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: (GATE CS 2004)
(A) n = D1D2….Dm-i and rev = DmDm-1…Dm-i+1
(B) n = Dm-i+1…Dm-1Dm and rev = Dm-1….D2D1
(C) n != rev
(D) n = D1D2….Dm and rev = DmDm-1…D2D1
Answer: (A)
Explanation: We can get it by taking an example like n = 54321. After 2 iterations, rev would be 12 and n would be 543.
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