C++Talk.NET C++ language newsgroups   Archives   FAQ   Search   Memberlist   Usergroups   Register   Profile   Log in to check your private messages   Log in  strange operator problem          C++Talk.NET Forum Index -> C Language (Moderated) View previous topic :: View next topic   Author Message dreamume Guest Posted: Thu Mar 29, 2007 6:23 am    Post subject: strange operator problem Hi, everyone! Here is my simple test code: int main() { int b = 3, c = 5; int a = ( 1 == 1 || b > c ? b : c ); // sentence 1 printf( "%d\n", a ); } I thought the result should be 1, but is 3, if I change the sentence 1 like this, int a = ( 1 == 0 || b > c ? b : c ); // the result is 5 or int a = ( 1 == 1 || ( b > c ? b : c ) ); // the result is 1 But I can't explain why, I guess it's a operator precedence level problem. Who can help me? -- comp.lang.c.moderated - moderation address: clcm (AT) plethora (DOT) net -- you must have an appropriate newsgroups line in your header for your mail to be seen, or the newsgroup name in square brackets in the subject line. Sorry. Back to top John Dallman Guest Posted: Mon Apr 16, 2007 8:18 pm    Post subject: Re: strange operator problem In article <clcm-20070328-0010 (AT) plethora (DOT) net>, dreamume (AT) gmail (DOT) com (dreamume) wrote: Quote: int main() { int b = 3, c = 5; int a = ( 1 == 1 || b > c ? b : c ); // sentence 1 printf( "%d\n", a ); } I thought the result should be 1, but is 3 The precedence rules mean that the compiler is reading your code as equivalent to int a = ( (1 == 1 || b > c) ? b : c ); Since 1==1 is true, the overall logical expressions is true at that point, and C doesn't bother comparing b to c. Since the logical expression is true, the conditional expression returns b. Quote: int a = ( 1 == 0 || b > c ? b : c ); // the result is 5 1==0 is false, so C looks at the "b > c". That's false too, so the result of the conditional expression is c, and 5 is correct. Quote: int a = ( 1 == 1 || ( b > c ? b : c ) ); // the result is 1 "1==1" is logically true. "( b > c ? b : c )" returns either b or c, and either of those, being non-zero, is logical true. So the expression is equivalent to true OR true, which is true, and C true converted to an int is 1. -- John Dallman, jgd (AT) cix (DOT) co.uk, HTML mail is treated as probable spam. -- comp.lang.c.moderated - moderation address: clcm (AT) plethora (DOT) net -- you must have an appropriate newsgroups line in your header for your mail to be seen, or the newsgroup name in square brackets in the subject line. Sorry. Back to top Thomas Richter Guest Posted: Mon Apr 16, 2007 8:18 pm    Post subject: Re: strange operator problem dreamume wrote: Quote: Hi, everyone! Here is my simple test code: int main() { int b = 3, c = 5; int a = ( 1 == 1 || b > c ? b : c ); // sentence 1 printf( "%d\n", a ); } I thought the result should be 1, but is 3, if I change the sentence 1 like this, int a = ( 1 == 0 || b > c ? b : c ); // the result is 5 or int a = ( 1 == 1 || ( b > c ? b : c ) ); // the result is 1 The ternary operator ? binds weaker than the logical ||, so your above expression is equivalent to: int a = ( (1 == 1 || b > c) ? b : c ); So long, Thomas -- comp.lang.c.moderated - moderation address: clcm (AT) plethora (DOT) net -- you must have an appropriate newsgroups line in your header for your mail to be seen, or the newsgroup name in square brackets in the subject line. Sorry. Back to top Jonas Guest Posted: Mon Apr 16, 2007 8:18 pm    Post subject: Re: strange operator problem "dreamume" <dreamume (AT) gmail (DOT) com> wrote in message news:clcm-20070328-0010 (AT) plethora (DOT) net... Quote: Hi, everyone! Here is my simple test code: int main() { int b = 3, c = 5; int a = ( 1 == 1 || b > c ? b : c ); // sentence 1 printf( "%d\n", a ); } I thought the result should be 1, but is 3, if I change the sentence 1 like this, int a = ( 1 == 0 || b > c ? b : c ); // the result is 5 or int a = ( 1 == 1 || ( b > c ? b : c ) ); // the result is 1 But I can't explain why, I guess it's a operator precedence level problem. Operator precedence can be found e.g. at http://www.difranco.net/cop2220/op-prec.htm. Note that ?: has the lowest precedence in the expression, and is right-to-left-associative. Using parenthesis to illustrate the evaluation of 1 == 1 || b > c ? b : c gives (((1 == 1) || (b > c)) ? b : c) I guess what you expected was ((1 == 1) || ((b > c) ? b : c)) Jonas -- comp.lang.c.moderated - moderation address: clcm (AT) plethora (DOT) net -- you must have an appropriate newsgroups line in your header for your mail to be seen, or the newsgroup name in square brackets in the subject line. Sorry. Back to top Siddhartha Gandhi Guest Posted: Mon Apr 16, 2007 8:19 pm    Post subject: Re: strange operator problem On Mar 28, 9:23 pm, "dreamume" <dream...