With below code sample why the first addition (1/2+1/2) prints 0 but the second addition prints 00.
System.out.println(1/2+1/2+"=1/2+1/2");
System.out.println("1/2+1/2="+1/2+1/2);
Output:
0=1/2+1/2
1/2+1/2=00
4 Answers
Integer math (int 1 divided by int 2 is int 0, if you want a floating point result cast one, or both, of 1 and 2 to a floating point type) and order of operations, the second example is String concatenation. The compiler turns that into
System.out.println(new StringBuilder("1/2+1/2=").append(1/2).append(1/2));
and then you get
System.out.println(new StringBuilder("1/2+1/2=").append(0).append(0));
Comments
The first statement "System.out.println(1/2+1/2+"=1/2+1/2");" prints 0 because an the integer value obtained from 1/2 is zero. The remainder is dropped and since 1/2 equals 0.5 the .5 is dropped. The second statement "System.out.println("1/2+1/2="+1/2+1/2);" prints out 00 because of the concatenation sign. In the second statement the first integer 1 is shown as +1 so the statement is actually being read as (+1/2 +1/2) which is why it returns 00. If the second statement was set up like this:
System.out.println("1/2+1/2="+ (1/2+1/2));
The output would be the same as the first statement.
Comments
Expression is evaluated from left to right. In the first case it does int+int (which is 0), then int + "= String" which is a String tmp = "0= String". In the other case you have '"String =" + intwhich becomes"String =int"to which you append one moreint`. Thus you print String, "0" and "0".
Comments
java assumes that the result of the division is integer , since its members are integers. For the floating result ( 0.5 ) of each division , the divisor or the dividend should be of type float
System.out.println("1/2+1/2="+(1/2.0+1/2.0));
1/2which is0(integer division) and then1/2which is 0 again. So0+0 = 0then it will concat that onto the String1/2+1/2. For the second one it will do concat(1/2) onto theString"1/2+1/2=` which (1/2) is 0 again. and then again for the next one