddress getAddress()b. Public int getPort()c. Public byte[] getData()d. Public int getLength()
Cu 11: Khi nhn c mt DatagramPacket t"mt qu trnh khc gi n, ta c th
ly thng tin c!ng ca qu trnh g%i trn DatagramPacket ny b(ng phng thc?
a. Public InetAddress getAddress()b. Public int getPort()c. Public byte[] getData()d. Public int getLength()
Cu 12: Khi nhn c mt DatagramPacket t"mt qu trnh khc gi n, ta c thly thng tin dliu t"gi tin trn DatagramPacket ny b(ng phng thc?
a. Public InetAddress getAddress()b. Public int getPort()c. Public byte[] getData()d. Public int getLength()
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Cu 13: Khi nhn c mt DatagramPacket t"mt qu trnh khc gi n, ta c th
ly thng tin chiu di ca dliu trong gi tin trn DatagramPacket ny b(ng
phng thc?
a. Public InternetAddress getAddress()b. Public int getPort()c. Public byte[] getData()d. Public int getLength()
Cu 14: Trc khi g%i mt DatagramPacket i, ta c tht thng tin a chmy
nhn trn DatagramPacket ny b(ng phng thc ?
a. Public void setAddress(InternetAddress dis)b. Public void setPort(int port)c. Public void setData(byte buffer[])d. Public void setLength(int len)
Cu 15: Trc khi g%i mt DatagramPacket i, ta c tht c!ng qu trnh nhn trn
DatagramPacket ny b(ng phng thc ?
a.
Public void setAddress(InternetAddress dis)b. Public void setPort(int port)c. Public void setData(byte buffer[])d. Public void setLength(int len)
Cu 16: Trc khi g%i mt DatagramPacket i, ta c tht dliu g%i i trn
DatagramPacket ny b(ng phng thc ?
a. Public void setAddress(InternetAddress dis)b. Public void setPort(int port)c. Public void setData(byte buffer[])d. Public void setLength(int len)
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Nhm Lp trnh mng Khoa CNTT trng i hc Kinh doanh v Cng nghH Ni
Cu 17: Trc khi g%i mt DatagramPacket i, ta c tht chiu di dliu g%i
trn DatagramPacket ny b(ng phng thc ?
a. Public void setAddress(InternetAddress dis)b. Public void setPort(int port)c. Public void setData(byte buffer[])d. Public void setLength(int len)
Cu 18: Trong lp DatagramSocket phng thc kh%i dng to kt ni UDP l?
a. Public DatagramSocket() throws SocketExceptionb. Public DatagramSocket(int port) throws SocketExceptionc. Public void synchronized send(DatagramPackage p) throws IOExceptiond. Public void synchronized receive(DatagramPackage p) throws IOException
Cu 19: Trong lp DatagramSocket phng thc kh%i dng to kt ni UDP vi
shiu c!ng l?
a. Public DatagramSocket() throws SocketExceptionb. Public DatagramSocket(int port) throws SocketExceptionc. Public void synchronized send(DatagramPackage p) throws IOExceptiond. Public void synchronized receive(DatagramPackage p) throws IOException
Cu 20: Trong lp DatagramSocket phng thc gi gi dliu i l?
a. Public DatagramSocket() throws SocketExceptionb. Public DatagramSocket(int port) throws SocketExceptionc. Public void synchronized send(DatagramPackage p) throws IOExceptiond. Public void synchronized receive(DatagramPackage p) throws IOException
Cu 21: Trong lp DatagramSocket phng thc nhn gi dliu vl?
a. Public DatagramSocket() throws SocketExceptionb. Public DatagramSocket(int port) throws SocketException
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Nhm Lp trnh mng Khoa CNTT trng i hc Kinh doanh v Cng nghH Ni
c. Public void synchronized send(DatagramPackage p) throws IOExceptiond. Public void synchronized receive(DatagramPackage p) throws IOExceptionCu 22: Trong lp DatagramPackage, phng thc kh%i to ra gi tin c dliu
trong bnhm v chiu di gi dliu l?
a. Public DatagramPackage (byte buf[], int len)b. Public DatagramPackage (byte buf[], int len, InetAddress I addr, int port)c. Public Datagram Package (byte buf[], int len, I addr, port, number)d. Public DatagramPackage(byte buf[]).
