General Tech Learning Aids/Tools . 1 year ago

Does Java Operators precedence is working when compare to C language perspective? [closed]

Does Java Operators precedence is working when compare to C language perspective? [closed]

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manpreet Tuteehub forum best answer Best Answer 1 year ago

 

Actually i am a kickbhut in C. I just started to learn Java. And directly preparing for OCJP6 Certification. In Kathy-Sierra book, and as well as in exam syllabus also, there's no Operator's precedence matter. But i was in fully confused when i saw the Java Operators precedence table from orablce-sun Documentation.

From my C-Language: Operators(44) precedence table
1) () [] .(dot)
2)Unary:  ++pre/--pre, -, +, (cast)
3)Arithmetic: 
4)bitshift 
5)relational: 
6)bitwise:
7)logical:
8)ternary Operator ?:
9)Assignment operators  = += -== etc
10) unary post++/post--

//In C if i say..
int main()
{
int i = 1, j;

j = i++;

/*
      The above expression is solved based on the operators precedence!

      2 operators i have used!   

       one is unary post, another is assignment.

       here, assignment is higher precedence then unary post.

       So, first i value is assigned to j.

       then, i value is incremented because of post increment operator!
 */ 

 printf("i = %d, j = %d", i, j);// i = 2, j = 1

 return 0;
}

Java Operators precedence table
1)postfix   expr++ expr--
2)unary ++expr --expr +expr -expr ~ !
3) arithmetic
4)shift << >> >>>
5)relational    < > <= >= instanceof
  equality  == !=
6)bitwise AND   &
   bitwise exclusive OR ^
   bitwise inclusive OR |
7)logical AND   &&
  logical OR    ||
8)ternary   ? :
9)assignment    = += -= *= /= %= &= ^= |= <<= >>= >>>=


//What my biggest doubt is...!
class Test
{
  public static void main(String[] args)
  {
      int i = 1, j;

      j = i++;

      /*
          according to the operators precedence table!
          unary post operator is 1st precedence than assignment operator!


          This way first i value should be increment!
          then after assignment should happen!

          How come here also i am getting the same values as in C language?
       */

       //System.out.printf("i = %d, j= %d", i, j); //i = 2, j = 1
       System.out.println("i = "+i+", j = "+j); // i = 2, j = 1   
    } 
   }

Please some one clarify me!

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manpreet 1 year ago

 

I think you are confused as to what precedence means. It doesn't mean order of execution in all cases. It means order of nested or association e.g. it means that

j = i++

is the same as

j = (i++)

not

(j = i)++

like

a = b + c * d

is

(a = (b + (c * d))) 

As Jacob notes: the ++ means increment this value after using or saving the original value. In Java this is always done at the end, whereas in C and C++ it is not defined as to when this will happen.


EDIT: A more complex example is as follows

int i = 3;
int j = 4;
int k = i-- * j++; // same as int k = i * j; i--; j++;
System.out.println(k);

prints

12

and it is not the same as

int i = 3;
int j = 4;
int k = (i = i - 1) * (j = j + 1);
System.out.println(k);

which prints

10

 
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