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+ operator doubt

 
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Hello



The output of above code is
1234
46
I know the concept of '+' operator behaviour when used with strings.
Can someone please explain what happened here to the '+' operator precedence by just changing the string literal location.

Thanks
 
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in line 1 the compiler first encouters "", as it scans from left to right, hence it converts everything that follows it into string..

in line 2 it first encounters an integer hence performs addition and then int + string-->string
 
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int a1 = 12;
int a2 = 34;
System.out.println(""+a1+a2);
System.out.println(a1+a2+"");

""+a1+a2 -> ""+a1 = 12 [String] so "12"+a2 = 1234 {Stirng concantenation}
If one of the operands to + is a string it performs concantenation

a1+a2+"" -> a1+a2 = 46 [int] and now 46+"" = 46 {String}
Thanks
Deepak
 
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You need to understand when string conversion happens.
http://java.sun.com/docs/books/jls/third_edition/html/conversions.html#5.4

Basically if any of the operands of + is a string then other one will be converted to a string. And also + is left associative. So an expression with + gets evaluated from left to right.

System.out.println(""+a1+a2);
System.out.println(a1+a2+"");



So for the first statement, a1 gets converted in a string, and then a2.
But in the second statement, first a1 and a2 get added, and then the result is converted to a string.
In fact following statement will give same result as 1st one

System.out.println(a1+(a2+""));



Thanks.
 
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when you put System.out.println(""+12+34) - it take the final output as String.

when you put System.out.println(12+34+"") - it takes the final output as String but before that it adds both the integers.
 
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Operators are evaluated based on precedence.
Operators of same precedence are evaluated from left to right.

 
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