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Consider the following program fragment for reversing the digits in a given integer to obtain a new integer. Let n = D1D2…Dmint n, rev;rev = 0;while (n > 0){ rev = rev*10 + n; 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+1b)n = Dm-i+1…Dm-1Dm and rev = Dm-1….D2D1c)n != revd)n = D1D2….Dm and rev = DmDm-1…D2D1Correct answer is option 'A'. Can you explain this answer? for Computer Science Engineering (CSE) 2025 is part of Computer Science Engineering (CSE) preparation. The Question and answers have been prepared according to the Computer Science Engineering (CSE) exam syllabus. Information about Consider the following program fragment for reversing the digits in a given integer to obtain a new integer. Let n = D1D2…Dmint n, rev;rev = 0;while (n > 0){ rev = rev*10 + n; 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+1b)n = Dm-i+1…Dm-1Dm and rev = Dm-1….D2D1c)n != revd)n = D1D2….Dm and rev = DmDm-1…D2D1Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for Computer Science Engineering (CSE) 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Consider the following program fragment for reversing the digits in a given integer to obtain a new integer. Let n = D1D2…Dmint n, rev;rev = 0;while (n > 0){ rev = rev*10 + n; 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+1b)n = Dm-i+1…Dm-1Dm and rev = Dm-1….D2D1c)n != revd)n = D1D2….Dm and rev = DmDm-1…D2D1Correct answer is option 'A'. Can you explain this answer?.

Solutions for Consider the following program fragment for reversing the digits in a given integer to obtain a new integer. Let n = D1D2…Dmint n, rev;rev = 0;while (n > 0){ rev = rev*10 + n; 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+1b)n = Dm-i+1…Dm-1Dm and rev = Dm-1….D2D1c)n != revd)n = D1D2….Dm and rev = DmDm-1…D2D1Correct answer is option 'A'. Can you explain this answer? in English & in Hindi are available as part of our courses for Computer Science Engineering (CSE). Download more important topics, notes, lectures and mock test series for Computer Science Engineering (CSE) Exam by signing up for free.

Here you can find the meaning of Consider the following program fragment for reversing the digits in a given integer to obtain a new integer. Let n = D1D2…Dmint n, rev;rev = 0;while (n > 0){ rev = rev*10 + n; 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+1b)n = Dm-i+1…Dm-1Dm and rev = Dm-1….D2D1c)n != revd)n = D1D2….Dm and rev = DmDm-1…D2D1Correct answer is option 'A'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of Consider the following program fragment for reversing the digits in a given integer to obtain a new integer. Let n = D1D2…Dmint n, rev;rev = 0;while (n > 0){ rev = rev*10 + n; 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+1b)n = Dm-i+1…Dm-1Dm and rev = Dm-1….D2D1c)n != revd)n = D1D2….Dm and rev = DmDm-1…D2D1Correct answer is option 'A'. Can you explain this answer?, a detailed solution for Consider the following program fragment for reversing the digits in a given integer to obtain a new integer. Let n = D1D2…Dmint n, rev;rev = 0;while (n > 0){ rev = rev*10 + n; 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+1b)n = Dm-i+1…Dm-1Dm and rev = Dm-1….D2D1c)n != revd)n = D1D2….Dm and rev = DmDm-1…D2D1Correct answer is option 'A'. Can you explain this answer? has been provided alongside types of Consider the following program fragment for reversing the digits in a given integer to obtain a new integer. Let n = D1D2…Dmint n, rev;rev = 0;while (n > 0){ rev = rev*10 + n; 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+1b)n = Dm-i+1…Dm-1Dm and rev = Dm-1….D2D1c)n != revd)n = D1D2….Dm and rev = DmDm-1…D2D1Correct answer is option 'A'. Can you explain this answer? theory, EduRev gives you an ample number of questions to practice Consider the following program fragment for reversing the digits in a given integer to obtain a new integer. Let n = D1D2…Dmint n, rev;rev = 0;while (n > 0){ rev = rev*10 + n; 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+1b)n = Dm-i+1…Dm-1Dm and rev = Dm-1….D2D1c)n != revd)n = D1D2….Dm and rev = DmDm-1…D2D1Correct answer is option 'A'. Can you explain this answer? tests, examples and also practice Computer Science Engineering (CSE) tests.