Cu 23: Trong lp DatagramPackage, phng thc kh%i to ra gi tin c dliu
trong bnhm cng vi chiu di gi dliu a chmy ch v shiuc!ng?
a. Public DatagramPackage (byte buf[], int len)b. Public DatagramPackage (byte buf[], int len, InetAddress I addr, int port)c. Public Datagram Package (byte buf[], int len, I addr, port, number)d. Public DatagramPackage(byte buf[]).
Cu 24: Trong DatagramSocket phng thc dng ng kt ni l?
a. Public void synchronized close()b. Public void synchronized () closec. Public close()d. Public () close
Cu 25: to i tng kt ni Socket theo giao thc UDP ti c!ng 1234 tachn cch sau?
a. DatagramSocket serverSocket= new DatagramSocket(1234)b. Datagram serverSocket= new DatagramSocket(1234)c. Datagram serverSocket= DatagramSocket(1234)
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Nhm Lp trnh mng Khoa CNTT trng i hc Kinh doanh v Cng nghH Ni
d. DatagramSocket serverSocket=DatagramSocket(1234)
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Nhm Lp trnh mng Khoa CNTT trng i hc Kinh doanh v Cng nghH Ni
Cu 1: Phm vi truy cp ca mt i tng khi c khai bo public l g!"
a# C$ th% c truy cp t& t't c( m)i n*i#
b# C$ th% c truy cp t& cc l+p trong c,ng packag-#
c# C$ th% c truy cp t& cc l+p trong c,ng packag- v l+p con n.m trong packag-
khc#/# Ch0 c$ th% truy cp t& cc ph*ng thc khc trong cla22 $#
Cu 3: 4ava 5irtual 6achin- l g!"
Cu 7: 8hi bi9n /ch g;p lc-ption in thr-a/ main
?ava#lang#@oCla22A-Boun/=rror: myprogram# D
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Nhm Lp trnh mng Khoa CNTT trng i hc Kinh doanh v Cng nghH Ni
c# Y< tr ch*ng tr!nh chy tr9n iKn thoi /i ngZ my tTnh b[ tSi PA\ hay Palm#
/# Y< tr pht tri%n cc ch*ng tr!nhZ ph]n m^m nhSng#
Cu 11: Cu lKnh khai bo chu_n cho ph*ng thc main nh thI no"
a# public 2tatic int main`tring arg2 db# public 2tatic voi/ main`tringef a d
c# public 2tatic main`tringef arg2 d
/# public 2tatic Binal voi/ main`tringef arg2 d
Cu 13: D+p l g!"
Cu 17: % khai bo l+p -/ap1 kI th&a l+p -/ap ph(i lm nhthI no"
Cu 1F: il- cha m ngj "n ?ava 2au khi c bi9n /ch c$ uVi l g! "
Cu 1J: Cch ;t t9n no 2au y l 2ai"
a# 2tring
b# /3or/
c# 6ain/# 3or/
Cu 1M: @gVn ng lp tr!nh 4ava cung c'p cc tTnh nng no 2au y"
a# Gh viKn tTch hp
b# iao /iKn lp tr!nh ng /Lng
c# q cVng cL giao /iKn ngXi /,ng
/# G't c( p n tr9n
Cu 1N: Phm vi truy cp ca mt i tng khi c khai bao prot-ct-/ l g!"
a# C$ th% c truy cp t& cc l+p trong c,ng packag- v l+p con n.m trong packag-
khc#
b# C$ th% c truy cp t& cc l+p trong c,ng packag-#
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Nhm Lp trnh mng Khoa CNTT trng i hc Kinh doanh v Cng nghH Ni
c# C$ th% c truy cp t& t't c( m)i n*i#
/# Ch0 c$ th% truy cp t& cc ph*ng thc khc trong cla22 $#
Cu 1O: 4ava chy tr9n hK i^u hnh no 2au y:
Cu 1: Cch ;t t9n no 2au y l khVng chTnh >c"
a# Binal
b# /-m
c# Binal
/# Binal
Cu 3U: % chuy%n si gi tr ca mt biIn bool-anZ ta /,ng ton t no"a#
b#
c# w
/#
Cu 31: Ph*ng thc ha2@->t` trong l+p cann-r c$ tc /Lng"
Cu 33: DKnh iB Cx@AGx@ 8hi lKnh d c$ tc /Lng g!"
