Concatenation of Strings and characters

Question

The following statements,

String string = "string";   

string = string +((char)65) + 5;
System.out.println(string);

Produce the output stringA5.

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The following however,

String string = "string";

string += ((char)65) + 5;
System.out.println(string);

Produce string70.

Where is the difference?

Answer 1

Case 1

string = string +((char)65) + 5;

everything is treated as String but in second case

Sequence of operation performed:

• string +((char)65 = stringA
• stringA + 5 = stringA5

Case 2

 string += ((char)65) + 5;

first right hand side is calculated means first operation will be like ((char)65) + 5, So result of ((char)65) + 5 is 70 and after that += operation.

Sequence of operation performed:

• (char)65 + 5 = 70
• string + 70 = string70

Lets see 1 more example

String string = "string";
string += ((char)65) + 5 + "A";
System.out.println(string); 

Output string70A

Reason Same first right hand side is calculated and sequesce of opertion performed is

• (char)65 + 5 = 70
• 70 + "A" = 70A
• string + 70A = string70A

Answer 2

string = string +((char)65) + 5;

This means "Set string to the concatenation of string, ((char)65), and 5." This gets evaluated from left-to-right (so first string + ((char)65), then that +5, and concatenation of a string and an integer converts that integer to a string.

string += ((char)65) + 5;

This means "Calculate the result of ((char)65)+5 and add it to string" - the entire right-hand side is evaluated before adding the result to the string. Since a char is really just a 16-bit integer, it adds them together as integers - giving 70 - and then it appends that to string.

Answer 3

When you write:

string = string + ((char)65) + 5;

It is like writing:

String string =
     new StringBuilder(string).append((char)65).append((int)5).toString();

Which is like appending string to A (The char representation of the decimal 65) and 5.

In the latter, you're first calculating the right hand, and then adding the result to string, it's like writing:

string = string + 70;

Answer 4

You see this behavior as a result of the combination of operator precedence and string conversion.

JLS 15.18.1 states:

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.

Therefore the right hand operands in your first expression are implicitly converted to string: string = string + ((char)65) + 5;

For the second expression however string += ((char)65) + 5; the += compound assignment operator has to be considered along with +. Since += is weaker than +, the + operator is evaluated first. There we have a char and an int which results in a binary numeric promotion to int. Only then += is evaluated, but at this time the result of the expression involving the + operator has already been evaluated.

Answer 5

It is the same case as the example:

System.out.println("abc"+10 + 5);

produces abc105 (add as String)

and

System.out.println(5 + 10);

produces 15 (add as Number)

The String included in the (first place of the) operation forces all the elements to be treated as Strings while executing the operation. In your case with += however, the operation is executed first in the right part (treating the elements as int) because of the operator precedence and then is concatenated with the String.