Following questions have been asked in GATE CS exam.
1. Consider the following C program segment:
The output of the program is (GATE CS 2004)
a) gnirts
b) gnirt
c) string
d) no output is printed
Answer(d)
Let us consider below line inside the for loop
p[i] = s[length — i];
For i = 0, p[i] will be s[6 — 0] and s[6] is '\0'
So p[0] becomes ‘\0’. It doesn't matter what comes in p[1], p[2]..... as P[0] will not change for i >0. Nothing is printed if we print a string with first character '\0'
2. Consider the following C function
In order to exchange the values of two variables x and y. (GATE CS 2004)
a) call swap (x, y)
b) call swap (&x, &y)
c) swap (x,y) cannot be used as it does not return any value
d) swap (x,y) cannot be used as the parameters are passed by value
Answer(d)
Why a, b and c are incorrect?
a) call swap (x, y) will not cause any effect on x and y as parameters are passed by value.
b) call swap (&x, &y) will no work as function swap() expects values not addresses (or pointers).
c) swap (x, y) cannot be used but reason given is not correct.
3. Consider the following C function:
The value returned by f(1) is (GATE CS 2004)
a) 5
b) 6
c) 7
d) 8
Answer (c)
Since i is static, first line of f() is executed only once.
Execution of f(1)
i = 1
n = 2
i = 2
Call f(2)
i = 2
n = 4
i = 3
Call f(4)
i = 3
n = 7
i = 4
Call f(7)
since n >= 5 return n(7)
4. Consider the following program fragment for reversing the digits in a given integer to obtain a new integer. Let n = D1D2...Dm
The loop invariant condition at the end of the ith iteration is:(GATE CS 2004)
a) n = D1D2....Dm-i and rev = DmDm-1...Dm-i+1
b) n = Dm-i+1...Dm-1Dm and rev = Dm-1....D2D1
c) n ≠ rev
d) n = D1D2....Dm and rev = DmDm-1...D2D1
Answer (a)
5. Consider the following C program
