Assignment vs. comparison in an if condition: determine the compile-time outcome for this Java snippet.\n\nint x = 3;\nint y = 1;\nif (x = y) // uses assignment, not comparison\n{\n System.out.println("x =" + x);\n}

Introduction / Context:
This checks whether you can distinguish assignment from comparison in Java conditions. Unlike some languages, Java requires a boolean expression inside if. Assigning an int inside the condition is not allowed.

Given Data / Assumptions:

• Two integers: x = 3, y = 1.
• The condition uses x = y (assignment) instead of x == y (equality comparison).

Concept / Approach:
The assignment expression x = y yields an int value, not a boolean. Java will not convert non-zero integers to true automatically; thus the compiler reports “incompatible types: int cannot be converted to boolean”.

Step-by-Step Solution:
Compilation: the parser accepts the syntax, but type checking fails.Error message: assignment returns int; if expects boolean.No runtime output is possible.

Verification / Alternative check:
Change to if (x == y) and the code will compile and run. Alternatively, explicitly test a boolean condition, e.g., if ((x = y) == 1) (though that is poor style).

Why Other Options Are Wrong:
They assume execution proceeds; it cannot due to the type error.

Common Pitfalls:
Accidentally writing = instead of ==; always enable compiler warnings or use linters/IDE inspections.

Final Answer:
Compilation fails.