Given: public class Msg { public static String doMsg(char x) { return "Good Day!"; } public static String doMsg(int y) { return "Good Luck!"; } public static void main(String[] args) { char x = '8'…

The correct answer is A. Good Day! Good Luck! Option A is correct because Java resolves overloaded methods at compile time based on the declared type of the argument, not its value. x is declared as char, so doMsg(x) binds to doMsg(char) → "Good Day!"; z is declared as int (Java widened '8''s Unicode value 56 into it), so…

Question

Given:
public class Msg {
public static String doMsg(char x) {
return "Good Day!";
}
public static String doMsg(int y) {
return "Good Luck!";
}
public static void main(String[] args) {
char x = '8';
int z = '8';
System.out.println(doMsg(x));
System.out.println(doMsg(z));
}
}
What is the result?

Options

AGood Day! Good Luck!
BGood Day! Good Day!
CGood Luck! Good Day!
DGood Luck! Good Luck!
ECompilation fails

How the community answered

(62 responses)

A
71% (44)
B
5% (3)
C
16% (10)
D
6% (4)
E
2% (1)

Explanation

Option A is correct because Java resolves overloaded methods at compile time based on the declared type of the argument, not its value. x is declared as char, so doMsg(x) binds to doMsg(char) → "Good Day!"; z is declared as int (Java widened '8''s Unicode value 56 into it), so doMsg(z) binds to doMsg(int) → "Good Luck!".

Why the distractors fail:

B & D assume both calls resolve to the same overload, ignoring that x and z have different declared types.
C reverses the logic - it would require char to resolve to the int overload and vice versa, which contradicts Java's type-matching rules.
E is wrong because the code is perfectly valid; both calls have an exact type match to an available overload.

Memory tip: Think "declared type drives dispatch." The line int z = '8' may look like a char, but once stored in an int, the compiler only sees int when choosing which overload to call.

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