AppletTalk.com Java discussions newsgroups   Archives   FAQ   Search   Memberlist   Usergroups   Register   Profile   Log in to check your private messages   Log in  priority of operators          AppletTalk.com Forum Index -> Java Help View previous topic :: View next topic   Author Message alias Guest Posted: Mon May 16, 2005 3:52 pm    Post subject: priority of operators Why do i get 2 different results for the 'same' expression (priority of parentheses) ? I notice that although (x = 3) is between parentheses in a), the first term of the addition (x) is evaluated to 10 before working out the second term (x=3). int x = 10; a) x = x + (x = 3); -> result x = 13; b) x = (x = 3) + x; -> result x = 6; Does anyone know the answer ? Thanks Back to top klynn47@comcast.net Guest Posted: Mon May 16, 2005 3:58 pm    Post subject: Re: priority of operators Expressions are evaluated from left to right. In the first case, originally x is 10. As an expression, an assignment statement has the value of what is assigned into the left-hand side. So the expression assigns 3 to x, (x=3) has the value 3, and this is added to the original 10. In the second expression, (x=3) has the value 3 and assigns 3 to x, so you get 6. Back to top alias Guest Posted: Mon May 16, 2005 4:06 pm    Post subject: Re: priority of operators But the priority-rules are : 1) ( ) 2) + 3) = So in a) normally you first do (x = 3), this evaluates to 3, then x + 3, this evaluates to 6 because x has been set to 3 and finaly x = 6. Or am I wrong ? <klynn47 (AT) comcast (DOT) net> schreef in bericht news:1116259109.479209.165470 (AT) g43g2000cwa (DOT) googlegroups.com... Quote: Expressions are evaluated from left to right. In the first case, originally x is 10. As an expression, an assignment statement has the value of what is assigned into the left-hand side. So the expression assigns 3 to x, (x=3) has the value 3, and this is added to the original 10. In the second expression, (x=3) has the value 3 and assigns 3 to x, so you get 6. Back to top klynn47@comcast.net Guest Posted: Mon May 16, 2005 4:22 pm    Post subject: Re: priority of operators In the binary operation, +, the left-hand operand appears to be fully evaluated before any part of the right-hand operand is evaluated. This is according to the Java Language Specification. Back to top alias Guest Posted: Mon May 16, 2005 4:29 pm    Post subject: Re: priority of operators Thanks. This will be the answer to my problem. <klynn47 (AT) comcast (DOT) net> schreef in bericht news:1116260555.642693.290920 (AT) g49g2000cwa (DOT) googlegroups.com... Quote: In the binary operation, +, the left-hand operand appears to be fully evaluated before any part of the right-hand operand is evaluated. This is according to the Java Language Specification. Back to top Virgil Green Guest Posted: Wed May 18, 2005 5:50 pm    Post subject: Re: priority of operators alias wrote: Quote: Thanks. This will be the answer to my problem. [email]klynn47 (AT) comcast (DOT) net[/email]> schreef in bericht news:1116260555.642693.290920 (AT) g49g2000cwa (DOT) googlegroups.com... In the binary operation, +, the left-hand operand appears to be fully evaluated before any part of the right-hand operand is evaluated. This is according to the Java Language Specification. "will be the answer"? Was this homework? -- Virgil Back to top Simon Guest Posted: Fri May 20, 2005 6:56 pm    Post subject: Re: priority of operators Hello, You can check out www.eassignment.net for help with your assignments. They are really helpful. Simon Tanner alias wrote: Quote: Why do i get 2 different results for the 'same' expression (priority of parentheses) ? I notice that although (x = 3) is between parentheses in a), the first term of the addition (x) is evaluated to 10 before working out the second term (x=3). int x = 10; a) x = x + (x = 3); -> result x = 13; b) x = (x = 3) + x; -> result x = 6; Does anyone know the answer ? 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