Questions in: C

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

d is the right answer
# should not be present at starting.

it will print p2=Name

ANS: NULL
It is a bit tricky question. If u observe carefully then we
are incrementing the pointers p1,p2. When it reached the end
of the string, *p2 points to NULL. We have lost the address
of the starting position.

Hi all,
#1 Mannucse's ans is wrong. cos as mentioned "Name" will
not be the output.
#2 Sanath's ans is wrong. Cos at the end of the while loop,
p2 will not point to NULL. It will point to the next byte
to the NULL termination ie., 6th byte.
#3 Shruti's ans is wrong. cos i think she got confused
between assignment(=) and comparisonal(==) operators. And
the statement given as "we cannot copy the value of p1 in
p2, the way its mentioned here" is absolutely wrong.
So,
Lets Analyse the program and how to get the required output.
hav a look on th program again.
main()
{
char *p1="Name";
char *p2;
p2=(char *)malloc(20);
while(*p2++ = *p1++)
{
printf("TEST n");
}
printf("%sn",p2);
getch();
}
Here, in every iteration of while loop, we are assigning
*p1 to *p2, and incrementing both pointers p1 and p2, After
completion(when *p1 value would be '')of the while loop,
first 5 bytes of p2 holds the
characters 'N','a','m','e' ''. At the end of while loop
p2 points to the 6th byte in the memory.
So, now printf("%sn",p2); shall start print the values
from the 6th byte to 20th bytes of the memory which was
allocated dynamically.
----------------------------------
To get the desired output change the printf statement to
printf("%sn",p2-5);
Now (p2-5) points to the starting address of p2 and will
print the values in the memory till it encounters ''
termination. ie., The output would be -> Name

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.

array of pointers

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

6

it is x for hexadecimal

c

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

answer: c
here in do-while:
fist the statements inside the block gets executed.then
when it checks the condition ,it will be come out of loop
because conditions goes wrong

c.free()

here malloc(),calloc(),realloc)() are the memory allocation
function after allocation of memory free is used to freed
the memory which is allocated by the above function.
so the free is odd term.

d.not available is d right ans.

Hi All,
Upto my knowledge exponentiation operator is not available
in C. Because ^ operator is used for XOR operation. So we
cannot use that. But its possible by using Bitwise
operator to perform Exponentiation.
Thanks & Regards
Sathish Kumar

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

the value of y is not changing so i think y wll reamin 5.

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

this code depends upon the value of 'n' actually....... if
the value is 4 the operation is differewnt... if 3 its
different..... so give the value of 'n'!

Here a is declared as static,so it can't be re-initialized..

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.

infinite coz there is no condition to stop it

infinite loop, bcoz there is no condition to break the loop.

d) Fail 1, Pass 2

Fail 1 , Pass 2.
Some explanation,
1. Fail 1
first of all, to compare strings in C, you use this strcmp
function, so this WOULD give PASS 1 :
if(strcmp(p,"String") == 0)
but
if(p=="String")
will fail because this line means :
if address of p is the same as address of some temporary
storage for literals, where literal "String" is stored,
which is very rarely true, because storing literals is
compiler specific and is very hard to estimate at runtime.
2. Pass 2
sizeof(p) gives 7, because sizeof(char) is 1 byte, and we
have 7 chars in array storing "String", which are :
[0]S
[1]t
[2]r
[3]i
[4]n
[5]g
[6] (EOS)
now, clearly sizeof(p) - 2 is [5] which is "g"
thats why
if(p[sizeof(p)-2]=='g')
is true.

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,.

I assume above code to be correctly written as below:
struct s
{
int a;
long b;
}
Union u
{int a;
long b;
}
sizeof(s)= 8
sizeof(u) = 4

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

oldstring

b) The String is :OldString
Because ,for Static var memory is only one time created.
Eventhough the fn is multiple times called,
so
consider addr of xxx is 4444,
first strcpy copy the string "string" to addr 4444,
then g=4444;
then oldstring overwrites to location 4444.

a)the string is string

Run time error/Core dump.