Assignment vs. comparison in an if condition: determine the compile-time outcome for this Java snippet. int x = 3; int y = 1; if (x = y) // uses assignment, not comparison { System.out.println("x =" + x); }
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Ax = 1
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Bx = 3
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CCompilation fails.
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DThe code runs with no output.
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EIt prints x = 0
Answer
Correct Answer: Compilation fails.
Explanation
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 ofx == 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.