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What can a switch statement expression evaluate to?
byte short char int or as of Java 5 enum
int x = 2; final int a; a = 2; switch(x) { default: System.out.println(""Default""); case a: System.out.print(""a""); } What prints?
Compiler error. Variable a while final is not a compile time constant. final int a = 2; (IS compile constant) final int a; a = 2; (is NOT) COMPILE CONSTANT REQUIREMENTS: 1) Declared final 2) Assignment at same time as declaration 3) The assigned value is a literal OR another compile time constant
Are duplicate labels allowed in switch statements? ... case 10: case 10: ...
byte b = 1; switch(b) { case 128: System.out.println(""Default""); } What prints?
Compiler error. The case label 128 is of a value too large for the byte type.
Name 5 ways to exit out of a loop.
1. break statement 2. System.exit() 3. Exception thrown 4. return 5. meet the conditional requirement
When does a while statement end with a semicolon?
When it's part of a do-while statement. i.e. int x = 0; do{ x++; } while (x<2);
int[] arr = {1 2 3}; int element; for (element : arr) { System.out.println(element); } Output?
Complier error. The variable ""element"" is already declared. A for loop that loops through arrays requires that a newly created block variable be declared. i.e. for (int x : arr);
for (int i=0; i<0; i++); Legal?
Yup. This for loop instead of a block simply loops on the first statement following it. In this case that statement is simply a semicolon.
for (int i; i<0; i++); Legal?
No. i is a local variable and so has not been explicitly initialized. You cannot begin using it as an operand a relation operator or increment operator without first assigning it a value. i.e. for (int i=0; i<0; i++);
When does a while statement end with a semicolon?
When it's part of a do-while statement. i.e. int x = 0; do{ x++; } while (x<2);
Can a try block exist without a catch or finally block?
No. A try block must have at least a catch block or a finally block. Otherwise the code will cause a compiler error.
try{ throw new Exception(); } System.out.println(""End Try.""); catch (Exception e) { System.out.println(""Exception.""); } What prints?
Compiler error. You cannot have any statements between a try block and a catch block. Compiler assumes a try block has been declared without a catch or finally block when it gets to the intruding statement and complains that the try statement needs one of its buddies to be legal.
try{} catch(Exception e) {System.out.println(""A"");} catch(RuntimeException e){System.out.println(""B"");} Will ""B"" ever be printed?
Compiler error! It's a compiler error to specify a catch clause based on a Throwable object that's will be caught in a preceeding catch clause.
Can you nest a try/catch/finally block inside a catch block?
Yes. If you take the error caught by the initial catch block and then throw it in a nested try block you can then catch it in an accompanying NESTED catch block. NOTE: Any other un-nested catch blocks (peers to our original outer catch block) will NOT be reconsidered by any exceptions thrown in the nested try/catch/finally blocks. Also. If any catch block turns around and throws any error - it will not be caught by itself or any of it's peer catch blocks.
try{ throw new Error(); } catch(Error e) { System.out.print(""Error ""); try { throw new RuntimeException(); } finally { System.out.print(""Finally Inner ""); } } catch (RuntimeException e) { System.out.print(""RuntimeException ""); } finally{ System.out.print(""Finally Outer ""); } What prints?
Error Finally Inner Finally Outer
The curly braces are optional in the body of an if statement? True or False?
boolean doStuff() {return true;} ... if (doStuff()){} Compiles? Runs?
Yes and yes. The expression in an if statement must simply be a boolean expression.
What is the format of a switch statement?
switch(expression) { case const1: 0 to N statements and/or blocks case const2: 0 to N statements and/or blocks default: 0 to N statements and/or blocks } When the following statement is encountered the switch ends: break;
You can use expressions that evaluated to a byte short char int or long in a switch statment. True or False?
False. Longs cannot be used. As of Java 5 enums can be used.
final int label1 = 2; final int label2 = 2 * 2; final int label3 = label1 * label2; final int label4 = label3 + 2; final int label5 = -1; switch(1) { default: case label1: case label2: case label3: case label4: case label5: System.out.println(""Works""); } What prints?
Works. All the case labels qualify as compile time constants so this switch statement compiles and runs. Whena a variable qualifies as a compile time constant it then itself can be treated as if it were a literal
switch(1) { default: case 2: { break; System.out.println(""Works""); } } What prints?
Compiler error. The statement after break; can never be reached and compiler will throw error.
switch(1) { default: 2: { System.out.println(""Works""); break; } } What prints?
Complier error. Cases actually require the keyword ""case"". So in this situation 2: should read case 2: It would then compile run and print ""Works""
switch(1) { default: case 2: { System.out.println(""Works""); break; } } What prints?
while (true); Compiles?
Yes. And it will run as an endless loop (boo). Curly braces (block) are not required. while(boolean) block or statment
while(1){} Compiles?
No. The expression in a while statement must be of type boolean not int.
int i=0; for (int i=0; i<10; i++) {} Compiles?
