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Probably too obvious.
public static void main(String[] varargs) throws Exception{ char a, b = (char)Integer.parseInt("000d",16); // Chars have \u000d as value, so they're equal if(a == b){ throw new Exception("This should be thrown"); } } What?
Throws a syntax error after
\u000d.\u000dis the unicode for a new line. Even though it is commented out, the Java compiler treats what is after this as code since it isn't commented out anymore.
#Java
Probably too obvious.
public static void main(String[] varargs) throws Exception{ char a, b = (char)Integer.parseInt("000d",16); // Chars have \u000d as value, so they're equal if(a == b){ throw new Exception("This should be thrown"); } } What?
Throws a syntax error after
\u000d.\u000dis the unicode for a new line. Even though it is commented out, the Java compiler treats what is after this as code since it isn't commented out anymore.
Probably too obvious.
public static void main(String[] varargs) throws Exception{ char a, b = (char)Integer.parseInt("000d",16); // Chars have \u000d as value, so they're equal if(a == b){ throw new Exception("This should be thrown"); } } What?
Throws a syntax error after
\u000d.\u000dis the unicode for a new line. Even though it is commented out, the Java compiler treats what is after this as code since it isn't commented out anymore.
#Java
Probably too obvious.
public static void main(String[] varargs) throws Exception{ char a, b = (char)Integer.parseInt("000d",16); // Chars have \u000d as value, so they're equal if(a == b){ throw new Exception("This should be thrown"); } } What?
Throws a syntax error after
\u000d.\u000dis the unicode for a new line. Even though it is commented out, the Java compiler treats what is after this as code since it isn't commented out anymore.
Probably too obvious.
public static void main(String[] varargs) throws Exception{ char a, b = (char)Integer.parseInt("000d",16); // Chars have \u000d as value, so they're equal if(a == b){ throw new Exception("This should be thrown"); } } What?
Throws a syntax error after
\u000d.\u000dis the unicode for a new line. Even though it is commented out, the Java compiler treats what is after this as code since it isn't commented out anymore.
#Java
Probably too obvious.
public static void main(String[] varargs) throws Exception{ char a, b = (char)Integer.parseInt("000d",16); // Chars have \u000d as value, so they're equal if(a == b){ throw new Exception("This should be thrown"); } } What?
Throws a syntax error after
\u000d.\u000dis the unicode for a new line. Even though it is commented out, the Java compiler treats what is after this as code since it isn't commented out anymore.
Probably too obvious.
public static void main(String[] varargs) throws Exception{ char a, b = (char)Integer.parseInt("000d",16); // Chars have \u000d as value, so they're equal if(a == b){ throw new Exception("This should be thrown"); } } What?
Throws a syntax error after
\u000d.\u000dis the unicode for a new line. Even though it is commented out, the Java compiler treats what is after this as code since it isn't commented out anymore.