java - Immutable type wrapper -


this question has answer here:

i'm confused type wrappers being immutable when executing following code

static void inc(integer nr) {     system.out.printf("1. inc() \t %d \n", nr);     nr++;     system.out.printf("2. inc() \t %d \n", nr); } // inc()  public static void main(string[] args) {     integer nr1 = 10;     system.out.printf("a. main() \t %d \n", nr1);     inc(nr1);     system.out.printf("b. main() \t %d \n", nr1); } // main()

executing creates following output

  a. main()     10    1. inc()      10    2. inc()      11    b. main()     10

if type wrapper immutable, why value increased between line "1. inc" , "2. inc" , line "b. main" print same value "1. main"?

thank you

chris

if type wrapper immutable, why value increased between line "1. inc" , "2. inc"

because you're not mutating existing integer object - you're creating new 1 (well, - it'll use common cached objects, point value of nr refer different object after nr++). think of this:

nr++;

as instead:

int tmp = nr.intvalue(); tmp++; nr = integer.valueof(tmp);

so fact see text representation of nr changing doesn't mean object refers has mutated - in case, cause nr has taken on new value, referring different object.

you can see more diagnostics, too:

static void inc(integer nr) {     integer original = nr;     system.out.printf("1. inc() \t %d \n", nr);     nr++;     system.out.printf("2. inc() \t %d \n", nr);     // print 10     system.out.printf("original: %d\n", original); }

and line "b. main" print same value "1. main"?

the reference passed value, is. means inc(nr1) doesn't modify nr1 @ - refers same object did before. said above, inc also doesn't modify object (because wrapper types immutable). therefore after call, nr1 refers same object wrapping same value (10).


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