Cu 37: DKnh khai bo
cann-r in z n- cann-r`y2t-m#in{
c$ tc /Lng g!"
Cu 3F: 8i%u / liKu byt- c$ phm vi trong kho(ng:
Cu 3J: Cho cla22 nh 2au:
cla22 8i-mGra
privat- int i{
privat- int n->t/{
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Nhm Lp trnh mng Khoa CNTT trng i hc Kinh doanh v Cng nghH Ni
public voi/ 2-t/`
i/ z n->t/{
n->t/||{
d
d@Iu kh}i to 3 i tng cla22 \ ->t-n/2 8i-mGra v cla22 q ->t-n/2 8i-mGraZ i/ ca
i tng q c$ gi tr l "
Cu 3M: Ph*ng thc n->tnt` trong l+p cann-r c$ tc /Lng"
Cu 3N: Ph*ng thc ha2@->tDin-` trong l+p cann-r c$ tc /Lng"
Cu 3O: DKnh char\t`n c$ tc /Lng g!"
Cu 3: Cho cla22 nh 2au:
cla22 8i-mGra
privat- int i{
privat- 2tatic int n->t/{
public voi/ 2-t/` i/ z n->t/{
n->t/||{
d
d
@Iu kh}i to 3 i tng cla22 \ ->t-n/2 8i-mGra v cla22 q ->t-n/2 8i-mGraZ i/ ca
i tng q c$ gi tr l "
Cu 7U: DKnh 2au tr( v^ kIt ~u( l bao nhi9u"
tring gr--ting2 z Y-llo{
char l-tt-r z gr--ting2#char\t`U{
Cu 71: DKnh r-2ult z con/ition " valu-1 : valu-3 c$ nghEa g!"
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Nhm Lp trnh mng Khoa CNTT trng i hc Kinh doanh v Cng nghH Ni
Cu F7: Cho on ch*ng tr!nh 2au:
public cla22 v/333
public 2tatic voi/ main`tringef arg2
/oubl- azNZbzF{
int >z`ab"a:b{y2t-m#out#print`@ghi-m la: | >{
d
d
8It ~u( khi chy ch*ng tr!nh l:
Cu FF: Cho on ch*ng tr!nh 2au:
public cla22 v/333 public 2tatic voi/ main`tringef arg2
/oubl- azNZbzF{
/oubl- >z`ab"a:b{
y2t-m#out#print`@ghi-m la: | >{
d
d
8It ~u( khi chy ch*ng tr!nh l:
Cu FJ: 6(ng l g!"
Cu FM: qi%u thc no 2au y khVng hp lK:
a# int >zM{iB`w`>F d
b# int >zM{ >zw>{
c# int >zM{>z>J{/# int >zM{>Rz>{
Cu FN: 8hai bo 2au c$ W nghEa g!:
public 2tatic \rrayDi2t=mploy--2 ob?= z n- \rrayDi2t=mploy--2`{
a# 8hai bo mt m(ng cc i tng =mploy--2 thuc l+p ob?=
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Nhm Lp trnh mng Khoa CNTT trng i hc Kinh doanh v Cng nghH Ni
b# 8hai bo mt m(ng cc i tng ob?= thuc l+p =mploy--2
c# 8hai bo i tng ob?= thuc l+p =mploy--2
/# 8hai bo i tng =mploy--2 thuc l+p ob?=
Cu FO: Cho on ch*ng tr!nh 2au:public cla22 v/111
public 2tatic voi/ main`tringef arg2
/oubl- azNZbzJ{
/oubl- >{
>zaQb{
y2t-m#out#print`@ghi-m la: | >{
dd
8It ~u( khi chy ch*ng tr!nh l:
Cu F: i( 2 c$ khai bo 2au:
public 2tatic \rrayDi2t=mploy--2 ob?= z n- \rrayDi2t=mploy--2`{
% th9m mt i tng vo m(ng cc i tng khai bo } tr9n ta /,ng lKnh:
a# ob?=#\//`ob?b# ob?=#a//`ob?
c# ob?=#a//n-`ob?
/# ob?=#\//@-`ob?
