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Say I have this String expression

String hi = "Tom" + "Brady" + "Goat" 

I know that the String pool "allows a runtime to save memory by preserving immutable strings in a pool" String Pool

How many strings will be created in the string pool?

My initial guess was 5 - "Tom", "Brady", "Goat", "TomBrady","TomBradyGoat", because of the order of operations of String concatenation (left to right?) or is it only the final result, "TomBradyGoat", that is stored in the String pool?

4 Answers 4

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What you have here is a constant expression, as defined by the JLS, Section 15.28.

A constant expression is an expression denoting a value of primitive type or a String that does not complete abruptly and is composed using only the following:

  • Literals of primitive type and literals of type String (§3.10.1, §3.10.2, §3.10.3, §3.10.4, §3.10.5)

  • Casts to primitive types and casts to type String (§15.16)

  • The unary operators +, -, ~, and ! (but not ++ or --) (§15.15.3, §15.15.4, §15.15.5, §15.15.6)

  • The multiplicative operators *, /, and % (§15.17)

  • The additive operators + and - (§15.18)

(other possibilities)

The compiler determines that the expression "Tom" + "Brady" + "Goat" is a constant expression, so it will evaluate the expression itself to "TomBradyGoat".

The JVM will have only one string in the string pool, "TomBradyGoat".

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Comments

3

At runtime, that piece of code will translate into a single String object. The compiler will take care of concatenation at compile time and add a single value in the constants pool.

Comments

2

The other answers explain well why only 1 String is added to the String pool. But if you want to check and make some tests by yourself you can take a look on the bytecode to see the number of String created and added to the string pool. Ex:

Ex1:

public static void main(String[] args) { String hi = "Tom" + "Brady" + "Goat"; } 

ByteCode:

 // access flags 0x9 public static main(String[]) : void L0 LINENUMBER 6 L0 LDC "TomBradyGoat" ASTORE 1 L1 LINENUMBER 7 L1 RETURN L2 LOCALVARIABLE args String[] L0 L2 0 LOCALVARIABLE hi String L1 L2 1 MAXSTACK = 1 MAXLOCALS = 2 

As you can see only 1 String is created

Ex2:

public static void main(String[] args) { String str1 = "Tom"; String str2 = "Brady"; String str3 = "Goat"; String str = str1 + str2 + str3; } 

Bytecode:

 // access flags 0x9 public static main(String[]) : void L0 LINENUMBER 6 L0 LDC "Tom" ASTORE 1 L1 LINENUMBER 7 L1 LDC "Brady" ASTORE 2 L2 LINENUMBER 8 L2 LDC "Goat" ASTORE 3 L3 LINENUMBER 9 L3 NEW StringBuilder DUP ALOAD 1: str1 INVOKESTATIC String.valueOf (Object) : String INVOKESPECIAL StringBuilder.<init> (String) : void ALOAD 2: str2 INVOKEVIRTUAL StringBuilder.append (String) : StringBuilder ALOAD 3: str3 INVOKEVIRTUAL StringBuilder.append (String) : StringBuilder INVOKEVIRTUAL StringBuilder.toString () : String ASTORE 4 L4 LINENUMBER 10 L4 RETURN L5 LOCALVARIABLE args String[] L0 L5 0 LOCALVARIABLE str1 String L1 L5 1 LOCALVARIABLE str2 String L2 L5 2 LOCALVARIABLE str3 String L3 L5 3 LOCALVARIABLE str String L4 L5 4 MAXSTACK = 3 MAXLOCALS = 5 

As you can see 4 Strings are created

1 Comment

@committedandroider: Try javap -c com.mypackage.MyClass; see stackoverflow.com/questions/3315938/…
-1

As other people have mentioned, the compiler is good enough to optimize that very simple example, but if you have more complicated functionality (which is probably what you care about), it might not. For example:

String a = "Tom"; String b = "Brady"; String c = a; c += b; 

This would result in 3 strings being created.

1 Comment

It's not a matter of 'good enough'. The compiler is required by the JLS to treat the OP's expression as a constant expression and create a single string.

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