[Java lista] ket alap kerdes

[Zoltan Mozes]

1) double i = 2.65d
2) 1.4 és 5.0 (1.5 másnéven) JAVA-ban a sima String-ekre egy úgynevezett string pool lett létrehozva. Magyarul

String a = "aa";
létrehoz a JVMben szereplő string pool-ban egy string ojjektumot és az "a" referenciádat erre az obejktumra állítja. Namost ezután ha mész tovább így:

String b = "aa";

akkor mivel már szerepel egy string a pool-ban azzal a tartalommal hoyg "aa" ezért a "b" referenciádat is ugyanarra az ojjektumra fogja ráállítani.

Így tehát az a == b true-t ad vissza mivel télleg konkrétan ugyanarra a (JVM string poolban lévő ) string ojjektumra fog mutatni a két referenciád.


" It is important to be aware of what happens when you use a string literal ("immutable" in
both examples). Every string literal is represented internally by an instance of String. Java
classes may have a pool of such strings. When a literal is compiled, the compiler adds an appropriate
string to the pool. However, if the same literal already appeared as a literal elsewhere in
the class, then it is already represented in the pool. The compiler does not create a new copy.
Instead, it uses the existing one from the pool. This process saves on memory and can do no
harm. Because strings are immutable,"

Complete Java(r) 2
Certification: Study Guide,
Fifth Edition
Philip Heller
Simon Roberts
SYBEX(r)


Best Regards,
Zoltán Mózes


EPAM Systems
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-----Original Message-----
From: javalist-bounces at javagrund.hu [mailto:javalist-bounces at javagrund.hu] On Behalf Of Fisha
Sent: Tuesday, February 13, 2007 12:31 PM
To: javalist at javagrund.hu
Subject: [Java lista] ket alap kerdes

Sziasztok!

Ket alap nyelvi kerdesem lenne.

1. Szamabrezolosdi:

a kod:
double i = 2.65f;
System.out.println(i);

az eredmeny:
2.6500000953674316

Gondolom, hogy a binaris abrazolas pontatlansagabol ered a problema.
De orulnek, ha valaki pontosan le tudna irni, hogy mi is tortenik a
hatterben valojaban.

2. Optimalizalosdi:

a kod:
String a = "aa";
String b = "aa";
System.out.println(a == b);

az eredmeny: true

Gondolom, hogy valami optimalizacios folyamat van a hatterben.
Tapasztalataim szerint a 1.3-as javaban meg biztosan nem volt ilyen.
Mit lehet errol tudni? Miota van? Vagy nem is verzio, hanem jvm fuggo?

Elore is koszi,
Fisha                            mailto:fisha at freemail.hu


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