java使用字符串vs字节[]作为值，映射中的内存使用情况

Question

我在很多地方读到，用byte[]代替string可以节省内存。我想用jol测试一下

这是我的测试：

public static void main (String[] args) throws java.lang.Exception{
    System.out.println(VMSupport.vmDetails());
    String StrByte   = GraphLayout.parseInstance(sizeOfStrByteMap(100000)).toFootprint();
    String ByteByte = GraphLayout.parseInstance(sizeOfByteByteMap(100000)).toFootprint();
    String StrStr     = GraphLayout.parseInstance(sizeOfStrStrMap(100000)).toFootprint();
    System.out.println(StrByte);
    System.out.println(ByteByte);
    System.out.println(StrStr);
}

public static HashMap<String, String> sizeOfStrStrMap(int size) {
    String value = "this is the sample value";

    HashMap<String, String> map = new HashMap<>();
    for (int i = 0; i < size; i++) {
        map.putIfAbsent(Integer.toString(i), value);
    }
    return map;
}

public static HashMap<String, byte[]> sizeOfStrByteMap(int size) {
    byte[] value = "this is the sample value".getBytes();

    HashMap<String, byte[]> map = new HashMap<>();
    for (int i = 0; i < size; i++) {
        map.putIfAbsent(Integer.toString(i), value);
    }
    return map;
}

public static HashMap<byte[], byte[]> sizeOfByteByteMap(int size) {
    byte[] value = "this is the sample value".getBytes();

    HashMap<byte[], byte[]> map = new HashMap<>();
    for (int i = 0; i < size; i++) {
        map.putIfAbsent(Integer.toString(i).getBytes(), value);
    }
    return map;
}

以下是我的结果：

Running 64-bit HotSpot VM.
Using compressed oop with 3-bit shift.
Using compressed klass with 3-bit shift.
Objects are 8 bytes aligned.
Field sizes by type: 4, 1, 1, 2, 2, 4, 4, 8, 8 [bytes]
Array element sizes: 4, 1, 1, 2, 2, 4, 4, 8, 8 [bytes]

// StrByteMap
java.util.HashMap@15327b79d footprint:
     COUNT       AVG       SUM   DESCRIPTION
         1       184       184   [B
    100000        31   3120000   [C
         1   1048592   1048592   [Ljava.util.HashMap$Node;
    100000        24   2400000   java.lang.String
         1        48        48   java.util.HashMap
    100000        32   3200000   java.util.HashMap$Node
    300003             9768824   (total)

// ByteByteMap
java.util.HashMap@a9d12ad footprint:
     COUNT       AVG       SUM   DESCRIPTION
    100001        24   2400184   [B
         1   1048592   1048592   [Ljava.util.HashMap$Node;
         1        48        48   java.util.HashMap
    100000        32   3200000   java.util.HashMap$Node
    200003             6648824   (total)


// StrStrMap
java.util.HashMap@716d90fad footprint:
     COUNT       AVG       SUM   DESCRIPTION
    100001        31   3120344   [C
         1   1048592   1048592   [Ljava.util.HashMap$Node;
    100001        24   2400024   java.lang.String
         1        48        48   java.util.HashMap
    100000        32   3200000   java.util.HashMap$Node
    300004             9769008   (total)

正如您所见，StrByteMap和strstrstrmap之间的内存使用几乎相同。我测试错了吗

更新： 请看下面的@Amod Pandey'squestion，我也想知道原因

Answer 1

我觉得很奇怪

// StrByteMap
java.util.HashMap@15327b79d footprint:
     COUNT       AVG       SUM   DESCRIPTION
         1       184       184   [B
    100000        31   3120000   [C

字节数组有一个实例，字符数组有100000个实例。这是StrStr案例中的100001字符数组

// StrStrMap
java.util.HashMap@716d90fad footprint:
     COUNT       AVG       SUM   DESCRIPTION
    100001        31   3120344   [C

因此，即使您存储了字节数组，内存占用也是字符数组

另一点是字节数组对象的平均大小是24，这比字节数组大小184（StrByteMap）小，并且这三种情况下的总计数不应该相同