How much memory does a string use in Java 8?

问题:

I read a lot about memory allocation for Strings lately and can't find any details if things are the same with Java 8.

How much memory space would a String like "Alexandru Tanasescu" use in Java 8? I use the 64bit version.

回答1:

Java7 or lower

Minimum String memory usage :

(bytes) = 8 * (int) ((((no chars) * 2) + 45) / 8)

So

80 = 8 * (int) ((((19) * 2) + 45) / 8)

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Understanding String memory usage (SOURCE)

To understand the above calculation, we need to start by looking at the fields on a String object. A String contains the following:

• a char array— thus a separate object— containing the actual characters;
• an integer offset into the array at which the string starts;
• the length of the string;
• another int for the cached calculation of the hash code.

This means even if the string contains no characters, it will require 4 bytes for the char array reference, plus 3*4=12 bytes for the three int fields, plus 8 bytes of object header. This gives 24 bytes (which is a multiple of 8 so no "padding" bytes are needed so far).

Then, the (empty) char array will require a further 12 bytes (arrays have an extra 4 bytes to store their length), plus in this case 4 bytes of padding to bring the memory used by the char array object up to a multiple of 16. So in total, an empty string uses 40 bytes.

If the String contains, say, 19 characters, then the String object itself still requires 24 bytes. But now the char array requires 12 bytes of header plus 19*2=38 bytes for the seventeen chars. Since 12+38=50 isn't a multiple of 8, we also need to round up to the next multiple of 8 (56). So overall, our 19-character String will use up 56+24 = 80 bytes.

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Java8.

Java 8 does not have the offset and length anymore. Only hash and the CharArray.
@Thomas Jungblut

• a char array— thus a separate object— containing the actual characters;
• an integer offset into the array at which the string starts;
• the length of the string;
• another int for the cached calculation of the hash code.

So, in Java8 the way to calculate memory for strings remains same but you must subtract 8 bytes less due to the missing offset and length.

回答2:

If you look at the Oracle Java 8 sources, you have:

A char value[] and an int hash. A char is 2 bytes, and an int is 4 bytes.

So wouldn't the answer be yourstring.length * 2 + 4?

No. Every object had overhead. An array stores its dimensions, for example. And both the array (an object) and the string will incur extra memory from the garbage collector storing information about them.

There is no reliable way to calculate this, because AFAIK each JRE and JDK has no obligation to the size of object overhead.

回答3:

"Alexandru Tanasescu" uses 104 bytes. This is how to get the size

    long m0 = Runtime.getRuntime().freeMemory();
    String s = new String("Alexandru Tanasescu");
    long m1 = Runtime.getRuntime().freeMemory();
    System.out.println(m0 - m1);

Note: run it with -XX:-UseTLAB option

回答4:

According to the following JEP: http://openjdk.java.net/jeps/254

The current implementation of the String class stores characters in a char array, using two bytes (sixteen bits) for each character.

In Java SE 9 this might change.

Note however, since this is a JEP not a JSR (and it mentions implementation), I understand, that this is implementations specific and not defined by the JLS.