The answer is that the compilation error occurs because the value 200 exceeds the range of byte, which is -128 to 127. In Java, the byte data type is an 8-bit signed integer, so any literal assigned directly to a byte variable must fall within that narrow window; when you try to assign 200, the compiler detects a possible loss of precision and refuses to compile the code without an explicit cast. On the Oracle Java Foundations 1Z0-811 exam, this question tests your understanding of primitive data type ranges and implicit narrowing conversion rules—a common trap is assuming that any integer literal can be assigned to a smaller type, but Java strictly enforces type safety. A helpful memory tip is to think of a byte as a tiny container that can only hold values from -128 to 127; anything larger will overflow and cause a compilation error.
1Z0-811 Primitives, Strings and Operators Practice Question
This 1Z0-811 practice question tests your understanding of primitives, strings and operators. Read the scenario carefully and evaluate each option against the stated constraints before committing to an answer. After answering, compare your reasoning against the explanation and wrong-answer breakdown below. Once you have made your selection, read the full explanation to reinforce the concept and understand why each distractor is designed to mislead on exam day.
Exhibit
Refer to the exhibit.
javac Test.java
Test.java:3: error: incompatible types: possible lossy conversion from int to byte
byte b = 200;
^
1 error
Answer the question above first, then reveal the full breakdown to understand why each option is right or wrong.
Correct answer & explanation
✓
The value 200 exceeds the range of byte (-128 to 127)
The code attempts to assign the integer literal 200 to a variable of type byte. In Java, the byte data type has a range from -128 to 127. Since 200 exceeds this range, the compiler detects a possible loss of precision and raises a compilation error. Java does not automatically narrow a larger integer literal to fit into a byte without an explicit cast.
Key principle: Answer the scenario, not the keyword: identify the specific constraint before choosing the most familiar-sounding option.
Answer analysis
Option-by-option breakdown
For each option: why learners choose it and why it is or isn't the right answer here.
✓
The value 200 exceeds the range of byte (-128 to 127)
Why this is correct
Correct: 200 > 127, so cannot be assigned to byte without cast.
Related concept
Read the scenario before looking for a memorised answer.
✗
The literal 200 is a long
Why it's wrong here
Incorrect: 200 is an int literal by default.
✗
The assignment operator is invalid
Why it's wrong here
Incorrect: = is a valid assignment operator.
✗
The variable b is not declared
Why it's wrong here
Incorrect: 'byte b' declares the variable.
Common exam traps
Common exam trap: answer the scenario, not the keyword
The trap here is that candidates often confuse the range of byte with that of short or int, or mistakenly think the error is due to the literal being a long, when in fact Java treats all unsuffixed integer literals as int.
Detailed technical explanation
How to think about this question
Under the hood, Java's integer literals are of type int unless suffixed. When assigning an int literal to a byte variable, the compiler performs a compile-time check for range compatibility. If the literal is outside the byte's range (-128 to 127), the compiler rejects it to prevent silent truncation. This is a form of type safety that avoids data loss; an explicit cast like (byte)200 would compile but would wrap the value to -56 due to overflow.
KKey Concepts to Remember
Read the scenario before looking for a memorised answer.
Find the constraint that changes the correct option.
Eliminate answers that are true in general but not in this case.
TExam Day Tips
→Watch for words such as best, first, most likely and least administrative effort.
→Review why wrong options are wrong, not only why the correct option is correct.
Key takeaway
Answer the scenario, not the keyword: identify the specific constraint before choosing the most familiar-sounding option.
Real-world example
How this comes up in practice
A practitioner preparing for the 1Z0-811 exam encounters this exact type of scenario on the job. The correct answer here is not the most general option — it is the best answer for the specific constraint described. Answer the scenario, not the keyword: identify the specific constraint before choosing the most familiar-sounding option. Real exam questions reward reading the full scenario before eliminating options, because the constraint defines which answer fits.
What to study next
Got this wrong? Here's your next step.
Identify which exam domain this question belongs to, review the core concept, then practise similar questions from the same domain.
Primitives, Strings and Operators — This question tests Primitives, Strings and Operators — Read the scenario before looking for a memorised answer..
What is the correct answer to this question?
The correct answer is: The value 200 exceeds the range of byte (-128 to 127) — The code attempts to assign the integer literal 200 to a variable of type byte. In Java, the byte data type has a range from -128 to 127. Since 200 exceeds this range, the compiler detects a possible loss of precision and raises a compilation error. Java does not automatically narrow a larger integer literal to fit into a byte without an explicit cast.
What should I do if I get this 1Z0-811 question wrong?
Identify which exam domain this question belongs to, review the core concept, then practise similar questions from the same domain.
What is the key concept behind this question?
Read the scenario before looking for a memorised answer.
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Question Discussion
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