What will be the value of the following pseudocode for k=150?fun(integer k)if(k>155)returnend ifprint kfun(k+2)print kEnd of function fun()
Question
What will be the value of the following pseudocode for k=150?fun(integer k)if(k>155)returnend ifprint kfun(k+2)print kEnd of function fun()
Solution
The pseudocode provided is a recursive function that prints the value of k and calls itself with k+2 until k is greater than 155. Here's how it would execute for k=150:
- Call
fun(150). Since150is not greater than155, it will print150and callfun(152). - Call
fun(152). Since152is not greater than155, it will print152and callfun(154). - Call
fun(154). Since154is not greater than155, it will print154and callfun(156). - Call
fun(156). Since156is greater than155, it will return and not print anything. - After each function call returns, it will print
kagain. So it will print154,152, and150in that order.
So, the output of the pseudocode for k=150 will be 150, 152, 154, 154, 152, 150.
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