precedence of relational operators

came across the foll code:

Code:

---

#include<stdio.h>
   
    main()
    {
      int i=4,j=7;
     
      j=j||(printf("you can")&&(i=5));
      printf("%d %d",i,j);
    }

---

output: 4 1
Athough I am specifying the braces for the && operator so that it gets executed first..Then also the value of i remains 4 only..Why doesnot it gets changed to 5??Also the printf doesnot execute??

Adding parenthesis does not force the program to evaluate that code... the compiler can make a determination and do any number of things to optimize that line. True or anything is equal to true. In your case, it appears that when your compiler evaluates the above code, it can see that the expression will always evaluate to true because j is non-zero (true) so the part in the parenthesis is never executed/evaluated at all (as you stated, the printf call does not execute) and i remains 4.

Short-circuit evaluation - Wikipedia, the free encyclopedia

Quote:

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Originally Posted by hk_mp5kpdw

Adding parenthesis does not force the program to evaluate that code... the compiler can make a determination and do any number of things to optimize that line. True or anything is equal to true. In your case, it appears that when your compiler evaluates the above code, it can see that the expression will always evaluate to true because j is non-zero (true) so the part in the parenthesis is never executed/evaluated at all (as you stated, the printf call does not execute) and i remains 4.

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Thank you for your explaination but as per the foll link it says && has higher precedence than ||..So why does in my case || gets evaluated first C Language Operator Precedence Chart

Because you put the && inside the parentheses it is not seen because j is true. Since j is true and you use the OR evaluation the rest of the statement doesn't need to be evaluated.

Jim

@jimblumberg thank you but even if I remove paranthesis then also the result is the same i.e j is considered first and then the other parts of the expression..As far as I know precedence means That operator will be considered first...then why doesnt in my case && is considered first and its operands like printf and j are considered first??

Quote:

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Originally Posted by Saurabh Mehta

As far as I know precedence means That operator will be considered first

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You are mistaken: precedence determines grouping, not order of evaluation. It so happens that in some cases the order of evaluation follows logically from the grouping, but that does not mean that precedence determines order of evaluation.

Quote:

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Originally Posted by laserlight

You are mistaken: precedence determines grouping, not order of evaluation. It so happens that in some cases the order of evaluation follows logically from the grouping, but that does not mean that precedence determines order of evaluation.

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Thank you it was great explaination ...I have a last doubt that does compiler decides order of evalauation or is it always left to right??

Normally the compiler decides, but in this case (see Matticus' link) it is left to right.

http://cboard.cprogramming.com/c-pro...increment.html
Read this to see how precedence, evaluation order and side-effects can conspire to make a mess of your day.

Cramming too much into one line is almost always more trouble than it's worth.