Quiz – 1

Explanation:
In the first step i=140 is passed through the function fun(), and if the condition becomes false when it is checked so else will get executed.

When else gets executed, function fun() is called again and i=1 is passed and now condition in IF gets true and 'i' whose value is 1 gets returned.

Explanation:
The operator ?: works as follows expr1 is evaluated first if it is true expr2 is evaluated otherwise expr3 is evaluated. hence in expressions the return statement can not be used in C-language. The return statement is used for returning from a function , you can't use inside ternary operator

Refer this code for the given pseudo code for better understanding, this pseudo code will print a pyramid of * with six rows each row having an odd number of * from 1 to 11 i.e the first row will have 1st star and the 6th row will have 11 starts

#include<stdio.h>

int main ()

{

int i, j, k;

for (i = 1; i <= 6; i++)

{

for (j = i; j < 6; j++)

{

printf (" ");

}

for (k = 1; k < (i * 2); k++)

{

printf ("*");

}

printf ("\n");

}}

Explanation:
Inside the loop with a value of I from 0 to 4, it will increment the value of 4 times that will be equal to 1, 2, 3, and 4.
but first inside the for loop hello will be printed when the value of I = 3
and at last, a will be printed i.e 4so the output will be hello4

Explanation:
There are no errors in the given code, the output produced by the code is 18

Explanation:
Step 1:
Condition true for the while loop
a = 2
Condition true for if
Value of an i.e 2 will get printed
Now b=10
and a=3

Step 2:
Condition true for the while loop
a = 3
Condition false for if, else will get printed
i.e. 9 will get printed
Now b=2
and a=4

Step 3:
Condition true for the while loop
a = 4
Condition true for if
Value of an i.e 4 will get printed
Now b=0
and a=5

Step 4:
Condition false for while. Exit

This pseudocode performs some simple mathematical operations until the value of y is greater than 0 and c is greater than 30 and prints the value of x, y, and z.
below is the Python code for this pseudo code:
x=3

x=3

y=90

while(y>0):

y=y//3

x=x+6

c=x+y

while(c>30):

if(c % 3 == 0):

print(x)

else:

print (y)

c=c//5

print(c)