Cu JU: Cho on ch*ng tr!nh 2au:
public cla22 v/333
public 2tatic voi/ main`tringef arg2 int azNZbzF{
int >z`ab"a:b{
y2t-m#out#print`@ghi-m la: | >{
d
d
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Nhm Lp trnh mng Khoa CNTT trng i hc Kinh doanh v Cng nghH Ni
8It ~u( khi chy ch*ng tr!nh l:
a# @ghiKm l: N#U
b# @ghiKm l: N
c# @ghiKm l: F#U
/# @ghiKm l: F
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Nhm Lp trnh mng Khoa CNTT trng i hc Kinh doanh v Cng nghH Ni
Cu 26: Multicast l g?
a. L m hnh truyn thng point-to-point m chng ta thng gi l
multicasting.
b. L m hnh truyn thng point-to-point tt cho nhiu ng dng hot ng
ng thi.c. L m hnh truyn thng one-to-all pht sng cho mt nhm a ch ch
khng phi mt a ch.
d. L m hnh truyn thng one-to-many s dng chin lc phn phi cc
thng ip chmt ln.
Cu 27: C mt snhm multicast c bit gi l well-known multicast. Trong
nhm 224.0.0.1 l nhm g?
a. L nhm all-systems.mcast.net, tt ccc host c kh nng multicast trn
mng phi tham gia nhm.
b. L nhm all-routers.mcast.net, tt c cc multicast router c kh nng
multicast phi tham gia nhm.
c. L nhm mobile-agents.mscast.net, tt c cc i l di ng u c giao
din c khnng multicast.
d. L nhm ntp.mcast.net, tt ccc giao thc thi gian mng.
Cu 28: C mt snhm multicast c bit gi l well-known multicast. Trong
nhm 224.0.0.2 l nhm g?
a. L nhm all-hosts, tt c cc host c kh nng multicast trn mng phi
tham gia nhm.
b. L nhm all-router, tt ccc multicast router c khnng multicast phitham gia nhm.
c. L nhm all-router DVMRP, tt ccc router u c giao din c khnng
multicast.
d. L nhm all-router OSPF, tt ccc router u c giao din c khnng
multicast
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Nhm Lp trnh mng Khoa CNTT trng i hc Kinh doanh v Cng nghH Ni
Cu 29: C mt snhm multicast c bit gi l well-known multicast. Trong
nhm 224.0.0.4 l nhm g?
a. L nhm all-hosts, tt c cc host c kh nng multicast trn mng phi
tham gia nhm.b. L nhm all-router, tt ccc multicast router c khnng multicast phi
tham gia nhm.
c. L nhm all-router DVMRP, tt ccc router u c giao din c khnng
multicast.
d. L nhm all-router OSPF, tt ccc router u c giao din c khnng
multicast
Cu 30: C mt snhm multicast c bit gi l well-known multicast. Trong
nhm 224.0.0.5 l nhm g?
a. L nhm all-hosts, tt c cc host c kh nng multicast trn mng phi
tham gia nhm.
b. L nhm all-router, tt ccc multicast router c khnng multicast phi
tham gia nhm.
c. L nhm all-router DVMRP, tt ccc router u c giao din c khnng
multicast.
d. L nhm all-router OSPF, tt ccc router u c giao din c khnng
multicast
Cu 31: C mt snhm multicast c bit gi l well-known multicast. Trong
nhm 224.0.0.13 l nhm g?a. L nhm all-router PIM, tt c cc router u c giao din c kh nng
multicast.
b. L nhm all-router, tt ccc multicast router c khnng multicast phi
tham gia nhm.
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Nhm Lp trnh mng Khoa CNTT trng i hc Kinh doanh v Cng nghH Ni
c. L nhm all-router DVMRP, tt ccc router u c giao din c khnng
multicast.
d. L nhm all-router OSPF, tt ccc router u c giao din c khnng
multicast.
Cu 32: Vic chon mt gi trtime-to-live (TTL) tt trn cc gi tin multicast l kh
xc nh. Gi trhp lca TTL n(m trong khong no?
a. 0 n 255.
b. 0 n 16.
c. 0 n 127.
d. Gi trty .
Cu 33: Lp java.net.MulticastSocket c sdng nhthno?
a. Sdng bn pha client lng nghe cc gi tin m server pht sng cho
nhiu client cng lc.
b. Sdng bn pha client gi cc gi tin m nhn c t"nhiu server
cng lc.
c. Sdng bn pha client pht sng cc gi tin cho nhiu client trong cng
mt mng.
d. Khng phi cc trn.
