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This might asked too many times before but...

String str = "74.0";

let say I have "74.05" and I only need the 74. Below are some of my ways..

Integer.parseInt(str); //would result to java.lang.NumberFormatException
(int) Double.parseDouble(str); //working
new BigDecimal(str).intValue(); //I like this

I want this string(str) converted to int or Integer, what is the best way to do it?

yesterdaysfoe
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4 Answers4

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The java.lang.NumberFormatException is due to the fact that you are trying to convert a double literal to a an int type.

There is not a best way but the usage of

Double.parseDouble("74.05");

instead of

Double.valueOf("74.05");

depends on your need of a double primitive type or a Double object type.

If you need an integer type, you can round the double value like this:

Math.round(Double.parseDouble("74.05"));

or simply cast the double to obtain the integral part only

(int)Double.parseDouble("74.05")

The new BigDecimal(str).intValue(); is surely the worst choiche because can lead to unespected results as stated in the Oracle documentation (see the bold):

public int intValue()

Converts this BigDecimal to an int. This conversion is analogous to the narrowing primitive conversion from double to short as defined in section 5.1.3 of The Java™ Language Specification: any fractional part of this BigDecimal will be discarded, and if the resulting "BigInteger" is too big to fit in an int, only the low-order 32 bits are returned. Note that this conversion can lose information about the overall magnitude and precision of this BigDecimal value as well as return a result with the opposite sign.

Unrelated to the question, there is a sometime faster way when you need to instantiate an Integer from an int value. From the java.lang.Integer class documentation:

public static Integer valueOf(int i)

Returns an Integer instance representing the specified int value. If a new Integer instance is not required, this method should generally be used in preference to the constructor Integer(int), as this method is likely to yield significantly better space and time performance by caching frequently requested values. This method will always cache values in the range -128 to 127, inclusive, and may cache other values outside of this range.

Community
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fantaghirocco
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  • Any value that is too big to fit into a `int` will not be good even if you don't use `BigDecimal`. – Jonathan Drapeau Aug 07 '15 at 15:54
  • **"if the resulting "BigInteger" is too big to fit in an int, only the low-order 32 bits are returned"** I think same thing happens when you cast Double to int – yesterdaysfoe Aug 07 '15 at 16:03
  • @Jonathan Drapeau and @yesterdaysfoe you both are right but my point was that `Integer.parseInt()` raises an exception while `new BigDecimal(str).intValue()` does not. Anyway I missed the point that `Double.parseDouble(str)` does not raise an exception too returning `+/-Infinity` – fantaghirocco Aug 24 '15 at 13:49
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int x = (int) Double.parseDouble(s);

ran in 3 nano seconds,

int y = new BigDecimal(s).intValue();

ran in 8 nano seconds

j.con
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a really somewhat unorthodox solution would be:

Integer.parseInt(str.split("\\.")[0]);

but it works with Strings like "74.09" and Strings like "74" and so on

BadK
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  • this may work but I think this will process two different strings. Since, String is immutable it will consume more memory. – yesterdaysfoe Aug 07 '15 at 15:59
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How about

Integer myNum = Integer.parseInt(str.substring(0, str.indexOf("."));

since the OP explicitly wanted NO rounding

Jens Greven
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