About parameters for Java methods

Just see the function of C + +, that is, in C + + In addition to reference types of formal parameters, the other is a copy of the argument (personal summary).

Vaguely remember that the parameters of the method in Java are the same thing, so manual testing.

Results

The parameters of a method in Java are value passing , even if it is a reference type, and a copy of the reference itself (pointing to the same object) is passed.

Personally, it is easier to understand variables of reference types in Java as pointers.

Test code

Importorg.junit.Test; Public classtestfunction {@Test Public voidTestint () {intA = 3; intb = 4;        Swap (A, b); System.out.println (A+ "---" +b); } @Test Public voidteststring () {String a= "hehe"; String b= "What";        Swap (A, b); System.out.println (A+ "---" +b); } @Test Public voidTestinner () {Inner a=NewInner (); A.setage (10); Inner b=NewInner (); B.setage (20);        Swap (A, b); System.out.println (A.getage ()+ "---" +b.getage ()); }    /*** Basic type, is value passed, so the original object is not modified * *@paramA *@paramb*/     Public voidSwapintAintb) {a+=b; b= A-b; A-=b; }    /*** String is also value passed *@paramA *@paramb*/     Public voidSwap (String A, string b) {string TMP=A; A=b; b=tmp; }    /*** Reference, pass the reference itself, similar to the address.     So it's just a copy that doesn't affect the original value. * @paramA *@paramb*/     Public voidswap (Inner A, Inner b) {Inner tmp=A; A=b; b=tmp; }    /*** Inner class*/    classinner{Private intAge ;  Public voidSetage (intAge ) {             This. age=Age ; }         Public intGetage () {return  This. Age; }    }}

Add

Most of the time, we are using members of incoming objects (reference types), so this is not a problem. Because the referenced copy still points to the same object.

About parameters for Java methods