NumberFormatExceprion for binary string

Question

I'm trying to get int from String "11010001110011000000000111111110" with code:

int n = Integer.parseInt("11010001110011000000000111111110", 2);

but i get an error:

Exception in thread "main" java.lang.NumberFormatException: For input string: "11010001110011000000000111111110"

Since the number is only 32 bits you can use `parseUnsignedInt`, but you will get a negative number. — David Conrad, Mar 16 '20 at 21:45

score 3 · Accepted Answer · answered Mar 16 '20 at 21:28

It is simply because the value represented by the string with binary content is too large for an int data type. You need to parse to long data type. The binary string represents a value of 3519807998. Integer maximum value (Integer.MAX_VALUE) is: 2147483647. Try this:

String binaryString = "11010001110011000000000111111110";
long n = Long.parseLong(binaryString, 2);

Or this way:

String binaryString = "11010001110011000000000111111110";
long lng = new BigInteger(binaryString, 2).longValue();
System.out.println(lng);

NumberFormatExceprion for binary string

1 Answers1