Programming
October 17, 2023Logic-Gates
October 18, 2023Programming
Question 31 |
Consider the following program fragment for reversing the digits in a given integer to obtain a new integer. Let n = d1d2…dm.
int n, rev; rev = 0; while (n > 0) { rev = rev*10 + n%10; n = n/10; }
The loop invariant condition at the end of the ith iteration is:
n = d1d2…dm-i and rev = dmdm-1…dm-i+1 | |
n = dm-i+1…dm-1dm or rev = dm-i…d2d1 | |
n ≠ rev | |
n = d1d2…dm and rev = dm…d2d1 |
Question 31 Explanation:
In each iteration the right most digit of n is going to make to the right end of the reverse.
Correct Answer: A
Question 31 Explanation:
In each iteration the right most digit of n is going to make to the right end of the reverse.
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