Computer Knowledge

Programming Output Evaluation

1,953 Questions

Programming output evaluation tests the ability to trace code execution in languages like C, Java, and SAS. It focuses on arrays, loops, pointers, and data type conversions. These technical questions are standard in computer knowledge sections for IT officer and bank exams.

Java string bufferC language pointersLoop execution outputsData type conversionsMacro variable evaluation

Programming Output Evaluation Questions

Multiple choice

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]; printf("%s",p);The output of the program is

  1. gnirts

  2. string

  3. gnirt

  4. no output is printed
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

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′   

Multiple choice

Consider the following C program segment

The value returned by the function Do Something when a pointer to the proof of a non-empty tree is passed as argument is

  1. the number of leaf nodes in the tree

  2. the number of nodes in the tree

  3. the number of internal nodes in the tree

  4. the height of the tree
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Value initialized by 0 If any root node has left child then it adds 1 to the value & move to left child & if any mode has right child also then also calculated the value using recursion & take maximum of both left & right value is taken. So we know that height is the largest distance between root node & leaf. So this program calculates heights.

Multiple choice

Consider the following C-program:

double foo(double); /* Line 1 */
int main() {
  double da, db;
  // input da
  db = foo(da);
}

double foo(double a) {
  return a;
}

The above code complied without any error or warning. If Line 1 is deleted, the above code will show

  1. no compile warning or error

  2. some complier-warning not leading to unitended results

  3. Some complier-warning due to type-mismatch eventually leading to unitended results

  4. Complier errors
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Here if line 1 which is prototype declaration of the function foo, in C compilation processes this would give a compile warning, due to type-mismatch. Since then compiler won't know what the return type of foo is. So unintended results may occur

Multiple choice

Consider the following C-program fragment in which i, j and n are integer variables.
for( i = n, j = 0; i &gt; 0; i /= 2, j +=i ); Let val ( j ) denote the value stored in the variable j after termination of the for loop. Which one of the following is true?

  1. $val(j)=\Theta(\log n)$
  2. $val(j)=\Theta (\sqrt{n})$
  3. $val(j)=\Theta( n)$
  4. $val(j)=\Theta (n\log n)$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$j = n/2+n/4+n/2+\ldots +1$

number of iteration will be $2^k = n$ or $k = \log n$

this is in GP find sum till $\log n $,= $\Theta(n)$

Multiple choice

Consider the following C code segment:

int Is Prime (n) 
{
int i, n;
for (i = 2; i &lt;= sqrt (n) ; i ++)
 if (n% i == 0) { 
   print f (“ Not Prime  n”);
   return 0; 
  } 
  return 1;
}

Let T (n) denote the number of times the for loop is executed by the program on input n. Which of the following is TRUE?
  1. T (n) = O($\sqrt n$) and T (n) =$\Omega$($\sqrt n$)
  2. T (n) = O($\sqrt n$) and T (n) = $\Omega$ (1)
  3. T (n) = O(n) and T (n) = $\Omega$ ($\sqrt n$)

  4. None of the above
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Multiple choice

What will be the output of the following set of codes after execution?

void main() { int m=32769; unsigned int n=65540; printf(“%d %u”,m,n); }

  1. Garbage value
    • 32767 4
    • 32768 0
    • 32767 3
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Range of int is -32768 to +32767. Value of m exceeds the limit of int. So after 32767, it goes to the other side of the range and 32769 will convert to -32767. Range of unsigned int is 0 to 65535. Value of n exceeds the limit of int. So after 65535, it goes to the other side of the range and 65535 will convert to 4.