This question already has answers here:
Converting an int to a binary string representation in Java?
(19 answers)
Closed 3 years ago.
Let's say i have a binary number initialized like this:
int y=0b110101;
How could i convert 110101 to a String?
I would like this:
String str1 = Integer.toString(y);
System.out.println(str1);
to give result 110101 or 0b110101 and not 53.
Integer.toBinaryString(y) would give you 110101 in your case and you can prepend the 0b to the result if you'd like that.
Related
This question already has answers here:
How to get 0-padded binary representation of an integer in java?
(17 answers)
Closed 1 year ago.
I am currently trying to convert integers into a binary string. However, I need each binary "group" to be exactly 8 bits or 1 byte.
For example, Integer.toBinaryString(17) will give me "10001" where instead I need "00010001"
Is there some method I can use, or possibly a simpler solution outside of hard-coding the outputs?
Easiest is to use format strings
String.format("%08s",Integer.toBinaryString(i));
Actually this tells the formatter to pad the string with zeroes to make 8 digits
This question already has answers here:
Why is 08 not a valid integer literal in Java?
(6 answers)
Closed 4 years ago.
when put 08 in array then it shows a error "The literal 08 of type int is out of range"
Integer literals starting with 0 in Java are parsed as octal, and 0-7 are the only valid octal digits.
This question already has answers here:
How can I pad an integer with zeros on the left?
(18 answers)
Closed 5 months ago.
I want to make an integer value from the range 1-9999.
But the integer values should be always in 4 digits.
Like 0001, 0002, 0015, 0156, 1578
You can probably use String format of the integer you willing to, like this:
String.format("%04d", your integer));
This will always show missing 0 on the left on digits like 01 or 1.
System.out.format("%04d%n", n);
If you want to print 4 digit integer:
System.out.printf("%04d%n",your integer);
If you want to store in a variable:
String str=String.format("%04d", your integer));
Read more details about String format()
This question already has answers here:
Adding binary numbers
(22 answers)
Closed 7 years ago.
I needed to write a program which adds binary numbers as if they were decimal. But it isn't working like I expected.
int i = 0101, j=0001;
System.out.println(i+j);
I expected the answer to be either 6 (i.e decimal of sum of 0101 and 0001) or maybe 0102 (as I am adding them as simple decimal numbers). But unexpectedly, I am getting 66. Can anybody kindly explain this? Or may be help me with the code to add two binary numbers as decimal numbers.
You're using octal literals, i.e. 0101 = 65 = 1 * 8² + 1. To use binary literals use the following notation:
int i = 0b101, j = 0b1;
If you want to print a int as binary, use Integer.toBinaryString to get the string representation of a int in binary.
int num1 = Integer.parseInt(Integer.toString(i),2);
int num2 = Integer.parseInt(Integer.toString(j),2);
System.out.println(num1+num2);
First we need to convert the number from binary to decimal. And for that you need to parse it using the parse commands for the Integer wrapper class. Then you can add.
The format Integer.parseInt(String,radix) can be used to convert any string of digits to the base radix.
This question already has answers here:
Why are integer literals with leading zeroes interpreted strangely?
(8 answers)
Java int zero prefix? [duplicate]
(2 answers)
Closed 8 years ago.
why output so strange 0022 =18?
hello everyone to day i have test.the question is "what is the output of
System.out.println(0022);
i so confuse why Answer is 18
explain needed
If the first digit of an int literal is 0, Java interprets the number in octal (base 8).