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'When you concatenate a String and an integer, the result is a String. Internally, the code is inefficient because there are intermediate objects that are created behind the scene'.
Java Int to String
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String s = ""; s = Integer.toString(n);
that's it.
You don't even need the String s = "". You can just do
right?
Yes.
String s=""+n;
I did this!. I'll try the parser, never seen it before... Ok its constant
String s = Integer.toString(n); is showing an error
toString() used for objects
Can't you just use String s = String.valueOf(n);
Yes we can.
really smart way
thanks sir
Yep, this is my favorite way.
Or, String s = String.valueOf(n);
I wonder if there's any difference between static String.valueOf and static Integer.toString...
Ok, looking at the source code,
Integer.toString
String.valueOf
String.valueOf just calls Integer.toString, so the only difference is an extra method invocation.
Yeah! Good point. And it's only a convenience feature where we can pass any datatyped-value I guess..
It's better to use valueOf, because you will never get NPE.
Yes! Integer.toString(); Is always prefered here
yes
That's good content. I like it much abcya
Also could be: String s = ""; s = s.valueOf(n);
or simply, String s = "" + n;
This was my solution as well.
String str = String.valueOf(12);
The code is insufficient as explained in the link http://javadevnotes.com/java-integer-to-string-examples
'When you concatenate a String and an integer, the result is a String. Internally, the code is inefficient because there are intermediate objects that are created behind the scene'.
Perfect. if someone is about to start their career in Java can join this forum for getting a lot of free resources.
What a great resource thanks!
String s = null; if(n>=-100 && n<=100){ s = String.valueOf(n);
}