6 Responses to siddharth sir-problem on precedence

1. manoj says:

   September 4, 2012 at 6:02 pm

   Hi,

   printf maintains a stack of values to be printed. It does not print each value separately, first it will complete it’s stack than starts printing.

   In case of ++j and j printf will store the last updated value in the stack. and for j++ it stores the current value in the stack and then increments it.

   now u will be able to solve this.

   Reply
   • shalini singhal says:

     September 8, 2012 at 1:22 pm

     hi manoj,

     i am having a query regarding your answer that why is it not updateing value of j++……

     Reply
     • manoj says:

       September 10, 2012 at 6:16 pm

       Hi shalini,

       It is updating… But it will update the values only at the end of the expression..

       for eg. : if u use : printf(p++ + p++) nd if p = 5
       Then result will be 10.
       But in next statement print the value of p . It would have become 7. It is updating but at the end of the expression.

       Thanks

       Reply
     • manoj says:

       September 10, 2012 at 6:23 pm

       Hi,

       Other comment is related to the expression..
       But if u see, in case of printf(++j,j++,j) then u have to start from Right to left : j=1 (not stored in stack , waiting for the updated value), j++ = 1 (used).. j becomes 2 and then ++j = 3 .. LAst updated value is 3..

       So the output is 3,1,3

       Cheers..

       Reply
2. arjumand says:

   September 4, 2012 at 8:24 pm

   its due to the property of pre increment operator. It updates the value of j in ++j and j

   Reply
3. abhijeet.panwar says:

   September 8, 2012 at 11:13 am

   post increment store the valu 1st and then pass pre increment and final the pass to other valu like(j pass 1 not to store,j++store 1and pass 2, ++j have 3 then final store 3 and pass to j and j also stor 3) all these in stack…

   Reply