How many bytes does a specific integer require at runtime?

Question

This is likely to be a duplicate question, but I was unable to find a similar one when searching.

I'm looking for an easy, efficient method to determine how many signed bytes an int requires at runtime.

For example, consider an int with the following values:

1     - Requires 1 Byte
10    - Requires 1 Byte
128   - Requires 2 Bytes
1024  - Requires 2 Bytes
32768 - Requires 3 Bytes
...
Integer.MAX_VALUE - Requires 4 Bytes

Edit: It's obvious to me that an int requires 4 bytes of memory regardless of its value. Nevertheless, I'm looking for the amount of bytes that the value would take up if that wasn't the case.

Ideally the answer I'm looking for utilizes bit manipulation and returns a value of 1 for an input of 0.

Bytes are 8 bits. You need to identify how many bits your number needs. — Thorbjørn Ravn Andersen
– Thorbjørn Ravn Andersen, Commented Dec 20, 2017 at 20:30
@ThorbjørnRavnAndersen I'm looking to identify how many bytes my number needs. — Jacob G.
– Jacob G., Commented Dec 20, 2017 at 20:30
In Java, an int always needs 4 bytes of memory, regardless of its contents. — Alnitak
– Alnitak, Commented Dec 20, 2017 at 20:30
@Alnitak I know, but I'm looking for the exact amount of bytes the number takes up regardless. — Jacob G.
– Jacob G., Commented Dec 20, 2017 at 20:31
When you've added 4 1-byte values to the buffer, how do you know if it's 4, 2 or 1 value? How can you use those values? — Kayaman
– Kayaman, Commented Dec 20, 2017 at 20:55

Oleksandr Pyrohov · Accepted Answer · 2017-12-20 23:38:04Z

4

One line solution according to your requirements:

public int bytesCount(int n) {
    return n < 0 ? 4 : (32 - Integer.numberOfLeadingZeros(n)) / 8 + 1;
}

Where 32 - Integer.numberOfLeadingZeros(n) returns the position of the highest one-bit. After that you can easily calculate the number of required bytes.

edited Dec 20, 2017 at 23:38

answered Dec 20, 2017 at 21:25

Oleksandr Pyrohov

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2 Comments

Jacob G. Over a year ago

This was what I was looking for, as it works with signed byte values. Thank you!

Thorbjørn Ravn Andersen Over a year ago

You may want to check the implementation of numberOfLeadingZeros() to ensure that it is better than what you had already. grepcode.com/file/repository.grepcode.com/java/root/jdk/openjdk/…

scigs · Accepted Answer · 2017-12-20 20:41:41Z

0

I'm sure someone has a much more efficient way to do this but a simple and easy to understand this is see what the upper bound power of 2 is higher than it, thus you now how many bits it would fit in, then you can divide by 8 to figure out how many bytes you would need.

int bytesInInt(int i) {
    int exp = 0;
    while (Math.pow(2,exp) < i) {
        exp++;
    }
    return ((exp + 1) / 8) + 1;
}

answered Dec 20, 2017 at 20:41

scigs

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Comments

apilat · Accepted Answer · 2017-12-20 21:03:33Z

0

You can check it against the boundaries of numbers which can be stored in 1 byte (byte), 2 bytes (short), 3 bytes.

int bytes_needed(int n){
    if(n >= -128 && n <= 127){
        return 1;
    } else if(n >= -32768 && n <= 32767){
        return 2;
    } else if(n >= -8388608 && n <= 8388607){
        return 3;
    } else {
        return 4;
    }
}

edited Dec 20, 2017 at 21:03

answered Dec 20, 2017 at 20:41

apilat

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