Java rounding functions: count how many elements satisfy Math.round(x + 0.5) == Math.ceil(x).\n\npublic class Example {\n public static void main(String [] args) {\n double values[] = {-2.3, -1.0, 0.25, 4};\n int cnt = 0;\n for (int x = 0; x < values.length; x++) {\n if (Math.round(values[x] + .5) == Math.ceil(values[x])) {\n ++cnt;\n }\n }\n System.out.println("same results " + cnt + " time(s)");\n }\n}

Introduction / Context:
This problem compares two rounding-style expressions across an array of doubles: rounding x + 0.5 using Math.round, and taking Math.ceil of x. You must evaluate each element carefully, considering IEEE rounding rules, negative numbers, and boundary behavior.

Given Data / Assumptions:

• values = {-2.3, -1.0, 0.25, 4}.
• Test condition: Math.round(v + 0.5) == Math.ceil(v).
• Math.round rounds to nearest long with halves rounded up; Math.ceil returns the least integer ≥ v as a double.

Concept / Approach:
Work element by element, converting Math.round’s result to a long and Math.ceil’s result to a double; the == comparison promotes long to double for equality testing. Focus on how negative values behave with rounding and ceiling.

Step-by-Step Solution:
v = -2.3 → v + 0.5 = -1.8 → round(-1.8) = -2; ceil(-2.3) = -2 → equal (count++). v = -1.0 → v + 0.5 = -0.5 → round(-0.5) = 0; ceil(-1.0) = -1 → not equal. v = 0.25 → v + 0.5 = 0.75 → round(0.75) = 1; ceil(0.25) = 1 → equal (count++). v = 4.0 → v + 0.5 = 4.5 → round(4.5) = 5; ceil(4.0) = 4 → not equal. Total count = 2.

Verification / Alternative check:
Try v = 1.2 (round 1.7 = 2, ceil 1.2 = 2) to see another equality case; explore edge cases like v = -0.1.

Why Other Options Are Wrong:
0/1/4 miscount the matches; the code compiles fine.

Common Pitfalls:
Misunderstanding rounding with negatives; forgetting that round(4.5) becomes 5.

Final Answer:
same results 2 time(s)