In C, how does == behave when comparing an int and a float with equal numeric value?\n\n#include<stdio.h>\nint main()\n{\n int x = 3;\n float y = 3.0;\n if(x == y)\n printf("x and y are equal");\n else\n printf("x and y are not equal");\n return 0;\n}

Introduction / Context:
This item checks the rules of the usual arithmetic conversions in C when comparing different types (int and float). It probes whether the integer is converted to a floating point value before the comparison.

Given Data / Assumptions:

• x is 3 (type int) and y is 3.0 (type float).
• The program uses == to compare them and prints a message accordingly.

Concept / Approach:
For the relational and equality operators, C applies the usual arithmetic conversions so both operands share a common type. With int and float, the int is converted to float. Therefore comparison becomes 3.0f == 3.0f which is true.

Step-by-Step Solution:
Convert x to float: (float)3 → 3.0f.Compare 3.0f == 3.0f → true.The if branch executes and prints “x and y are equal”.

Verification / Alternative check:
Changing y to 3.0000001f would likely make the comparison false after conversion. As written, exact equality holds with small integers representable exactly in float.

Why Other Options Are Wrong:
“Not equal” contradicts conversion rules. “Unpredictable/No output/Compilation error” are not applicable.

Common Pitfalls:
Believing that int and float are never equal; confusing floating-point rounding issues with integers exactly representable in float (all integers up to 2^24 are exactly representable in IEEE 754 single precision).

Final Answer:
x and y are equal