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I have a file with many hex numbers (for eg - 0X3B4 ). Im trying to parse this file as assign these numbers to integers, but dont seem to get Integer.parseInt to work.
int testint = Integer.parseInt("3B4",16); <- WORKS
int testint = Integer.parseInt("0X3B4",16);
gives error:
Exception in thread "main" java.lang.NumberFormatException: For input string: "0x3b4"
at java.lang.NumberFormatException.forInputString(NumberFormatException.java:48)
What is the right way to assign the value 0XB4 to an int ?
Do I have to get rid of the 0X - its not unusual to represent hex nos this way...
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For Hexadecimal, the 0x or 0X is to be placed in the beginning of a number. Note − Digits 10 to 15 are represented by a to f (A to F) in Hexadecimal. Here are some of the examples of hexadecimal integer literal declared and initialized as int.
In Java code (as in many programming languages), hexadecimal nubmers are written by placing 0x before them. For example, 0x100 means 'the hexadecimal number 100' (=256 in decimal). Decimal values of hexadecimal digits.
The 0x is literal representation of hex numbers. And L at the end means it is a Long integer. If you just want a hex representation of the number as a string without 0x and L , you can use string formatting with %x . Or {0:x}.
Python: hex() function For the hexadecimal representation of a float value, use the hex() method of floats. The hexadecimal representation is returned as string and starts with the prefix '0x'. To remove the prefix, use slices to return the string starting from index 2: [2:].
You can do
int hex = Integer.decode("0x3b4");
You are right that parseInt and parseLong will not accept 0x or 0X
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