it will print p2=Name

choice d.
Reason:
The macro sign # is not applicable here.
Syntactical error!

int *a[5] means that there are 5 integer pointers[which
store adresess of only integers] whose adresses are in
continous locations and each pointer name is accesed by *a
[0],*a[1]….*a[4].for example
a[0]=n means a[0] points to n, a[1]=p means a[1] points to
p.[where n and p are integers]Let the adress of n and p be
some 625 and 254 respectively. then a[0] will hold 625
whose adress is(assume)1000 and a[1]will hold 254 whose
adress is 1002.similarly the a[2] will hold some adress(if
u make it to point) with its adress 1004 and a[3] will hold
some others adress with its adress 1006 and a[4] will hold
some others adress with its adress 1008.

6
becoz it(++*p) will evaluate like this-
1)
*(p)- gives value at this address that is 5
2)
now increment ++5
that is 6.

Answer is 6;
Sum being the static variale will retain its value state
between he function calls.

it is x for hexadecimal

6

It will lead to compilation error..
Note: # define prod(a,b)=a*b
‘=’ is not allowed with #define
regards,
Arun Raj

c.free()

c) Since expn2 becoming false will terminate do-while loop.

d.not available is d right ans.

infinite coz there is no condition to stop it

I’m getting “error C2181: illegal else without matching if”.
If that is corrected then the output will be y = 5.

Hello,
The answer is b.
a=0+1=1
a=1+1=2
a=2+1=3
a=3+1=4
a=4+1=5
While a variable is storing the 5 result a is nither less
then are equal to 5 hence the application will come out of
the loop.So while loop had run 5 times.

if this function is called from within a loop with n=0 to n<4
then cases 1,2,3,4 in switch will be called one by one
as int a is static , first time it will be initialized with the value p but afterwords
its memory location will be updated every time the fun is called with the latest value of a calculated within the case
but r should be a float value
example:
main()
int p=100;
float r=0.10;
{
for(n=1;na=100+100*.1
a=110

second iteration
a=110;
n=2
case(2)
a=a+a*r=>a=110+110*.1
a=121

third iteration
a=121
n=3
case(3)
a=a+a*r=> a=121+121*.1
a=133.1

fourth iteration
a=133
n=4
case(4)
a=133+133*.1
a=146

in this way, we can see this is the amount for compound interest for 1-4 years

Size fo strucure wil be the total bytes of the datatypes
inside it..
so,4+4=8;
For unions the size wi be the size of the datatype whose
memory is high,.
so,its 4,.

d) Fail 1, Pass 2

a)the string is string

b) The string is :Oldstring
as per the question static variable memory is created for one time so it will return Oldstring

Fail1, Pass2

Answer is e), since opening and closing flower braces do
not match in numbers and default do not have colon
following it. Assuming switch(i) has an opening flower
brace and default has colon after it “switch(i) {, …
default: i+= 4;”answer would be d).
All cases will be fall-through including default:

How is the answer BYE?

it will be an error as for as i know… since s1 is not a
pointer variable… but however array are implict
pointers…. here we cant assign the base address to
another arry which is capable of holding oly values not
addressses……………

Turbo c will give an answer
samco system
amco system
pointer ptr is at r
by(*ptr)++ we are incrementing it by one so we get s
by ptr ++ we are incrementing the addrestion of ptr so it
will give an output amco system;
but gcc compiler will not give any answer it will print
segmentation fault

5,20,1

the given program gives some meaningless output, with some
modification to the given program as
#include
#include
main()
{
char a[]=”ramesh”;
char *p1=”Name”;
char *p2=a;
while(*p2++=*p1++);/*copies contents of p1 to
p2*//* here it is not possible to use while(*a++=*p1++)
because a can not change its value*/
*p2=’’;
printf(“%sn”,a);
}
The output will be Name

it gives compilation error bcoz b is not defined

57
94

a