Cu 34: Multicast l g?
a. L truyn thng point-to-point.
b. L truyn thng cho tt ccc thnh vin c kt ni.
c. L truyn thng n mt nhm c chn ca cc thnh vin kt ni.d. Khng phi cc trn.
Cu 35: a chMulticast l di a chno?
a. 224.0.0.0 n 239.255.255.255
b. 225.0.0.0 n 238.255.255.255
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Nhm Lp trnh mng Khoa CNTT trng i hc Kinh doanh v Cng nghH Ni
c. 224.0.0.0 n 224.255.255.255
d. 223.0.0.0 n 224.255.255.255
Cu 36: Mt nhm multicast l mt tp hp cc host cng chia s# mt a ch
multicast. to mt nhm multicast ta lm nhthno?a. Chn mt a chng)u nhin t"225.0.0.0 n 238.255.255.255
b. Chn a chhthng trn mng con a phng l 224.0.0.1
c. Chn a chrouter trn mng con a phng l 224.0.0.2
d. Chn a chpht sng trn mng c dng l 224.2.X.X
Cu 37: Vic gi dliu multicast gm cc c im no?
a. Truyn thng l khng cn kt ni trc (1).
b. Cn phi to ra mt MulticastSocket (2).
c. Cn phi thit lp gi trtime-to-live ng cho datagram (3).
d. L cc UDP datagram gi cho mt a chmulticast (4).
e. (1), (2) v (3).
f. (2), (3) v (4).
g. (2) v (4).
Cu 38: Lp java.net.MulticastSocket l lp con ca lp no?
a. DatagramSocket.
b. SocketException.
c. Socket.
d. MulticastObject.
Cu 39: Mt Multicast API khng h$trhot ng no?
a. Cho php mt quy trnh tham gia vo mt nhm multicast cth.
b. Cho php quy trnh ri i nhm multicast.
c. Cho php mt quy trnh gi thng ip n tt c cc quy trnh ca mt
nhm multicast.
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Nhm Lp trnh mng Khoa CNTT trng i hc Kinh doanh v Cng nghH Ni
d. Cho php mt quy trnh nhn thng ip gi n t"mt nhm multicast.
e. Khng phi cc trn.
Cu 40: C 4 lp chnh trong Multicast API l cc lp no?
a. InetAddress (1).b. DatagramPacket (2).
c. DatagramSocket (3).
d. MulticastSocket (4).
e. SocketAddress (5).
f. (1), (2), (3) v (4).
g. (2), (3), (4) v (5).
h. (1), (3), (4) v (5)
Cu 41: S khc nhau c bn gia truyn thng multicast v truyn thng
broadcast l g?
a. Broadcast l gi cho mi nt mng c ng k nhm v/hoc mng
con.
b. Multicast l gi cho nhiu mng c nh ngha vi multicast.
c. Multicast l gi cho tt cthnh vin ca nhm multicast.
d. Multicast v broadcast vcbn l ging nhau.
e. Khng phi cc trn.
Cu 42: Pht biu no ng vmulticasting?
a. Multicasting cho php phn phi mt datagram n tt c cc host trong
mng ni b.b. Cc router gia cc mng cn cho php cc multicast datagram i qua.
c. Multicast khng gi c n mt a chc bit, gi l mt nhm.
d. Nhm Multicast khng cn phi ng k trc v cng bt ngun t"mt
mng/host a chcth.
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Nhm Lp trnh mng Khoa CNTT trng i hc Kinh doanh v Cng nghH Ni
Cu 43: Trong giao thc IGMPv1, mt nt mng c phi chmt khong thi gian
truy vn xc nhn thnh vin trc khi gia nhp mt nhm multicast hay khng?
a. C, cc router ang kt ni vi mng LAN nh k*thm d yu cu cc nt
mng thnh vin mun tham gia nhm.
b. C, cc nt multicast nh k*gi cc cu lnh tham gia multicast (m$i 60giy) trn LAN. tham gia vo mt nhm multicast, bm thi gian phi c
tt trc khi pht sng/ qung b mong mun tham gia vo mt nhm multicast.
c. Khng, khi mt nt mng mun tham gia vo mt nhm multicast, nt gi
mt yu cu lm thnh vin ca host n a chnhm all-router.
d. Khng, khi mt nt mng mun tham gia vo mt nhm multicast, nt thit
lp nhm a chIP m router cc bs&sdng. Cc nt tip theo yu cu lm thnh
vin nhm multicast t"router.
