Difficulty: Hard
Correct Answer: k=39
Explanation:
Introduction / Context:
This classic C puzzle checks understanding of nested function calls, right-associative assignments, and the fact that assignment is an expression whose value is the assigned value. The helper function increments its argument and returns the incremented value.
Given Data / Assumptions:
Concept / Approach:
Work from the innermost parentheses outward. Each func1 call adds 1 to its argument and returns that value. Each k = ... not only stores into k but also yields that stored value, which then becomes the argument to the next outer call.
Step-by-Step Solution:
Start k = 35.Innermost: func1(k) = 36; assign k = 36.Middle: need k = func1(k): with k = 36, func1(36) = 37; assign k = 37; the middle call is func1(37) = 38; then the middle assignment sets k = 38.Outer: evaluate func1(38) = 39; final outer assignment sets k = 39.Prints “k=39”.
Verification / Alternative check:
Replace nested form with sequential statements to verify: k=func1(k); k=func1(k); k=func1(k); would give 38; however the given expression performs an extra func1 due to the outer call applied to the already-assigned middle result, hence 39.
Why Other Options Are Wrong:
“k=38” overlooks that the outermost func1 runs after the middle assignment has already set k to 38. “k=35/36/37” ignore one or more increments.
Common Pitfalls:
Forgetting that an assignment returns the assigned value; miscounting the number of func1 calls; assuming chained comparisons or left-to-right argument evaluation semantics that do not apply to nested expressions.
Final Answer:
k=39
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