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Question

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:

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).n = D1D2….Dm-i and rev = DmDm-1…Dm-i+1

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Q. 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 =...

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