Cu 44: IP multicasting c nh ngh+a l g?
a. IP multicasting l mt lot cc giao thc thng c s dng vi cc
broadcast cbn gi i cc thng ip n ngn (1).
b. IP multicasting l mt lot cc giao thc cho php la chn chuyn tip lu
lng da trn lp 2 v lp 3 ca mng chn nhm cc nt mng (2).
c. IP multicasting vcbn ging nhIP broadcasting nhng vi mt schc
nng b!sung ca lp 3 ca mng (3).
d. IP multicasting l mt cng nghv hng lot cc giao thc cho php cc
broadcast la chn xuyn qua lp 3 nh tuyn cc bin mng theo cc thng s
multicast cth(4).
e. (1), (2) v (4).
f. (1), (3) v (4).g. (1), (2) v (3).
Cu 45: Giao thc no sdng a chIP l 224.0.0.5 v 224.0.0.6 cho truyn thng
multicast?
a. EIGRP.
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Nhm Lp trnh mng Khoa CNTT trng i hc Kinh doanh v Cng nghH Ni
b. OSPF.
c. PIM.
d. Khng phi cc trn
Cu 46: Sdng Internet Group Management Protocol version 2 (IGMPv2), mt ntmng lm thno thot ra hoc ng"ng tham gia vo mt nhm multicast?
a. N phi tthot ra v khng c cu lnh thot ra nhm.
b. Nt phi a ra mt cu lnh nghcth.
c. Nt khng kim sot c vic tham gia nhm multicast; router hon ton
qun l nhm multicast.
d. Cnt v router phi c cu hnh vi quyn admin khng tham gia
vo mt nhm multicast cth.
Cu 47: Sdng Internet Group Management Protocol version 1 (IGMPv1), mt nt
mng lm thno thot ra hoc ng"ng tham gia vo mt nhm multicast?
a. N phi tthot ra v khng c cu lnh thot ra nhm.
b. Nt phi a ra mt cu lnh nghcth.
c. Nt khng kim sot c vic tham gia nhm multicast; router hon ton
qun l nhm multicast.
d. Cnt v router phi c cu hnh hnh vi quyn admin khng tham
gia vo mt nhm multicast cth.
Cu 48: Giao thc multicast thc hin trong lp no ca m hnh OSI?
a. Lp 3 v lp 4.
b. Lp 2 v lp 3.c. Lp 4 v lp 7.
d. Lp 3 v lp 7.
e. Tt ccc lp tr"lp 5.
f. Khng phi cc trn.
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Nhm Lp trnh mng Khoa CNTT trng i hc Kinh doanh v Cng nghH Ni
Cu 49: Hnh ng no phi c thc hin b%i mt my chnu mun tham gia
vo mt nhm multicast?
a. Gi bo co thnh vin IGMPv2 sdng unicast n router mc nh trn
mng con a phng.
b. Gi bo co thnh vin IGMPv2 sdng unicast n im h,n cho nhm.c. Gi bo co thnh vin IGMPv2 s dng multicast n nhm multicast
"All-PIM-Router" (224.0.0.2), trn mng con a phng.
d. Gi bo co thnh vin IGMPv2 s dng multicast trn mng con a
phng vi a chIP ch thit lp c tham gia nhm multicast.
Cu 50: a ch IP no dnh ring c sdng b%i Internet Group Management
Protocol (IGMP) khm ph v/ hoc cung cp dch vkeep-alive cho nt mng
tham gia multicast?
a. 224.0.0.1
b. 239.0.0.1
c. 224.0.1.0
d. 239.0.0.255
e. 224.0.1.255
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Nhm Lp trnh mng Khoa CNTT trng i hc Kinh doanh v Cng nghH Ni
Cu 2.Mc ch chnh ca cc cu lnh import l.
Cu 3.Kt quthc hin ca on chng trnh sau:
public class MyClass
{
public static void main(String args[]) {
System.out.println("In first main()"); }
public static void main(char args[]) {
System.out.println('a'); }
}
Cu 4.Cho on m: int a = 9/0. Kt quca on m l g?
Cu 5.Lp DatagramPackage trong lp trnh mng c sdng vi mc ch g ?