@gmail.com> wrote: Quote: Hi, everyone! Here is my simple test code: int main() { int b = 3, c = 5; int a = ( 1 == 1 || b > c ? b : c ); // sentence 1 printf( "%d\n", a ); } I thought the result should be 1, but is 3, if I change the sentence 1 like this, int a = ( 1 == 0 || b > c ? b : c ); // the result is 5 or int a = ( 1 == 1 || ( b > c ? b : c ) ); // the result is 1 But I can't explain why, I guess it's a operator precedence level problem. Who can help me? -- comp.lang.c.moderated - moderation address: c...@plethora.net -- you must have an appropriate newsgroups line in your header for your mail to be seen, or the newsgroup name in square brackets in the subject line. Sorry. int a = ( 1 == 1 || b > c ? b : c ); //original expression int a = ( true || false ? b : c ); //simplifed int a = ( true ? b : c ); //true OR false is true int a = b; //so it is 3 int a = ( 1 == 0 || b > c ? b : c ); int a = ( false || false ? b : c ); int a = ( false ? b : c ); int a = c; //it is 5 int a = ( 1 == 1 || ( b > c ? b : c ) ); int a = ( 1 == 1 || ( false ? b : c ) ); //you have parantheses int a = ( true || (c) ); int a = true; //true will become 1 int a = 1; //I'm not completely sure about this last example, but I'm pretty certain The ternary operator is evalulated last. First the == , then the || afaik. -- comp.lang.c.moderated - moderation address: clcm (AT) plethora (DOT) net -- you must have an appropriate newsgroups line in your header for your mail to be seen, or the newsgroup name in square brackets in the subject line. Sorry. Back to top seni.yin Guest Posted: Mon Apr 16, 2007 8:19 pm    Post subject: Re: strange operator problem you are right, iti si operator precedence level problem a = ( 1 == 1 || b > c ? b : c ); equals a = ( (1 == 1 || b > c) ? b : c ); because (1 == 1 || b > c) is TRUE, so a = b -- comp.lang.c.moderated - moderation address: clcm (AT) plethora (DOT) net -- you must have an appropriate newsgroups line in your header for your mail to be seen, or the newsgroup name in square brackets in the subject line. Sorry. Back to top º­×Ó(Hanzi) Guest Posted: Mon Apr 16, 2007 8:19 pm    Post subject: Re: strange operator problem On Mar 29, 9:23 am, "dreamume" <dream...@gmail.com> wrote: Quote: Hi, everyone! Here is my simple test code: int main() { int b = 3, c = 5; int a = ( 1 == 1 || b > c ? b : c ); // sentence 1 printf( "%d\n", a ); } I thought the result should be 1, but is 3, if I change the sentence 1 like this, int a = ( 1 == 0 || b > c ? b : c ); // the result is 5 or int a = ( 1 == 1 || ( b > c ? b : c ) ); // the result is 1 But I can't explain why, I guess it's a operator precedence level problem. Who can help me? -- comp.lang.c.moderated - moderation address: c...@plethora.net -- you must have an appropriate newsgroups line in your header for your mail to be seen, or the newsgroup name in square brackets in the subject line. Sorry. In "sentence 1" a = ( 1==1 || b>c ? b : c) Since trinary operator ?: has the lowest precedence, the expression 1==1 || b>c will be evaluated first. And because 1==1 returns 1, the value of the expression is 1, and no judgement of b>c is processed. So it equals to a=b, which has a value of 3. But if you write a = (1==0 || b>c ? b : c) The program must test whether b>c to find the value of expression 1==0 || b>c, for 1==0 yields a 0. Now that b is 3 and c is 5, we can know that b > c will also yield a 0. Thus 1==0 || b>c yields a 0. So that the expression equals to a=c, which has a value of 5. And as regards 1==1 || (b>c ? b:c) the condition before || (i.e. 1==1) will be evaluated first. Because it yields a 1, no judge for (b>c ? b:c) will be performed. As a result, the value of the expression is 1, so after assignment, the value of a is 1. -- comp.lang.c.moderated - moderation address: clcm (AT) plethora (DOT) net -- you must have an appropriate newsgroups line in your header for your mail to be seen, or the newsgroup name in square brackets in the subject line. Sorry. 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