Use Long.parseLong()
Copy Long.parseLong("0", 10) // returns 0L
Long.parseLong("473", 10) // returns 473L
Long.parseLong("-0", 10) // returns 0L
Long.parseLong("-FF", 16) // returns -255L
Long.parseLong("1100110", 2) // returns 102L
Long.parseLong("99", 8) // throws a NumberFormatException
Long.parseLong("Hazelnut", 10) // throws a NumberFormatException
Long.parseLong("Hazelnut", 36) // returns 1356099454469L
Long.parseLong("999") // returns 999L
Answer from Mike Christensen on Stack OverflowUse Long.parseLong()
Copy Long.parseLong("0", 10) // returns 0L
Long.parseLong("473", 10) // returns 473L
Long.parseLong("-0", 10) // returns 0L
Long.parseLong("-FF", 16) // returns -255L
Long.parseLong("1100110", 2) // returns 102L
Long.parseLong("99", 8) // throws a NumberFormatException
Long.parseLong("Hazelnut", 10) // throws a NumberFormatException
Long.parseLong("Hazelnut", 36) // returns 1356099454469L
Long.parseLong("999") // returns 999L
To convert a String to a Long (object), use Long.valueOf(String s).longValue();
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Have a close look at the return types:
Long.parseLong(String)returns a primitivelong, so there will be re-boxing in this case:Long a = Long.parseLong(str);.new Long(String)will create a newLongobject in every case. So, don't do this, but go for 3)Long.valueOf(String)returns aLongobject, and will return cached instances for certain values -- so if you require aLongthis is the preferred variant.
Inspecting the java.lang.Long source, the cache contains the following values (Sun JDK 1.8):
private static class LongCache {
private LongCache(){}
static final Long cache[] = new Long[-(-128) + 127 + 1];
static {
for(int i = 0; i < cache.length; i++)
cache[i] = new Long(i - 128);
}
}
The best approach is Long.valueOf(str) as it relies on Long.valueOf(long) which uses an internal cache making it more efficient since it will reuse if needed the cached instances of Long going from -128 to 127 included.
Returns a
Longinstance representing the specified long value. If a new Long instance is not required, this method should generally be used in preference to the constructorLong(long), as this method is likely to yield significantly better space and time performance by caching frequently requested values. Note that unlike the corresponding method in the Integer class, this method is not required to cache values within a particular range.
Generally speaking, it is a good practice to use the static factory method valueOf(str) of a wrapper class like Integer, Boolean, Long, ... since most of them reuse instances whenever it is possible making them potentially more efficient in term of memory footprint than the corresponding parse methods or constructors.
Excerpt from Effective Java Item 1 written by Joshua Bloch:
You can often avoid creating unnecessary objects by using static factory methods (Item 1) in preference to constructors on immutable classes that provide both. For example, the static factory method
Boolean.valueOf(String)is almost always preferable to the constructorBoolean(String). The constructor creates a new object each time it’s called, while the static factory method is never required to do so and won’t in practice.
You can create rather complex regular expression but it isn't worth that. Using exceptions here is absolutely normal.
It's natural exceptional situation: you assume that there is an integer in the string but indeed there is something else. Exception should be thrown and handled properly.
If you look inside parseLong code, you'll see that there are many different verifications and operations. If you want to do all that stuff before parsing it'll decrease the performance (if we are talking about parsing millions of numbers because otherwise it doesn't matter). So, the only thing you can do if you really need to improve performance by avoiding exceptions is: copy parseLong implementation to your own function and return NaN instead of throwing exceptions in all correspondent cases.
From commons-lang StringUtils:
public static boolean isNumeric(String str) {
if (str == null) {
return false;
}
int sz = str.length();
for (int i = 0; i < sz; i++) {
if (Character.isDigit(str.charAt(i)) == false) {
return false;
}
}
return true;
}