Given a float f = 43.20, what do these printf conversions produce?\n\n#include<stdio.h>\nint main()\n{\n float f = 43.20;\n printf("%e, ", f);\n printf("%f, ", f);\n printf("%g", f);\n return 0;\n}\n\nAssume typical IEEE-754 and default precisions for %e, %f, and %g.

Introduction / Context:
The format specifiers %e, %f, and %g print floating-point values differently. Understanding each specifier and the effect of binary rounding explains the exact output.

Given Data / Assumptions:

• %e prints in exponential notation with six fractional digits by default.
• %f prints fixed-point with six fractional digits by default.
• %g chooses either %f or %e in a compact form, trimming trailing zeros.
• float values are promoted to double when passed to printf.

Concept / Approach:
Compute how each format represents 43.20. Due to binary rounding, the closest representable float near 43.20 may be slightly larger than 43.2, leading to 43.200001 with %f.

Step-by-Step Solution:
%e → "4.320000e+01".%f → prints six decimals → "43.200001" on many systems because 43.20 is not exactly representable in binary.%g → compact form → "43.2".

Verification / Alternative check:
Printing with higher precision (e.g., "%.9f") will reveal the tiny rounding difference more clearly.

Why Other Options Are Wrong:
They contain impossible suffixes (like "f" in output) or arbitrary rounding that does not match default printf behavior.

Common Pitfalls:
Expecting %f to always show exactly the decimal literal; forgetting argument promotion of float to double in variadic functions.

Final Answer:
4.320000e+01, 43.200001, 43.2