Build a String with String Operators [duplicate]

Question

In this example I thought that the result was true. I thought that the variable stored in the string pool. The answer was: returns false because the two String objects are not the same in memory. One comes directly from the string pool and the other comes from building using String operations.

String a = "";
 a += 2;
 a += 'c';
 a += false;
 if ( a == "2cfalse") System.out.println("==");

I do not understand where the variable a was stored

Answer 1

Okay, so two responses to this. First, the ethically correct one, do never test strings with ==, always use .equals() or .equalsIgnoreCase().

Secondly, it's true that indeed, "a" == "a" because the strings are stored in, as you call it, the same pool. The problem here is that you append to it. Appending to a string causes it to become a different string, which is not stored in the string pool. The string pool is only generated on-compile, and as the second string is calculated on runtime, it won't match the one generated on-compile.

Imagine a string-pool to work like this:

a = "test";
b = "te";
c = "st";
d = "test";

The compiler translates this into

sp1 = "test";
sp2 = "te";
sp3 = "st";
a = sp1;
b = sp2;
c = sp3;
d = sp1;

Now == will check if two variables refer to the same sp. If you run b + c java will not go back and check if any of the sp's is the same as that. It only does that on compile.