Sometimes java puzzles me.
I have a huge amount of int initializations to make.
What's the real difference?
Integer.toString(i)
new Integer(i).toString()
Integer.toString
calls the static method in the class Integer
. It does not need an instance of Integer
.
If you call new Integer(i)
you create an instance of type Integer
, which is a full Java object encapsulating the value of your int. Then you call the toString
method on it to ask it to return a string representation of itself.
If all you want is to print an int
, you'd use the first one because it's lighter, faster and doesn't use extra memory (aside from the returned string).
If you want an object representing an integer value—to put it inside a collection for example—you'd use the second one, since it gives you a full-fledged object to do all sort of things that you cannot do with a bare int
.
Another option is the static String.valueOf
method.
String.valueOf(i)
It feels slightly more right than Integer.toString(i)
to me. When the type of i changes, for example from int
to double
, the code will stay correct.
new Integer(i).toString();
This statement creates the object of the Integer and then call its methods toString(i)
to return the String representation of Integer's value.
Integer.toString(i);
It returns the String object representing the specific int (integer), but here toString(int)
is a static
method.
Summary is in first case it returns the objects string representation, where as in second case it returns the string representation of integer.
In terms of performance measurement, if you are considering the time performance then the Integer.toString(i); is expensive if you are calling less than 100 million times. Else if it is more than 100 million calls then the new Integer(10).toString() will perform better.
Below is the code through u can try to measure the performance,
public static void main(String args[]) {
int MAX_ITERATION = 10000000;
long starttime = System.currentTimeMillis();
for (int i = 0; i < MAX_ITERATION; ++i) {
String s = Integer.toString(10);
}
long endtime = System.currentTimeMillis();
System.out.println("diff1: " + (endtime-starttime));
starttime = System.currentTimeMillis();
for (int i = 0; i < MAX_ITERATION; ++i) {
String s1 = new Integer(10).toString();
}
endtime = System.currentTimeMillis();
System.out.println("diff2: " + (endtime-starttime));
}
In terms of memory, the
new Integer(i).toString();
will take more memory as it will create the object each time, so memory fragmentation will happen.
Here
Integer.toString
calls the static method in the class Integer. It does not require the object to call.
If you call new Integer(i)
you first create an instance of type Integer, which is a full Java object encapsulating the value of your int i. Then you call the toString
method on it to ask it to return a string representation of itself.
1.Integer.toString(i)
Integer i = new Integer(8);
// returns a string representation of the specified integer with radix 8
String retval = i.toString(516, 8);
System.out.println("Value = " + retval);
2.new Integer(i).toString()
int i = 506;
String str = new Integer(i).toString();
System.out.println(str + " : " + new Integer(i).toString().getClass());////506 : class java.lang.String
"" + 42
, but hey, that's me. Form #2 will result in a new integer object (for most values of i) as an intermediate. It is also more typing. – user166390"" + 42
isn't any better, as it requires an intermediateStringBuilder
which is quite a lot heavier than anInteger
. – Joachim Sauer"" + 42
because in the case of appending strings of integers you may end up adding the values and not realizing it. – BoltzmannBrain