Difficulty: Medium
Correct Answer: lice ice ce e
Explanation:
Introduction / Context:
This problem explores pointer arithmetic on C strings and how %s prints from the first \0 encountered onward. It also contains a deliberate twist: writing to a string literal via *x = .... That write is undefined behavior in standard C and can crash; however, many textbook problems assume it “works” for demonstration purposes. We follow that intent here.
Given Data / Assumptions:
x initially points to the literal "Alice" (bytes: A l i c e \0).
Concept / Approach:
*x = x[n] sets the first byte to \0. Thus, at the original location, the string becomes empty. The remainder of the bytes still contain l i c e \0. Advancing the pointer by one then points to "lice", which prints fully; further increments produce suffixes.
Step-by-Step Solution:
Before assignment: [A][l][i][c][e][\0]After *x = x[n]: [\0][l][i][c][e][\0]i = 0 → prints "" (empty), pointer moves to li = 1 → prints "lice"i = 2 → prints "ice"i = 3 → prints "ce"i = 4 → prints "e"i = 5 → prints ""Ignoring the empty prints, the visible output is: lice ice ce e
Verification / Alternative check:
If char x[] = "Alice"; (array, not literal), the write is valid and the same visible output results.
Why Other Options Are Wrong:
(a) and (c) would require the first byte to remain 'A'. (b) is reversal, which this code does not do. (e) can happen in real systems but the exercise assumes otherwise.
Common Pitfalls:
Forgetting that %s prints until \0 and that pointer increments shift the starting point; overlooking undefined behavior of modifying string literals.
Final Answer:
lice ice ce e
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