last exam we had the exercise to determine the output of the following code:
System.out.println(2 + 3 + ">=" + 1 + 1);
My answer was 5 >= 2 but now I realize that this is the wrong answer. It should be 5 >= 11.
But why?
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That is not valid Java code. Please correct the syntax error.cdhowie– cdhowie2010-12-08 18:56:28 +00:00Commented Dec 8, 2010 at 18:56
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2I'm not certain, but I think it may have to do with the fact that you switched over to a string part-way through the print. 2+3 it will treat as an integer, but once you hit the ">=" I think it switches over to considering the rest of the line as a string. so if you take the string ">=" plus the string "1" and "1" then you get >= 11. Again, I'm not certain of this, but it seems to make sense.guildsbounty– guildsbounty2010-12-08 19:00:10 +00:00Commented Dec 8, 2010 at 19:00
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Upvoted just for your nickname!GhostCat– GhostCat2017-08-28 11:37:16 +00:00Commented Aug 28, 2017 at 11:37
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5 Answers
Assuming that your syntax is :
System.out.println(2 + 3 + ">=" + 1 + 1);
expressions are evaluated from left to right, in this case 2 + 3 get summed to 5 and when "added" to a string result in "5 >=", which when added to 1 gives "5 >= 1", add another 1 and your result is: "5 >= 11"
2 Comments
Victor Parmar
yes!! one of those things that you just can't remember when you really need to :p
Karl Knechtel
No, this isn't the result of "precedence"; that's when you have two different operators (e.g.
+ and *) and one of them is handled first. What's at work here is left-to-right associativity.Because "adding" a string to anything results in concatenation. Here is how it gets evaluated in the compilation phase:
((((2 + 3) + ">=") + 1) + 1)
The compiler will do constant folding, so the compiler can actually reduce the expression one piece at a time, and substitute in a constant expression. However, even if it did not do this, the runtime path would be effectively the same. So here you go:
((((2 + 3) + ">=") + 1) + 1) // original
(((5 + ">=") + 1) + 1) // step 1: addition (int + int)
(("5>=" + 1) + 1) // step 2: concatenation (int + String)
("5>=1" + 1) // step 3: concatenation (String + int)
"5>=11" // step 4: concatenation (String + int)
You can force integer addition by sectioning off the second numeric addition expression with parentheses. For example:
System.out.println(2 + 3 + ">=" + 1 + 1); // "5>=11"
System.out.println(2 + 3 + ">=" + (1 + 1)); // "5>=2"
Comments
Number+number=number
number+string=string
string+number=string
etc.
Comments
It is evaluated from left to right. You concatenate"1" to "5 >=" and finally "1" to "5 >= 1".
Comments
Let's read it one token at a time from left to right:
The first literal encountered is an integer, 2, then a +, then another integer, 3. A + between two integers is addition, so they are added together to be 5.
Now we have 5, an integer, then a +, then a String ">=". A + between an integer and a String is a concatenation operator. So the Strings are combined to form "5>=".
Then we have "5>=", a String, a +, and then an integer, 1. This is String concatenation again. So the result is "5>=1".
Finally we have "5>=1", a String, a +, and the a 1. his is String concatenation again. So the result is "5>=11".
