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numbers, Strings and + operator

 
Greenhorn
Posts: 12
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Hello all:
Please help me understand this:
Given the code
class Foo {
static byte a;
static short b;
static char c;
static int d;
static long e;
static String s;
public static void main(String[] args) {
System.out.println(a+b+c+d+e+s);
}
}
on jdk 1.4 the line (String comes last)
System.out.println(a+b+c+d+e+s);
prints: 0null
and the line (String comes first)
System.out.println(s+a+b+c+d+e);
prints: null00
Why two ceros?
Also, why is i use local variables (initialized variables) the operator behaives different?
String a = "String";
int b = 3;
int c = 7;
System.out.println(a + b + c);
prints: String37
Thanks.
 
Ranch Hand
Posts: 63
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For the first question,
When you put string variable at the first, it should print null followed by 4 zeros not two because all the rest of variables are concatenated and treated as string.
for the second question, I dont quite follow what you want to know. It prints string and the rest as string.
 
Ranch Hand
Posts: 1090
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Hi Carlos
It's not to do with the local variable the thing is that when you say a+b+c you convert all the operands to String first and they are concatnated and printed. When you do c+b+a you first add the two numbers then concatnate the result to the String.
 
mister krabs
Posts: 13974
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and the line (String comes first)
System.out.println(s+a+b+c+d+e);
prints: null00

Actually it prints:

null00
00

The default char value causes a new line to print.
 
Kaz Yosh
Ranch Hand
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I dont think char default causes new line.
It's just a unicode null and should be ignored when concatenated.
 
Ranch Hand
Posts: 2120
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It prints
null00 00
thus it is not ignored when concatenated but when printed.
 
Consider Paul's rocket mass heater.
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