Cu 6.Thvin cn thit c dung lp trnh mng?
Cu 7.Phng thc getFile() trong lp URL c ngh+a g ?
Cu 9.Phng thc receive( DatagramPackage p) trong lp DatagramSocket c s
dng vi mc ch g ?
Cu 10.Phng thc accept ca lp ServerSocket c thgy ra ngoi lg
Cu 11.Chn cu trli chnh xc nht trong cc cu trli sau:
A.Mt lp c thkth"a nhiu lp cha b(ng t"kha extends
B.Mt lp c thkth"a nhiu lp tr"u tng b(ng t"kha extends
C.Mt lp chth"a kmt giao din v mt lp tr"u tng
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Nhm Lp trnh mng Khoa CNTT trng i hc Kinh doanh v Cng nghH Ni
D.Mt lp c thkth"a mt hoc nhiu giao din b(ng t"kha implements
Cu 13.Constructor l g?
A.Hm c tn trng vi tn lp, kiu dliu ca hm l voidB.Hm c tn trng tn lp v c du ng '~' trc tn hm.
C.Cc hm c tn trng vi tn lp v khng c tham s.
D.Cc hm c tn trng vi tn lp v khng c kiu dliu ca hm.
Cu 14.u l cch khai bo to ra chu$i k t
A.Chai phng n u ng
B.Chai phng n u sai
C.String s = new String("Test")
D.String s;
Cu 15.Phng thc getHostAddress() trong lp InetAddress c sdng vi mc
ch g ?
Cu 16.Cc my tnh trn Internet giao tip vi nhau thng qua giao thc g ?
Cu 17.Phng thc getInetAddress() trong lp Socket c dng lm g ?
A.ly vsc!ng c kt ni n my ch
B.ly vlung xut my khch c thgi dliu n my ch
C.ly vlung nhp my khch c thc dliu gi trvt"pha my chD.ly a chca my chc kt ni n di dng kiu InetAddress
Cu 18.Phng thc replaceFirst(char old, char new) c dng lm g ?
A.Thay thk told u tin b(ng k tnew trong mt chu$i no
B.Thay thk told cui cng b(ng k tnew trong mt chu$i no
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Nhm Lp trnh mng Khoa CNTT trng i hc Kinh doanh v Cng nghH Ni
C.Thay thtt ck tnew b(ng k told trong mt chu$i no
D.Thay thtt ccc k told b(ng k tnew trong mt chu$i no
Cu 19.Lp no sau y khng sdng trong ng dng giao tip mng Socket s
dng giao thc UDP ?
Cu 20. in phn cn thiu cho c php hm sau DatagramPacket( byte buf[],
. len, InetAddress host, int port) trong lp DatagramPackage
Cu 21.Khi chy ng dng theo m hunh Client/Server, th chng trnh no phi
chy trc
Cu 22.in phn cn thiu cho c php hm sau DatagramPacket( buf[], int
len, InetAddress host, int port) trong lp DatagramPackage
Cu 23. Lp DatagramSocket trong lp trnh mng c sdng vi mc ch g
Cu 24. Phng thc DatagramPacket( byte buf[], int len, InetAddress host, int port)
trong lp DatagramPackage c sdng vi mc ch g ?
Cu 25.Trong lp trnh Socket b(ng giao thc UDP/IP, chng trnh client phi to
i tng Socket no giao tip vi chng trnh server?
Cu 26. Cho on chng trnh bn pha Server, chng trnh giao tip qua mng
dng cchTCP. Chn cu lnh ng nht in vo cu lnh cn thiu?
ServerSocket ssk = new ServerSocket( 1234);
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Nhm Lp trnh mng Khoa CNTT trng i hc Kinh doanh v Cng nghH Ni
Cu 30.Chng trnh c vit b(ng ngn ngJava c thchy trn bt k*nn no
b%i v
A.My o JVM c nhiu nn tng h$tr
B.Lp trnh Java c kth"a t"C++
C.Hm API thc hin cc thao tcD.Trnh bin dch c xc nh b%i trnh bin dch ca C++
Cu 31.Gi trmu no l sai trong cc mu lit k sau khi dng thay !i mu v&
trn Applet
A.Color.red
B.Color.black
C.Color.blue
D.Color.Red
Cu 32.Mt a chURL nh dng mt file ti nguyn gm my phn:
A.4 phn
B.5 phn
C.3 phn
D.2 phn
Cu 33.Phng thc getPort() trong lp URL c ngh+a g ?
