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I'm parsing a string of 15 digits like this:

String str = "100004159312045";
int i = Integer.parseInt(str);

I'm getting an exception when doing so. Why? What are the limitations for Integer.parseInt? What other options do I have for converting such long string to a number?

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Long or BigDecimal. –  Dave Newton Nov 6 '11 at 21:24
thx, I forgot about that :P –  Alon Nov 6 '11 at 21:29

5 Answers 5

up vote 15 down vote accepted

Your number is too large to fit in an int which is 32 bits and only has a range of -2,147,483,648 to 2,147,483,647.

Try Long.parseLong instead. A long has 64 bits and has a range of -9,223,372,036,854,775,808 to 9,223,372,036,854,775,807.

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because int's maximum value is a little above 2,000,000,000

you can use long or BigInteger

long has double the digits it can store (the max value is square that of int) and BigInteger can handle arbitrarily large numbers

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int is 32 signed bit, its max value is 2^31-1 = 2147483647

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The value you are trying to parse exceeds the maximum size of an Integer.

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The range of int is between Integer.MAX_VALUE and Integer.MIN_VALUE inclusive (i.e. from 2147483647 down to -2147483648). You cannot parse the int value of out that range.

You can use long m = Long.parseLong(str); or BigInteger b = new BigInteger(str); for large integer.

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