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I want to know which method is faster?

Integer.valueOf(String string) or Integer.parseInt(String string)?

Is there any Performance or Memory difference between the two approaches?

I have seen Difference between parseInt and valueOf in java? but this does not explains difference in terms of performance.

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    See [this][1]. Please search google and SO before asking questions. [1]: stackoverflow.com/questions/508665/… Commented Dec 18, 2012 at 12:44
  • you can see stackoverflow.com/questions/7355024/… Commented Dec 18, 2012 at 12:44
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    My question is about performance. not about difference between these to methods. Commented Dec 18, 2012 at 12:47
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    you should reopen the question. because it is about performance that is not discussed in mentioned answers. Commented Dec 18, 2012 at 12:52
  • Integer.valueOf("1") is functionally equivalent to Integer.valueOf(Integer.parseInt("1")); alternatively Integer.parseInt("2") is equivalent to Integer.valueOf("2").intValue(). Because they return different types, there is no situation where one is a direct replacement, so you need to ask about the performance of whichever one gives you the type you actually require, or better how to get the performance you require for a larger chunk of your code. Commented Dec 18, 2012 at 14:24

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I would not look at performance. The API says that Integer.valueOf(String) is interpreted the same as if it has been passed to Integer.parseInt(String), except it is wrapped into an Integer. I would look at what you need: an Integer or an int.

Integer.valueOf returns an Integer.

Integer.parseInt returns an int.

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Integer.valueOf() uses Integer.parseInt() internally and valueOf returns Integer Object whereas parseInt() returns int. So parseInt() is faster.

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valueOf(String) returns a new Integer() object whereas parseInt(String) returns a primitive int.

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This can be easily be shown to be false by running System.out.println(Integer.valueOf ( "1" ) == Integer.valueOf ( "1" ));
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Integer.valueOf(String string) returns a newly created wrapped object.

Integer i = Integer.valueOf("5");

Integer.parseInt(String string) returns the named primitive.

int i = Integer.parseInt("5");

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6 Comments

So if I invoke Integer.valueOf("2") from MyClass, then it returns a new instance of MyClass, because that is the type which invoked the method?
why are so much concern, which one faster, even they are not same. I mean, their return type are different.
@PeteKirkham please refer kathy sierra book (SCJP Guide).And, i believe, It doesn't mean that, it means, it return Integer class object.
@PeteKirkham, now is this ok ?? or still require more improvement ??
@PeteKirkham, either tell me what's wrong in my answer or undo your downvote.
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