Question

Consider the following functions: (4) int hidden(int num1, int num2) { if (num1 > 20) num1 = num2 / 10; else if (num2 > 20) num2 = num1 / 20; else return num1 - num2; return num1 * num2; } int compute(int one, int two) { int secret = one; for (int i = one + 1; i <= two % 2; i++) secret = secret + i * i; return secret; } What is the output of each of the following program segments? a. cout << hidden(15, 10) << endl; b. cout << compute(3, 9) << endl; c. cout << hidden(30, 20) << " " << compute(10, hidden(30, 20)) << endl; d. x = 2; y = 8; cout << compute(y, x) << endl;

Answer 1

Answer:

a.  cout<<hidden(15,10)<<endl;

output = 5

b. cout << compute(3, 9) << endl;

output=3

c. cout << hidden(30, 20) << " " << compute(10, hidden(30, 20)) << endl;

output = 10

d. x = 2; y = 8; cout << compute(y, x) << endl;

output= 8

Explanation:

solution a:

num1= 15;

num2= 10;

according to condition 1, num1 is not greater than 20. In condition 2, num2 is also not greater than 20.

so,

the solution is

return num1 - num2 = 15 - 10 = 5

So the output for the above function is 5.

solution b:

as in function there are two  integer type variables named as one and two. values of variables in given part are:

one = 3

two = 9

secret = one = 3

for (int i= one + 1 = 3+1 = 4; i<=9%2=1; i++ )

9%2 mean find the remainder for 9 dividing by 2 that is 1.

//  there 4 is not less than equal to  1 so loop condition will be false and loop will terminate. statement in loop will not be executed.

So the answer will be return from function compute is

return secret ;

secret = 3;

so answer return from the function is 3.

solution c:

As variables in first function are num1 and num2. the values given in part c are:

num1 = 30;

num2= 20;

According to first condition num1=30>20, So,

num1= num2/10

so the solution is

num1 = 20/10 = 2

now

num1=2

now the value return from function Hidden is,

return num1*num2 = 2* 20 = 40

Now use the value return from function hidden in function compute as

compute(10, hidden(30, 20))

as the value return from hidden(30, 20) = 40

so, compute function becomes

compute(10, 40)

Now variable in compute function are named as one and two, so the values are

one= 10;

two = 40;

as

secret = one

so

secret = 10;

for (int i= one + 1 = 10+1 = 11; i<=40%2=0; i++ )

40%2 mean find the remainder for 40 dividing by 2 that is 0.

//  there 11 is not less than equal to  0 so loop condition will be false and loop will terminate. statement in loop will not be executed.

So the answer will be return from function compute is

return secret ;

secret = 10;

so answer return from the function is 10.

solution d:

Now variable in compute function are named as one and two, so the values are

one = y = 8

two = x = 2

There

Secret = one = 8;

So

for (int i= one + 1 = 8+1 = 9; i<=2%2=0; i++ )

2%2 mean find the remainder for 2 dividing by 2 that is 0.

//  there 9 is not less than equal to  0 so loop condition will be false and loop will terminate. statement in loop will not be executed.

So the answer will be return from function compute is

return secret ;

secret = 8;

so answer return from the function is 8.