output i want to generate:Code:public class test {
public static void main(String[] args) {
int i =0 ;
for( i =0 ; i<=25; i++)
{
System.out.printf("%d", i);
(i%5==0)?System.out.println():System.out.print(" ");
}
}
0 1 2 3 4
5 6 7 8 9
...
P.S. sorry the title is incorrect， it should be (i%5==0) instead
What exactly do you mean, "doesn't work"?
Errors? if so paste them, incorrect output? Post the given output.
sorry ,i don't express well .
this statement (i%5==0)?System.out.println():System.out.print(" ");
the compiler shows the message " The left-hand side of an assignment must be a variable"
The message you get is not very clear. And neither is Oracle's Tutorial statement that ?: "can be thought of as shorthand for an if-then-else statement" because it is a little more limited.
From the JLS, 15.25 Conditional Operator ? : "Note that it is a compile-time error for either the second or the third operand expression to be an invocation of a void method." So the bpttom line is that if you want to conditionally evaluate one of two void methods, then you have to use if/else.
(A general point - which you should feel free to ignore for now - is that ?: will form an expression which always has a type. if/else on the other hand forms a statement that has no type associated with it.)
In the statement [a ? b : c] (without [ ]), a most return boolean (which your does), and b and c most return then same data type (or same class),Quote:
Originally Posted by oszc
System.out.println and System.out.print does not return anything (void). (you should be able to use methods returning Void, but not void).
So you must use regular [outline] if-statements, or:
Code:System.out.print((i % 5 == 0) ? '\n' : ' ');
One comment I'd like to add:
With some thought you can accomplish this(hint: System.out.print(...), can take an argument). Although this works, you should favor clarity over brevity.
Using the ternary operator may indeed save you a couple of lines of code, however; it could also increase the complexity of the code. If you are insistent on using the clever styled ternary operator you should be sure to comment what the code is doing(which forfeits the brevity the operator saved you).
That being said, this is not a rule, but something to keep in mind when deciding whether or not to use the ternary operator.
I don't think ?: expressions are OS dependent are they? It would seem rather unjavalike if they were.
The second and third expressions don't have to be the same type - numeric promotion, boxing conversion, capture conversion can all go on (as detailed in the link I gave). And subclasses and nulls can be present. All of which gives plenty of scope: for good or ill.
(favor clarity over brevity)++
Well, they only need to be castable to the same type without loss if data, i.e. implicitly castable, which includes subclasses. But you can not write
? true : 1
? "hello" : 'w' //note that one of them is a String, and the other is a char (character).
And so on.
But why go into the complex when we are just describing the basics, implicitly castable is one of the things beginners usually have problems with.
And I think ?: is less OS dependent then +, which I believe is processor dependent and behaves exactly the same for all processors.
Perhaps I am wrong, but I thought '\n' was an os dependent character.
No, it is not.Quote:
Originally Posted by sunde887
\n = 10
\r = 13
\f = ? //I can't bother to look it up
Unix standard for new lines: \n (Mac OS X is Unix-like)
Mac standard for new lines: \r (Mac OS X is not included, see above)
Windows standard for new lines: \r\n
(There are more, but less common, variants.)
However all of them generally accepts \n as a new line character as it is programming standard, but text editor, might not.
So \n is not OS dependent.
However [the result of printing] \eE is shell dependent.
Yet another edit:
I'm not sure if this is for Java, but I think it is. In text mode (not the binary mode which is standard) in file writing/reading \n gets converted to the OS default new line character sequence.
Oh I see! Yes, to properly replicate the intent of being like println() you would have to append line.separator rather than \n.Quote:
Perhaps I am wrong, but I thought '\n' was an os dependent character.
My reason for citing the JLS in the first place was to give authority for the simple statement about the second and third expressions of the ternary operator. I referred to it again only to reject the erroneous assertion that those expressions have to have the same type. I'll cite it again for anyone interested in deciding whether it's a matter of castibility. The following compiles fine:Quote:
Well, they only need to be castable to the same type without loss if data, i.e. implicitly castable, which includes subclasses. But you can not write
? true : 1
? "hello" : 'w' //note that one of them is a String, and the other is a char (character).
And so on.
But why go into the complex when we are just describing the basics, implicitly castable is one of the things beginners usually have problems with.
I don't mean to be argumentative (and hope you don't take it that way, Hibernate) but I think what the ternary operator does is complex in a way that if-then-else statements are not. The ternary operator provides brevity and, used simply, is clear. In less simple circumstances its meaning may not be clear. I think Sunde had it right (without telling the OP what to do): favour clarity over brevity.Code:public class Test {
public static void main(String[] args) {
Object foo = true ? "hello" :'w';
System.out.println(foo);
foo = false ? true : 1;
System.out.println(foo);
}
}
-----
But where is the most important person in the thread? All the variants mentioned start with a new line: is that intended? Again whether they end with a new line depends on the number of things being printed. Perhaps it would make more sense to print the separators *before* the data elements.
Dude that is auto-boxing (ugly!), making both expressions Object.Quote:
Originally Posted by pbrockway2
But I do think you can do that without declaring foo, at least not in Java 6.
And no I do not that it as that you are argumentative.
But shouldn't you can't a compiler warning, or maybe error, from
foo = true ? "hello" :'w';
and
foo = false ? true : 1;
since they use constant booleans (in this case literals) as conditions, making it always give the same output?
The only warning you'll (probably) get is "dead code" but autoboxing makes those expressions perfectly valid (although highly cryptic). Had Gosling decided that Java should have no primitives he would've defend himself on attacks arguing about speed reasons; but I agree, autoboxing can be ugly and just a mixed blessing.Quote:
Originally Posted by Hibernate
kind regards,
Jos
Exactly. And whether ugly or not it is certainly complex (which was my only point really).Quote:
Dude that is auto-boxing (ugly!), making both expressions Object.
Using constants for the condition doesn't elicit any warning or anything. I only did it that way to make a brief example. Likewise declaring foo: you have to do that if the ?: is to be an expression-statement (or whatever the term is). But the ?: expressions you originally came up with can both be used without having to declare anything:
Running it by Eclipse, I see it does warn of "dead code" when boolean literals are used. But beware of thinking that true will always return the first alternative!Code:import java.util.Scanner;
public class Test {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
boolean cond = in.nextLine().isEmpty();
// we'll declare foo so we can assign the ternary expression to something
Object foo = cond ? "hello" :'w';
System.out.println(foo);
// here another ugly ternary is used without any declaration
System.out.println(cond ? true : 1);
}
}
(As discussed too many years ago at Sun's forum.)Code:public class Test {
public static void main(String[] args) {
boolean cond = 666 > 0;
System.out.println(cond ? 1 : 1.0);
System.out.println(cond ? 1 : "one");
}
}
Using ?: in multiple languages must have gotten me confused on this.