What is the value of i after the following code is executed? int i = 1; while (i i) j--; i + = j; }

Question

Victoria Hardy

Computers & Technology

5 December, 20:44

What is the value of i after the following code is executed? int i = 1; while (i i) j--; i + = j; }

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Answers (1)

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Jocelyn Valencia · Answer 1

The value of "i" after executing the entire code will be "16"

Explanation:

Initially the value of i will be 1. This value keeps changing after entering the while loop. Let me give you step by step result.

When i = 1, it checks whether ii. so before the outer while loop ends "i" is reinitialized using "i=i+j".

So When i = 1, the value of j after running inner while loop will be 1 and i will be reinitialized to 1+1 = 2, now i = 2

So When i = 2, j = 2, Reinitialized value of i = 4

So When i = 4, j = 4, Reinitialized value of i = 8

So When i = 8, j = 8, Reinitialized value of i = 16.

The loop will not get executed thereafter.

You can add the below statement, after "j--; " to understand better.

cout<<"i="<