Computer Science Engineering (CSE) Exam > Computer Science Engineering (CSE) Questions > int f(int n){ static int r = 0; if (n <=...
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int f(int n)
{
static int r = 0;
static int r = 0;
if (n <= 0) return1;
if (n > 3)
{ r = n;
return f(n-2) + 2;
}
return f(n-1) + r;
return f(n-1) + r;
}
What is the value of f(5)?
- a)5
- b)7
- c)9
- d)18
Correct answer is option 'D'. Can you explain this answer?
Verified Answer
int f(int n){ static int r = 0; if (n <= 0) return1; if (n > ...
f(5) = f(3)+2
The line "r = n" changes value of r to 5. Since r
is static, its value is shared be all subsequence
calls. Also, all subsequent calls don't change r
because the statement "r = n" is in a if condition
with n > 3.
f(3) = f(2)+5
f(2) = f(1)+5
f(1) = f(0)+5
f(0) = 1
So f(5) = 1+5+5+5+2 = 18
Most Upvoted Answer
int f(int n){ static int r = 0; if (n <= 0) return1; if (n > ...
Explanation:
The given code is recursively calculating the value of f(n), where n is an integer input. Let's understand the code step by step:
- The function f(n) is defined with a static variable r initialized to 0.
- If n is equal to 0, the function returns 1.
- If n is greater than or equal to 3, r is set to the value of n, and the function calls itself with n-2 as the input argument, and returns the result multiplied by 2.
- If n is less than 3, the function calls itself with n-1 as the input argument, and returns the value of r.
Now, let's calculate the value of f(5) by applying the above steps:
- The input argument is 5, which is greater than or equal to 3. So, r is set to 5, and the function calls itself with 3 as the input argument.
- The input argument is 3, which is equal to 3. So, r is not changed, and the function calls itself with 1 as the input argument.
- The input argument is 1, which is less than 3. So, the function calls itself with 0 as the input argument.
- The input argument is 0, which is equal to 0. So, the function returns 1.
- Now, the previous function call with input argument 1 returns the value of r, which is 5.
- So, the function call with input argument 3 returns 5.
- Now, the initial function call with input argument 5 returns the result of the previous function call multiplied by 2, which is 10.
Therefore, the value of f(5) is 10, which is option D.
The given code is recursively calculating the value of f(n), where n is an integer input. Let's understand the code step by step:
- The function f(n) is defined with a static variable r initialized to 0.
- If n is equal to 0, the function returns 1.
- If n is greater than or equal to 3, r is set to the value of n, and the function calls itself with n-2 as the input argument, and returns the result multiplied by 2.
- If n is less than 3, the function calls itself with n-1 as the input argument, and returns the value of r.
Now, let's calculate the value of f(5) by applying the above steps:
- The input argument is 5, which is greater than or equal to 3. So, r is set to 5, and the function calls itself with 3 as the input argument.
- The input argument is 3, which is equal to 3. So, r is not changed, and the function calls itself with 1 as the input argument.
- The input argument is 1, which is less than 3. So, the function calls itself with 0 as the input argument.
- The input argument is 0, which is equal to 0. So, the function returns 1.
- Now, the previous function call with input argument 1 returns the value of r, which is 5.
- So, the function call with input argument 3 returns 5.
- Now, the initial function call with input argument 5 returns the result of the previous function call multiplied by 2, which is 10.
Therefore, the value of f(5) is 10, which is option D.
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