Home >Java >javaTutorial >Java Integer Comparison: When Should I Use `==` vs. `equals()`?
Integer Comparison Conundrum in Java: == vs. equals()
In Java 1.5 and beyond, Integers can be treated as primitive int values in many scenarios. However, a recent code issue has raised questions about the validity of using the == operator for Integer comparisons.
Problem Statement
Consider the following code snippet:
Integer cdiCt = ...; Integer cdsCt = ...; ... if (cdiCt != null && cdsCt != null && cdiCt != cdsCt) mismatch = true;
This code appears to set mismatch to true when the values of cdiCt and cdsCt are equal. The issue becomes apparent when examining the boolean expression: it evaluates to false when the values are indeed equal, but somehow mismatch is still set to true when execution continues.
Resolution
The solution to this problem is to use the equals() method instead:
if (cdiCt != null && cdsCt != null && !cdiCt.equals(cdsCt))
Explanation
The == operator compares the memory references of two objects, while the equals() method compares their values. For primitive types like int, the == operator works as expected. However, for objects like Integer, the == operator only checks if the references point to the same cached value.
The JVM caches Integer values between -128 and 127 for performance reasons. Therefore, when comparing two Integers using ==, the operator will return true only if both values fall within this range. In the given scenario, the values are likely just outside this range, leading to the unexpected behavior.
Conclusion
While using == to compare primitives is acceptable, it is recommended to use the equals() method for comparing wrapped objects like Integer. This ensures consistent behavior regardless of the values involved.
Additional Resources
The above is the detailed content of Java Integer Comparison: When Should I Use `==` vs. `equals()`?. For more information, please follow other related articles on the PHP Chinese website!