Quiz-1a

The above code is converting an octal number into its decimal representation. Here we are treating 127 as an octal input and converting it into its decimal representation, that is 87. So, answer will be 87.

Loop will end up giving output as 2.7182818284590455 as value of f is initialized to 1 and then following BODMAS rule the next step "s = s +(i / f)" will be followed and i is incremented.

First iteration of while loop: count <= limit. 1<=6. So, we will enter while loop.  count=2, Now we will print n3. So value of n3 is 1. 1 will be printed. n3=n1+n2. So, n3=1,n1=1,n2=1

Second iteration: Count=3, Print n3. So 1 will be printed , n3=2, n1=1, n2=2

Third iteration: Count=4, Print n3. So 2 will be printed , n3=3, n1=2,  n2=3

Fourth iteration: Count=5 , Print 3, n3=5, n1=3, n2=5

Fifth iteration: Count=6, Print 5, n3=8, n1=5, n2=8

Sixth iteration: Count=7, Print 8, n3=13, n1=8, n2=13

Hence output will be 112358

First iteration: Digit: 0, Num_remainder: 263
second iteration: Digit: 3, Num_remainder: 26
Third iteration: Digit: 6, Num_remainder: 2
Fourth iteration: Digit: 2, We will enter  if. Even_counter: 1, Num_remainder: 0

Exit while loop

Even counter=1. So, 1 will be printed

In this program, we are continuously increasing the value of t by a till b is greater than 0.

We were asked to sort the array elements in ascending order and the code is written to sort the elements in the descending order. Change the sign at the line no. 7 to get the correct output.

a=9

b=7

c=2

b = b mod c =7 mod 2=1

a = a mod c= 9 mod 2=1

a+b=1+1=2

So, 2 will be returned by the function

a = 30, b = 60, C = 90

sum=180

((sum EQUALS 180) and (a NOT EQUALS 0) and (b NOT EQUALS 0) and (c NOT EQUALS 0))

So, (true) and (true) and (true) and (true)

So, output will be "success"

The “ptr” variable is a pointer which holds the address of variable “a”. And “*ptr” returns the value of “a” variable. “cho” is a reference variable to “ch”. So any change made to “cho” will be reflected to “ch”. As such, when “cho” is increased by 32, it adds to the ASCII value of “A”(which is 65), and this results to 97 which is the ASCII value of “a”(from the alphabet). So this “a” gets stored in “ch”. As for when “*ptr” is incremented by “ch”, it gives value 97+32=129.

For the first iteration, ch="b" then after while, ch="c".

again the while condition will hold true, ch="c" and ch will be incremented

the loop will continue till ch="z"

hence the loop will run for 25 times