About Question enthuware.ocajp.i.v7.2.1059 :
Posted: Wed Jan 23, 2013 1:15 am
I didn't know the return value of Math.random() can be represented like .05. Where is the specification?
Math.random()
HTH,
Paul.
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About Question enthuware.ocajp.i.v7.2.1059 :
Page 1 of 1
Re: About Question enthuware.ocajp.i.v7.2.1059 :
Posted: Wed Jan 23, 2013 6:23 am
Re: About Question enthuware.ocajp.i.v7.2.1059 :
Posted: Fri Sep 26, 2014 7:31 am
I wondered til now what the bounds were for the update clause.The third part (i.e. the update part) of the for loop does not allow every kind of statement. It allows only the following statements here: Assignment, PreIncrementExpression, PreDecrementExpression, PostIncrementExpression, PostDecrementExpression, MethodInvocation, and ClassInstanceCreationExpression.
Thank you, you've answered a nagging question for me
But where is this specified?
JLS §14.4 doesn't seem to cover it
EDIT-I'm looking now in JLS §15, Expressions to see if I can find the answer myself...
2nd EDIT-No luck in §15 but reading §2.2, Grammars has thrown light on what the grey syntax definition boxes are about
Maybe I need to take a day or two to read JLS from the start to get to grips with its layout... It's fairly terse
Am I on the right track Paul?
no luck answering my own question above
Posted: Sun Sep 28, 2014 3:37 pm
Think my original question has got a bit lost in among my own Edits/Postscripts
So just to re-ask:
JLS §14.4 doesn't appear to cover it
Thanks
So just to re-ask:
Where is this specified?The third part (i.e. the update part) of the for loop does not allow every kind of statement. It allows only the following statements here: Assignment, PreIncrementExpression, PreDecrementExpression, PostIncrementExpression, PostDecrementExpression, MethodInvocation, and ClassInstanceCreationExpression.
JLS §14.4 doesn't appear to cover it
Thanks
Re: About Question enthuware.ocajp.i.v7.2.1059 :
Posted: Sun Sep 28, 2014 7:38 pm
It is given in 14.14.1:
Paul.
14.8 defines StatementExpression:ForUpdate:
StatementExpressionList
StatementExpressionList:
StatementExpression
StatementExpressionList , StatementExpression
HTH,ExpressionStatement:
StatementExpression ;
StatementExpression:
Assignment
PreIncrementExpression
PreDecrementExpression
PostIncrementExpression
PostDecrementExpression
MethodInvocation
ClassInstanceCreationExpression
Paul.
Re: About Question enthuware.ocajp.i.v7.2.1059 :
Posted: Mon Sep 29, 2014 8:01 am
Great stuff Paul
Thank you
Thank you
Re: About Question enthuware.ocajp.i.v7.2.1059 :
Posted: Thu May 14, 2015 9:20 am
I do not understand, is this a right number " .05 "- What is this?
Re: About Question enthuware.ocajp.i.v7.2.1059 :
Posted: Thu May 14, 2015 9:38 am
Not sure what is your confusion but yes, .05 is a valid double. BTW, there is no primitive data type called "number" in Java.
Please see this: https://docs.oracle.com/javase/tutorial ... types.html
Please see this: https://docs.oracle.com/javase/tutorial ... types.html
Re: About Question enthuware.ocajp.i.v7.2.1059 :
Posted: Fri Jul 17, 2015 6:51 pm
Pretty tough question! I got it wrong even though I have actually read the for loop part of the JLS some time ago. I was wondering why some expressions/statements are valid in a for loop and some are not.
(I am currently banging my head against the wall for answering that "Math.random()<.05? break : continue" is valid...)
(I am currently banging my head against the wall for answering that "Math.random()<.05? break : continue" is valid...)
Re: About Question enthuware.ocajp.i.v7.2.1059 :
Posted: Mon Jan 18, 2016 5:35 pm
I'm not sure I understand the explanation for the following:
for(;;){ Math.random()<.05? break : continue; }
"This is an invalid use of ? : operator. Both sides of : should return the same type (excluding void)."
The code bellow works just fine, even if both sides of ":" do not return same type:
String s = "true";
boolean b = false;
for(int i = 0; i < 2; i++){
System.out.println(i==1 ? s : b);
}
Am I missing something here?
for(;;){ Math.random()<.05? break : continue; }
"This is an invalid use of ? : operator. Both sides of : should return the same type (excluding void)."
The code bellow works just fine, even if both sides of ":" do not return same type:
String s = "true";
boolean b = false;
for(int i = 0; i < 2; i++){
System.out.println(i==1 ? s : b);
}
Am I missing something here?
Re: About Question enthuware.ocajp.i.v7.2.1059 :
Posted: Mon Jan 18, 2016 9:34 pm
You are right. The explanation is incorrect.
As per JLS section 15.25:
The explanation has been updated.
thank you for your feedback!
Paul.
As per JLS section 15.25:
So basically, the type of the value returned by both the operands is actually converted to a common type by using one of the above rules. In your example, the second last line of the above quote applies.The type of a conditional expression is determined as follows:
If the second and third operands have the same type (which may be the null type), then that is the type of the conditional expression.
If one of the second and third operands is of primitive type T, and the type of the other is the result of applying boxing conversion (§5.1.7) to T, then the type of the conditional expression is T.
If one of the second and third operands is of the null type and the type of the other is a reference type, then the type of the conditional expression is that reference type.
Otherwise, if the second and third operands have types that are convertible (§5.1.8) to numeric types, then there are several cases:
If one of the operands is of type byte or Byte and the other is of type short or Short, then the type of the conditional expression is short.
If one of the operands is of type T where T is byte, short, or char, and the other operand is a constant expression (§15.28) of type int whose value is representable in type T, then the type of the conditional expression is T.
If one of the operands is of type T, where T is Byte, Short, or Character, and the other operand is a constant expression (§15.28) of type int whose value is representable in the type U which is the result of applying unboxing conversion to T, then the type of the conditional expression is U.
Otherwise, binary numeric promotion (§5.6.2) is applied to the operand types, and the type of the conditional expression is the promoted type of the second and third operands.
Note that binary numeric promotion performs value set conversion (§5.1.13) and may perform unboxing conversion (§5.1.8).
Otherwise, the second and third operands are of types S1 and S2 respectively. Let T1 be the type that results from applying boxing conversion to S1, and let T2 be the type that results from applying boxing conversion to S2.
The type of the conditional expression is the result of applying capture conversion (§5.1.10) to lub(T1, T2) (§15.12.2.7).
The explanation has been updated.
thank you for your feedback!
Paul.