Possible Duplicate:
Concatenating null strings in Java
Please find below the code snippet
String str = null;
str = str + "hi";
System.out.println(str)
The output of the above code is nullhi.
I thought the output will be hi, so kind of surprised with the output and not able to find the reason behind it.
Can someone please explain it.
This is because append method. + gets converted as either StringBuilder or StringBuffer append operations.
public AbstractStringBuilder append(String str) {
if (str == null) str = "null";
public static String valueOf(Object obj) {
return (obj == null) ? "null" : obj.toString();
}
Below is the generated byte code for your program
0 aconst_null
1 astore_1 [str]
2 new java.lang.StringBuilder [16]
5 dup
6 aload_1 [str]
7 invokestatic java.lang.String.valueOf(java.lang.Object) : java.lang.String [18]
10 invokespecial java.lang.StringBuilder(java.lang.String) [24]
13 ldc <String "hi"> [27]
15 invokevirtual java.lang.StringBuilder.append(java.lang.String) : java.lang.StringBuilder [29]
18 invokevirtual java.lang.StringBuilder.toString() : java.lang.String [33]
21 astore_1 [str]
22 getstatic java.lang.System.out : java.io.PrintStream [37]
25 aload_1 [str]
26 invokevirtual java.io.PrintStream.println(java.lang.String) : void [43]
29 return
So your code actully gets transformed into append(null) and then append("hi") which is why you get such output
Also it is clearly documented in 15.18.1. String Concatenation Operator +
An implementation may choose to perform conversion and concatenation in one step to avoid creating and then discarding an intermediate String object. To increase the performance of repeated string concatenation, a Java compiler may use the StringBuffer class or a similar technique to reduce the number of intermediate String objects that are created by evaluation of an expression.
JLSFrom JLS.
http://docs.oracle.com/javase/specs/jls/se7/html/index.html
§15.18.1. String Concatenation Operator + (reference)
If only one operand expression is of type String, then string conversion (§5.1.11) is performed on the other operand to produce a string at run-time.
§5.1.11. String Conversion (reference)
If the reference is null, it is converted to the string "null" (four ASCII characters n, u, l, l).
null is the value printed if you try to print a null reference in java.
When you concatenate values together that are not all strings, those that are not are converted as if by String.valueOf. String.valueOf(null) is the string "null".
str is a String, so this doesn't explain it.
null is not a String (indeed it is not any object).String.valueOf? Your implication is false: String.valueOf is always called for all reference-typed arguments to +.String.valueOf."That was a decision that the creators of Java made early on. Any null value that is concatenated to a string will be output as null. This sort of makes sense if you imagine that you are trying to output a log message:
String personName = null;
System.out.println("Person name is: " + personName);
Result:
Person name is: null
That behavior can help you when debugging because you know that the variable was actually null (as opposed to an empty string, e.g.).
They could have chosen to simply throw an exception when you tried to do this, but that would have caused a lot of things to break even when doing something like building a log message or exception string.
They could also have done as you expected and made null values get output as an empty string. This is what the .NET framework does. But ultimately, that's not what they decided to do.
In the example shown by you
you are concatenating the string null with hi so the output generated is
nullhi
String str = null;
str = str + "hi";
System.out.println(str);
as str=str+"hi "//shows one string is concatenating with the other
so the output you get is right.
