UGC NET Computer Science C Programming Language Previous Year Questions (PYQs) – Page 1 of 4
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Solution
Solution
In C, pointer multiplication and pointer division are not allowed.
Here,
j = j * 2;
and
k = k / 2;
are invalid operations.
Consider the following pseudo-code fragment in which an invariant for the loop is $m \times x^k=p^n$ and $k\ge 0$.
Pre-conditions: $p\ge 1$ and $n\ge 0$
Assume that overflow never occurs.
int x = p;
int k = n;
int m = 1;
while (k != 0) {
if (k is odd) then m = m * x;
x = x * x;
k = floor(k / 2);
}
Which of the following must be true at the end of the while loop?
Solution
Given loop invariant is
$m \times x^k=p^n$
At the end of the while loop, condition becomes false.
So,
$k=0$
Putting $k=0$ in invariant,
$m \times x^0=p^n$
Since,
$x^0=1$
Therefore,
$m=p^n$
Consider the following C-code fragment running on a $32$-bit x86 machine:
typedef struct {
union {
unsigned char a;
unsigned short b;
} U;
unsigned char c;
} S;
S B[10];
S *p = address of B[4];
S *q = address of B[5];
p->U.b = 0x1234;
If $M$ is the value of $q-p$ and $N$ is the value of address of $(p->c)$ minus address of $p$, then $(M,N)$ is:
Solution
Given that structure $S$ takes $32$ bits.
So,
$32\text{ bits}=4\text{ bytes}$
Now,
$p$ points to $B[4]$
$q$ points to $B[5]$
Since $p$ and $q$ are pointers of type $S$, pointer subtraction gives the difference in number of structures.
So,
$q-p=1$
Therefore,
$M=1$
Now, union $U$ contains $unsigned\ char\ a$ and $unsigned\ short\ b$.
Size of union $U$ is $2$ bytes.
After union $U$, variable $c$ is stored.
So, offset of $c$ from the beginning of structure is $2$ bytes.
Therefore,
$N=2$
Hence,
$(M,N)=(1,2)$
Solution
$num\ !=\ 65$
It means $num$ is not equal to $65$.
Option (a):
$num>65\ ||\ num<65$
This is true when $num$ is not equal to $65$.
Option (b):
$!(num==65)$
This also means $num$ is not equal to $65$.
Option (c):
$num-65$
In C language, any non-zero value is treated as true.
So, this is also true when $num\neq 65$.
Option (d):
$!(num-65)$
This becomes true when $num-65=0$, that is $num=65$.
So, it is opposite of $num\ !=\ 65$.
Solution
They can be used to simulate call by reference.
They are also used to manipulate dynamic data structures like linked lists, trees, etc.
But pointers do not reference values directly. They access values indirectly through addresses.
Consider the following C++ function $f()$:
unsigned int f(unsigned int n) {
unsigned int b = 0;
while (n) {
b = b + (n bitwise AND 1);
n = n shifted right by 1;
}
return b;
}
The function $f()$ returns the integer that represents the $P$ in the binary representation of positive integer $n$, where $P$ is:
Solution
The expression $n$ bitwise AND $1$ checks the last bit of $n$.
If the last bit is $1$, then $b$ increases by $1$.
After that, $n$ is shifted right by one bit.
So, the loop checks each bit of $n$ one by one.
Therefore, $b$ counts the total number of $1$'s in the binary representation of $n$.
Solution
p is a pointer variable and must store the address of x, written as:
p = &x;
But the program assigns:
p = x;
which means an integer is assigned to a pointer, causing a compilation error (type mismatch).
Solution
Solution
Outer loop runs for:
$i=0,1,2,3,4$
Inner loop runs only when $j<i$.
But inside inner loop, there is a $break$ statement, so inner loop executes only one time for each valid value of $i$.
For $i=0$:
$j<0$ is false, so inner loop does not execute.
For $i=1$:
$x=x+(1+0-1)=0+0=0$
For $i=2$:
$x=x+(2+0-1)=0+1=1$
For $i=3$:
$x=x+(3+0-1)=1+2=3$
For $i=4$:
$x=x+(4+0-1)=3+3=6$
So, final value of $x$ is:
$x=6$
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