1Z0-811 Primitives, Strings and Operators Practice Question
Which two of the following are valid ways to check if two String objects contain the same characters? (Assume s1 and s2 are non-null String references.)
Answer choices
Why each option matters
Answer the question above first, then reveal the full breakdown to understand why each option is right or wrong.
Correct answer & explanation
✓
if (s1.compareTo(s2) == 0) { ... }
The correct ways to compare two String objects for character equality are using the equals() method (option E) or compareTo() method returning 0 (option C). Option A (equalsIgnoreCase) ignores case, so it would treat strings with different cases as equal. Option B (==) compares object references, not content. Option D compares s1 with the interned version of s2; this only works if s1 is also interned, which is not guaranteed, making it unreliable.
Answer analysis
Option-by-option breakdown
For each option: why learners choose it and why it is or isn't the right answer here.
- ✗
if (s1.equalsIgnoreCase(s2)) { ... }
Why it's wrong here
equalsIgnoreCase() compares strings ignoring case differences, so it does not verify exact character equality.
- ✗
if (s1 == s2) { ... }
Why it's wrong here
The == operator compares memory references, not character content. It only returns true if both references point to the same object.
- ✓
if (s1.compareTo(s2) == 0) { ... }
Why this is correct
compareTo() returns 0 when the strings are lexicographically equal, indicating identical character content.
- ✗
if (s1 == s2.intern()) { ... }
Why it's wrong here
The intern() method returns a canonical representation, but using == is still reference comparison and not a reliable way to check content equality.
- ✓
if (s1.equals(s2)) { ... }
Why this is correct
The equals() method is designed to compare the character content of two strings.
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