working with right shift operators in C

can you please elaborate further. what i want to do is bring the msb 8 bits of the 16 bit value in the variable sum to the lsb position. so i have to shift right 8 times and so the loop is used

sum>>=8

where sum= 32 bit no.
what i want to do is bring the msb 8 bits of the 16 bit value in the variable sum to the lsb position.

I didn't catch the information whether the sum is a 32bit or 16bit number.And the result you want to get.

You'd better draw a simple figure to present your problem clearly. Just like:
+0 1 2 3 4 5 6 7 8 9 a b c d e f+
| | |

...............

My interpretation of the original post is that he wants to do something to each of the least significant 8 bits one at a time.

for (i = 0; i < 8; ++i)
    {
    bit = (sum >> i) & 1;
    DoSomething(bit); // called for bit 0, bit 1... bit 7
    }

This should work, though it's a little inefficient because it shifts too much. It also shifts a U32 unnecessarily, which will take time on an 8051. I'd probably write code more like:

U8 temp = (U8)sum;
for (i = 0; i < 8; ++i)
    {
    DoSomething (temp & 1);
    temp >>= 1;
    }

so that you only have to shift an 8-bit quantity, and you only do 8 1-bit shifts.

So, if my interpretation of the OP is correct, the question is why (sum >> i) doesn't work in the posted code, though (sum >> 1) does.

How about posting the assembly code output from the .cod file in both cases?

here sum is defined as long int in the code. hence it can have a 32 bit value. but in the present case 'sum' holds a 16 bit value only.
the problem is:
the instruction for right shift is not executed if the syntax is as follows:
for(i=0;i<=8;i++)
{ j=sum>>i;
} but the instruction is executed if the syntax is as follows:
for(i=0;i<=8;i++)
{ j=0x0258>>i;
} i.e. the instruction is not executed if i am using the variable name 'sum' but if i put a numerical value instead of 'sum' then it is easily executed.
how do i fix this problem?

i tried doing this but it doesnt work.

"it doesnt work"

You haven't actually stated what it is supposed to do yet!
What is it, exactly, that you are trying to achieve?

"How about posting the assembly code output from the .cod file in both cases?"

Before posting any more code, please take care to read the Notes immediately above the 'Message' window when you post - they tell you what to do to have your code appear properly formatted (as in Drew's post).

Note that the .cod file is generated by the LX51 Linker, and shows the result of Linker optimisations; you also need to show the Compiler listing (.lst) to see what the compiler actually generated...

this is the actual code i am referring to:

#include<stdio.h>
#include<intrins.h>
#include<absacc.h>
#include<math.h>
#include<reg51.h>

int count;          /* holds counter value*/
int arr[15];        /* array to store 15 nos.*/
int avg;            /* average of 15 nos. */
long int sum;       /* sum of 15 nos. (16 bit)*/

void main()

{

int i,j;            /* variable initialization */
sum=0x00;           /* initial value*/
count=0x00;         /* initial value*/

#pragma asm         /* entering 15 nos into memory locations*/
     mov r2, #0fh
     mov dptr, #3000h
      mov a, #21h
 up1:movx @dptr, a
        inc dptr
        inc a
       djnz r2,up1

             mov r3, #0fh     /* retreiving data from memory and storing 15 nos in array*/
         mov dptr, #3000h
 up2:     movx a, @dptr
              mov b, a
#pragma endasm

arr[count]=ACC;
count++;

#pragma asm
         inc dptr
         djnz r3,up2
#pragma endasm

for(count=0;count<=15;count++)  /* adding 15 nos. */
{
sum=sum+arr[count]; /* sum */
}

avg=sum / 0x0f;    /* average of 15 nos. */
P2=avg;

ACC=sum;           /*moving lsb 8 bits of 16 bit sum on port 0 */
P0=ACC;

for(i=0;i<=8;i++)
{
j=sum>>i;      /*moving msb 8 bits of 16 bit sum on port 1*/
}

B=j;
P1=B;

}

here i am not able to divide neither am i able to right shift the data when i use the variable name sum

this is the actual code i am referring to:

Good. However, you still have not explained what the code is supposed to do, what it is doing, and why you think it does not work properly.

for(count=0;count<=15;count++)  /* adding 15 nos. */

If that means that the loop should add 15 numbers, then you've got a bug - this loop will add 16 numbers. Could this be why you are not getting the result you expect ?

for(i=0;i<=8;i++)
{
j=sum>>i;      /*moving msb 8 bits of 16 bit sum on port 1*/
}

This code snippet does not move anything to P1. In fact, what it does is equivalent to

j = sum >> 8;

, but wastes a lot more cpu cycles.

the code is for adding 15 nos and then finding the average. the problem i am facing is:
if the synax is as follows:

j=sum>>8;

the instruction is not executed.
but if i put a numerical value instead of the variable name sum then the instruction is easily executed.
Similarly if the syntax is as follows:

avg=sum / 0x0f;

the above instruction is not executed using the variable name 'sum'.
Since i need to use only variable names in the code what can i do about the above mentioned problems.

"Since i need to use only variable names in the code"

Why is that?

Despite many requests, you still have not explained why you need to do this shift in a loop, rather than simply

j = sum >> 8;

"the above instruction is not executed"

How have you confirmed that it is actually "not executed"?

Have you examined the generated code?

Have you stepped through it in the simulator?

"the above instruction is not executed using the variable name 'sum'"

Just a thought: have you checked that 'sum' hasn't been defined with some special meaning, etc...

What if you call it 'my_sum' or something?