A.Trvtn ca my chtrong mt chu$i a chURLB.Trvtn ca file trong mt chu$i a chURL
C.Trvtn ca giao thc trong mt chu$i a chURL
D.Trvshiu c!ng sdng trong mt chu$i +a chURL
Cu 34.Phng thc getData() trong lp DatagramPacket c sdng vi mc
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Nhm Lp trnh mng Khoa CNTT trng i hc Kinh doanh v Cng nghH Ni
ch g ?
A.Trvkch thc ca gi cha trong gi dliu
B.Trvni dng dliu tht scha trong gi dliu
C.Trvshiu c!ng cha trong gi dliu
D.Trva chcha trong gi dliu
Cu 35.Lp String thuc package no?
A.java.lang
B.java.util
C.java.string
D.javax.util
Cu 36.Cho chu$i a chURL c to ra b(ng lnh sau
URL u = new URL("http://mail.google.com/mail/trinhxuan#inbox");
Kt quin ra mn hnh khi thc hin lnh:
System.out.println("The protocol is "+u.getProtocol());
A.The protocol is mail.google.com
B.The protocol is http
C.The protocol is /mail/trinhxuan
D.The protocol is -1
Cu 38.Phng thc getPort() trong lp DatagramPacket c sdng vi mc ch
g ?
A.Trvni dng dliu tht scha trong gi dliu
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Nhm Lp trnh mng Khoa CNTT trng i hc Kinh doanh v Cng nghH Ni
A.Sc!ng
B.Chiu di gi
C.Dliu
D.a chIP
Cu 41.Lp Socket c dng trong cchlp trnh no sau y?
A.Tt cu sai
B.Lp trnh Socket UDP/IP
C.Lp trnh Socket TCP/IP
D.Lp trnh IP
Cu 42.Cc nh ngh+a phng thc sau y nh ngh+a no sai:
A.protected float MyMethd(float r, String s){r=10;return r;}
B.private float MyMethd(float z, int i){z=10;return z;}
C.final float MyMethd(float y){}
D.static void My(float x){x=5;return x}
Cu 43.Trong cng mt my c nhiu ng dng mng cng chy, phn bit cc
ng dng vi nhau thng dng thng sno sau y?
Cu 45.in phn cn thiu cho c php hm sau DatagramPacket( buf[], int
len) trong lp DatagramPackage
Cu 46.Cho on m nhsau:
InetAddress host=InetAddress.getByName("127.0.0.1");
Kt quin ra mn hnh khi thc hin dng lnh:
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Nhm Lp trnh mng Khoa CNTT trng i hc Kinh doanh v Cng nghH Ni
System.out.println("Host name:" + host.getHostName());
Cu 47.Package l tp hp ca:
Cu 49.M$i gi dliu trong lp trnh mng cha thng tin g
Cu 50.Phng thc writeBytes(String v) trong lp DataOutputStream c sdng
vi mc ch g ?
Cu 51.Phng thc getPort() trong lp Socket c dng lm g
Cu 53.Phng thc ServerSocket(int port) trong lp ServerSocket c sdng vi
mc ch g ?
Cu 54.Phng thc Socket(String host, int port) trong lp Socket c sdng vi
mc ch g ?
Cu 55.Lp Soket trong lp trnh mng c sdng khi vit chng trnh bn pha
my no trong lp trnh theo m hnh Client/Server:
A.My chv my khch
B.My ch
C.Khng phi my chv my khch
D.My khch
Cu 57.Phng thc writeInt( int v)trong lp DataOutputStream c sdng vi
mc ch g ?
A.Gi mt chu$i k tv ti ngun xut bt k*
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B.Gi mt snguyn v ti ngun xut bt k*
C.Gi mt k tv ti ngun xut bt k*
D.Gi mt sthc v ti ngun xut bt k*
Cu 59.in phn cn thiu cho c php hm sau DatagramPacket( byte buf.., intlen) trong lp DatagramPackage
A." "
B.{ }
C.[ ]
D.()
Cu 59.in phn cn thiu cho c php hm sau DatagramPacket( byte buf[],
len) trong lp DatagramPackage
A. int
B. int[]
C. String
D. float
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