Consider the following C code using math.h:\n\n#include<stdio.h>\n#include<math.h>\n\nint main()\n{\n float i = 2.5;\n printf("%f, %d", floor(i), ceil(i));\n return 0;\n}\n\nWhat output is most consistent with typical platform behavior, noting the incorrect format specifier for ceil?

Introduction / Context:
This question highlights two concepts: what floor and ceil return (both return double), and the necessity of matching printf format specifiers to argument types. Mismatched specifiers cause undefined behavior.

Given Data / Assumptions:

• i = 2.5.
• floor(i) returns 2.0 as a double.
• ceil(i) returns 3.0 as a double.
• printf uses "%f" for the first value (correct for double) and "%d" for the second (incorrect; expects int).

Concept / Approach:
Because "%d" does not match the double argument, the behavior is undefined by the C standard. On many common ABIs, passing a double where an int is expected often results in printing 0 due to how arguments are placed in registers/stack. The first value prints as 2.000000 correctly.

Step-by-Step Solution:
floor(2.5) = 2.0 → "%f" prints 2.000000.ceil(2.5) = 3.0 → but "%d" reads raw bits as int → undefined; often appears as 0.Thus a commonly observed output is: 2.000000, 0.

Verification / Alternative check:
Correcting the code to printf("%f, %f", floor(i), ceil(i)) would portably print 2.000000, 3.000000. The provided code is intentionally mismatched to test understanding.

Why Other Options Are Wrong:
2, 3 and 2, 0: they do not match the used format string "%f, %d" for the first field.2.000000, 3: would require a correct "%f" for the second field; not the case here.

Common Pitfalls:
Relying on undefined behavior; assuming implicit conversions happen in printf arguments—they do not.

Final Answer:
2.000000, 0