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C Language | Set 2

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Following questions have been asked in GATE CS exam. 1. Consider the following C program segment:
char p[20];
char *s = "string";
int length = strlen(s);
int i;
for (i = 0; i < length; i++)
     p[i] = s[length — i];

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
void swap (int a, int b)
   int temp;
   temp = a;
   a = b;
   b = temp;

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:
int f(int n)
   static int i = 1;
   if (n >= 5)
      return n;
   n = n+i;
   return f(n);

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
int n, rev;
rev = 0;
while (n > 0) 
   rev = rev*10 + n%10;
   n = n/10;

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
   int x, y, m, n;
   scanf ("%d %d", &x, &y);
   /* x > 0 and y > 0 */
   m = x; n = y;
   while (m != n)
         m = m - n;
         n = n - m;
   printf("%d", n);

The program computes (GATE CS 2004) a) x + y using repeated subtraction b) x mod y using repeated subtraction c) the greatest common divisor of x and y d) the least common multiple of x and y Answer(c) This is an implementation of Euclid’s algorithm to find GCD

Last Updated : 27 Mar, 2017
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