No. Compiler will announce that i is already declared. You cannot shadow in this case.
for (int i=0 String s=""""; i<10;i++ s +=""+""); Runs?
Compiler error (so no it doesn't run either). In the initialization section you can either DECLARE (plus set values) or INITIALIZ (you have to do 1 or the other - can't mix). Here we are doing DECLARATION. In this case you can NOT mix type. If we had only been INITIALIZING variables defined outside for loop we could have mixed type.
int i=0; String s = """"; for (i=0 s=""""; i<10;i++ s +=""+""); Compiles?
What are the rules for the initialization section of a traditional for statement?
1. Must be a a) Declaration statement [where you can also optionally set initial values for the variablese declared] OR b) Initialization statement 2) If Initialization then type can be mixed 3) If Declaration only 1 type allowed 4) You can't mix Declaration and Initialization kinds of statements
for (int i=0; i<10;i++); int i = 10; System.out.println(i); What prints?
10. Notice that once for loop ends it's ""i"" goes out of scope so it's perfectly legal to then redeclare ""i"".
for (int i=0; i<10;i++); System.out.println(i);
Compiler error. ""i"" goes out of scope at end of for loop so attempting to print an undefined variable.
i: for (int i=0; i<3; i++){ System.out.println(i); continue i; } What prints?
0 1 2 Albeit poor style but it works.
How many test expressions can you have in a traditional for statement?
What is the definition of the new for statement?
for (declaration : expression) statement or block where delcaration is a NEWLY DECLARED block variable of a type compatible with the elements of the array being accessed and expression evaluates to an array
for (int i=0; i<0; i++){ System.out.println(i); next; } What prints?
Compiler error. next is not a known symbol. continue; is the statement used to skip the current itteration of the loop
Are these 2 the same? i: for (int i=0; i<3; i++){ System.out.println(i); continue i; } for (int i=0; i<3; i++){ System.out.println(i); }
for (int i=0; i<10; i++){ continue i; } Compiles?
No. By putting ""i"" after continue i will be treated like a label. Since there is no label with the title ""i"" (the string ""i"" - not the value of the variable i) compiler complains.
Can continue statement be written outside of a loop? How about a break statement?
No and yes. A break statement can also be used in a switch statement.
What constitutes a legal statement label?
A label is just another identifier. It follows the same rules as all other identifiers. It's followed by a colon and whitespace
What is they syntax of a label declaration? of a break statement using that label?
cats: for(int i=0; i<1;i++){ break cats; }
cats: for(int i=0; i<10;i++){break dogs;} dogs: for(int i=0; i<10;i++){} Compiles?
No. Labeled continue and break statements must be inside the loop that has the same label name.
What is the hierarchy of Object Throwable Error Exception and RuntimeException classes?
Object at top. Throwable descends from Object. Error and Exception descend from Throwable RuntimeException descends from Exception. NOTE 1: Anything that is a Throwable or a subtype can be thrown. NOTE 2: All Exceptions (except RuntimeException and its subclasses) are ""CHECKED"" exceptions and must be handled by the programmer.
In regards to checked exceptions what is ""handle or declare""?
A method that MIGHT produce a checked exception must either handle it using a try/catch/finally clause or must add a throws clause with the offending exception to its signature.
When do the following get thrown: ArrayIndexOutOfBoundsException ClassCastException IllegalArgumentException IllegalStateException NullPointerException NumberFormatException AssertionError ExceptionInitializerError StackOverflowError NoClassDefFoundError
Look it up buddy (page 370).
static String s; static public void main (String[] args) { System.out.println(s.length()); } Compiles?
Yes. But it does not run. At runtime a NullPointerException is thrown. We easily see that s is null but compiler doesn't. It never looks for what's actually IN s it's just checking to make sure a thing of type String behaves in ways the programmer is asking it to (in this case a programmer says that a String has a length() method which the compiler knows to be a perfectly okay behavior).
void go() { go(); } What happens?
Compiles fine. StackOverflowError is thrown at runtime.
What's the difference between Exceptions thrown by the JVM and progromatically thrown Exceptions?
programatically means Exceptions thrown by a programmer and/or the API. An example would be a call to a wrapper object's parseXxx method where the String argument was improperly formatted. (NumberFormatException gets thrown). JVM is just that - the Java Virtual Machine throws. An example would be if the JVM was asked to call the length() method on a String variable that was set to null (NullPointerException).
How do assertions work?
You assert something to be true. If it is no problem. If what you say is true is actually false then a AssertionError is thrown THAT SHOULD NEVER BE CAUGHT! (Whole point is that during development these errors scream out at you so they can get noticed - don't want to handle them away)
What is the syntax of an assertion?
assert(boolean expression); OR assert(boolean expression): statement that results in a value; In the latter form a value would be a primitive or object which will then get converted to a string (much like println() converts prims and objects)).
What happens to a string you add to an assert statement?
It gets added to the stack trace.
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