How are Strings created and stored in Java? -
to understand how string objects created , stored, tried following program , see output against have query. can please help?
package corejava.immutable; public class stringtester { public static void main(string[] args) { // todo auto-generated method stub string s1 = "omkar patkar"; string s2 = "omkar patkar"; string s3 = "omkar" + " patkar"; string s4 = "omkar"; string s5 = s4 +" patkar"; string s6 = new string("omkar patkar"); system.out.println("hashcode s1 = "+s1.hashcode()); system.out.println("hashcode s2 = "+s2.hashcode()); system.out.println("hashcode s3 = "+s3.hashcode()); system.out.println("hashcode s4 = "+s4.hashcode()); system.out.println("hashcode s5 = "+s5.hashcode()); system.out.println("hashcode s6 = "+s6.hashcode()); system.out.println("~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~"); system.out.println("indentity hashcode s1 = "+system.identityhashcode(s1)); system.out.println("indentity hashcode s2 = "+system.identityhashcode(s2)); system.out.println("indentity hashcode s3 = "+system.identityhashcode(s3)); system.out.println("indentity hashcode s4 = "+system.identityhashcode(s4)); system.out.println("indentity hashcode s5 = "+system.identityhashcode(s5)); system.out.println("indentity hashcode s6 = "+system.identityhashcode(s6)); system.out.println("indentity hashcode intern s6 = "+system.identityhashcode(s6.intern())); system.out.println("~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~"); system.out.println("s1 == s2 : - "+(s1 == s2)); system.out.println("s1 == s3 : - "+(s1 == s3)); system.out.println("s1 == s4 : - "+(s1 == s4)); system.out.println("s1 == s5 : - "+(s1 == s5)); system.out.println("s1 == s6 : - "+(s1 == s6)); system.out.println("\ns2 == s3 : - "+(s2 == s3)); system.out.println("s2 == s4 : - "+(s2 == s4)); system.out.println("s2 == s5 : - "+(s2 == s5)); system.out.println("s2 == s6 : - "+(s2 == s6)); system.out.println("\ns3 == s4 : - "+(s3 == s4)); system.out.println("s3 == s5 : - "+(s3 == s5)); system.out.println("s3 == s6 : - "+(s3 == s6)); system.out.println("\ns4 == s5 : - "+(s4 == s5)); system.out.println("s4 == s6 : - "+(s4 == s6)); system.out.println("\ns5 == s6 : - "+(s5 == s6)); system.out.println("\ns1 == s6.intern() : - "+(s1 == s6.intern())); system.out.println("s2 == s6.intern() : - "+(s2 == s6.intern())); system.out.println("s3 == s6.intern() : - "+(s3 == s6.intern())); system.out.println("s4 == s6.intern() : - "+(s4 == s6.intern())); system.out.println("s5 == s6.intern() : - "+(s5 == s6.intern())); } } and see following output: -
hashcode s1 = 2062602683 hashcode s2 = 2062602683 hashcode s3 = 2062602683 hashcode s4 = 76311326 hashcode s5 = 2062602683 hashcode s6 = 2062602683 ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ indentity hashcode s1 = 4337374 indentity hashcode s2 = 4337374 indentity hashcode s3 = 4337374 indentity hashcode s4 = 18019860 indentity hashcode s5 = 31054905 indentity hashcode s6 = 605645 indentity hashcode intern s6 = 4337374 ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ s1 == s2 : - true s1 == s3 : - true s1 == s4 : - false s1 == s5 : - false s1 == s6 : - false s2 == s3 : - true s2 == s4 : - false s2 == s5 : - false s2 == s6 : - false s3 == s4 : - false s3 == s5 : - false s3 == s6 : - false s4 == s5 : - false s4 == s6 : - false s5 == s6 : - false s1 == s6.intern() : - true s2 == s6.intern() : - true s3 == s6.intern() : - true s4 == s6.intern() : - false s5 == s6.intern() : - false identity hashcodes of s5 , s6 different s1, s2, s3...why so?
in area of memory these objects created? ... object graph understand ...
you have created string literals , string objects. string literals s1 , s2 stored in string pool. same string have same reference. efficient.
string objects created using new keyword result in object stored on heap. treated in same way other object. creating 2 objects same string value result in 2 objects each it's own reference. using new not have same efficiencies string literals in string pool. interning string object moves string pool.
when compare 2 objects using == comparing references. thus, comparing 2 string literals same value result in test being true. however, testing 2 objects created new keyword not. why should use equals method compare objects.
edit
strings created concatenation of 2 string literals result in string literal, example s3. thus, s3 has same identity hashcode s1 , s2. however, s5 created string literal , reference string literal, resulting in new object. explains why has different identity hashcode.
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