How to Convert String to int in Java?
While operating upon strings, there are times when we need to convert a number represented as a string into an integer type in Java. The method generally used to convert String to Integer in Java is parseInt().
Variants of parseInt() Method
There are two variants of this method:
Variant 1: This function parses the string argument as a signed decimal integer.
public static int parseInt(String s) throws NumberFormatException
Variant 2: This function parses the string argument as a signed integer in the radix specified by the second argument
public static int parseInt(String s, int radix) throws NumberFormatException
- A string that needs to be converted to an integer. It can also have the first character as a minus sign ‘-‘ (‘\u002D’) or plus sign ‘+’ (‘\u002B’) to represent the sign of the number.
- The radix is used while the string is being parsed.
Note: This parameter is only specific to the second variant of the method.
Return Type: Both methods convert the string into its integer equivalent. The only difference being is that of the parameter radix. The first method can be considered as an equivalent of the second method with radix = 10 (Decimal).
Note: If your String has leading zeroes, the parseInt() method will ignore them.
Exception Thrown: NumberFormatException is thrown by this method if any of the following situations occur.
Now, geeks do remember certain keypoints associated with both the variants:
- The string is null or of zero-length
- The value represented by the string is not a value of type int
- Specifically for the parseInt(String s, int radix) variant of the function:
- The second argument radix is either smaller than Character.MIN_RADIX or larger than Character.MAX_RADIX
- Any character of the string is not a digit of the specified radix, except that the first character may be a minus sign ‘-‘ (‘\u002D’) or plus sign ‘+’ (‘\u002B’) provided that the string is longer than length 1
String str=”0111”; int t=Integer.parseInt(str); System.out.println(t);
The output for above code snippet will be 111. So in cases where leading zeroes are important(in binary representations) avoid using parseInt() method.
parseInt("20") returns 20 parseInt("+20") returns 20 parseInt("-20") returns -20
parseInt("20", 16) returns 16, (2)*16^1 + (0)*16^0 = 32 parseInt("2147483648", 10) throws a NumberFormatException
parseInt("99", 8) throws a NumberFormatException as 9 is not an accepted digit of octal number system
parseInt("geeks", 28) throws a NumberFormatException
parseInt("geeks", 29) returns 11670324, Number system with base 29 can have digits 0-9 followed by characters a,b,c... upto s
parseInt("geeksforgeeks", 29) throws a NumberFormatException as the result is not an integer.
How to use parseInt() Method?
Let us take an example straightaway in order to get usage so do the working of parseint() method and implementing the above mentioned as follows:
20 20 -20 32 11670324
Similarly, we can convert the string to any other primitive data type:
- parseLong(): parses the string to Long
- parseDouble(): parses the string to Double
If we want to convert the string to integer without using parseInt(), we can use valueOf() method. It also has two variants similar to parseInt()
- Difference between parseInt() and valueOf(): parseInt() parses the string and returns the primitive integer type. However, valueOf() returns an Integer object.
Note: valueOf() uses parseInt() internally to convert to integer.
20 20 -20 32 11670324
This article is contributed by Shikhar Goel. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to email@example.com. See your article appearing on the GeeksforGeeks main page and help other Geeks. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
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