SIEMENS Job Placement Paper
Paper Type : Aptitude - General
THIS PAPER CONSISTS 6 PARTS. all are multiple choice q's
1 general
2 c/unix
3 c++/motif
4 database
5 x-windows
6 ms-windows
we have written q's not acc. to each part.total 50. q's. time is sufficient.
if u have basic idea about all of the u can easily answer the
which of following operator can't be overloaded.
a)== b)++ c)?! d)<=
#include<iostream.h>
main()
{
printf("Hello World");
}
the program prints Hello World without changing main() the o/p should be
initialization Hello World Destruct the changes should be
a)iostream operator<<(iostream os, char*s) os<<'intialisation'<<(Hello
World)<<Destruct
b) c) d)none of the above
CDPATH shell variable is in(c-shell)
a) b) c) d)
semaphore variable is different from ordinary variable by
swap(int x,y)
{
int temp;
temp=x;
x=y;
y=temp;
}
main()
{
int x=2;y=3;
swap(x,y);
}
after calling swap ,what are yhe values x&y?
static variable will be visible in
a)fn. in which they are defined
b)module " " " "
c)all the program
d)none
unix system is
a)multi processing
b)multi processing ,multiuser
c)multi processing ,multiuser,multitasking
d)multiuser,multitasking
x.25 protocol encapsulates the follwing layers
a)network
b)datalink
c)physical
d)all of the above
e)none of the above
TCP/IP can work on
a)ethernet
b)tokenring
c)a&b
d)none
A node has the ip address 138.50.10.7 and 138.50.10.9.But it is transmitting
data from node1 to node2only. The reason may be
a)a node cannot have more than one address
b)class A should have second octet different
c)classB " " " " "
d)a,b,c
for an application which exceeds 64k the memory model should be
a)medium
b)huge
c)large
d)none
The condition required for dead lock in unix sustem is
set-user-id is related to (in unix)
wrong statement about c++
a)code removably
b)encapsulation of data and code
c)program easy maintenance
d)program runs faster
struct base
{
int a,b;
base();
int virtual function1();
}
struct derv1:base{
int b,c,d;
derv1()
int virtual function1();
}
struct derv2 : base
{int a,e;
}
base::base()
{
a=2;b=3;
}
derv1::derv1(){
b=5;
c=10;d=11;}
base::function1()
{return(100);
}
derv1::function1()
{
return(200);
}
main()
base ba;
derv1 d1,d2;
printf("%d %d",d1.a,d1.b)
o/p is
a)a=2;b=3;
b)a=3; b=2;
c)a=5; b=10;
d)none
for the above program answer the following q's
main()
base da;
derv1 d1;
derv2 d2;
printf("%d %d %d",da.function1(),d1.function1(),d2.function1());
o/p is
a)100,200,200;
b)200,100,200;
c)200,200,100;
d)none
struct {
int x;
int y;
}abc;
you can not access x by the following
1)abc-->x;
2)abc[0]-->x;
abc.x;
(abc)-->x;
a)1,2,3
b)2&3
c)1&2
d)1,3,4
automatic variables are destroyed after fn. ends because
a)stored in swap
b)stored in stack and poped out after fn. returns
c)stored in data area
d)stored in disk
variable DESTDIR in make program is accessed as
a) $(DESTDIR)
b) ${DESTDIR}
c) DESTDIR
d) DESTDIR
The keystroke mouse entrie are interpreted in ms windows as
a) interrupt
b) message
c) event
d) none of the above
link between program and out side world (ms -win)
a)device driver and hardware disk
b)application and device driver
c)application and hardware device
d)none
after logout the process still runs in the background by giving the command
a)nohop
b)
process dies out but still waita
a)exit
b)wakeup
c)zombie
d)steep
n dynamic memory allocation we use
a) doubly linked list
b) circularly linked
c) B trees
d) L trees
e) none
To find the key of search the data structure is
a)hask key
b)trees
c)linked lists
d)records
data base
--------------------------------
employ_code salary employ_code leave from to
--------------------------------------
1236 1500 1238 --- ---237
1237 2000 1238 --- --- 1237
1238 2500 1237 --- 1237
-------------------------- --------------------------------------
select employ_code,employ_data ,leave
the number of rows in the o/p
a)18
b)6
c)7
d)3
which is true
a)bridge connects dissimiler LANand protocol insensitive
b)router " " " " "
c)gateway " " " " "
d)none of the above
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