Jshell is not Printing properly with println method

Question

why Jshell is considering r , s as string when used without brackets ?

PS C:\Users\saiko\OneDrive\Learning\java> Jshell                                                                                                                                             
|  Welcome to JShell -- Version 11.0.6
|  For an introduction type: /help intro

jshell> int p = 10;  int q =20; int r=5; int s=10;
p ==> 10
q ==> 20
r ==> 5
s ==> 10

jshell> if( (p+q) > (r+s) ) System.out.println( p+q+" is greater than  "+r+s );
30 is greater than  510

jshell> if( (p+q) > (r+s) ) System.out.println( (p+q) +" is greater than  "+ (r+s) );
30 is greater than  15

jshell>

Because "+" is left-associative. First you add the string literal to r, and that becomes a string, then you add that string to s, so r and s aren't added together, just concatenated — user, May 10 '20 at 23:00
So You mean that after adding " 30 is greater than" + 5 becomes a string " 30 is greater than 5" and java considered 10 adding value to a String "30 is greater than 5" and concatenates again — Gandikota Saikoushik, May 10 '20 at 23:09

score 0 · Answer 1 · answered Dec 09 '20 at 16:35

As said in the comments and in the Java tutorials:

All binary operators except for the assignment operators are evaluated from left to right

In the expression p+q+" is greater than "+r+s, the first evaluation is p+q results in an int, then p+q+" is greater than " results in String, and all subsequent evaluation results also to a String.

This is why you obtain:

30 is greater than  510

Jshell is not Printing properly with println method

1 Answers1