For some reason I can't figure this one out.

Say I have a for loop

for (int i=0; i<=3; i++)
{
     setBits(i);

}

Say my SFR has a value of 0b11000100

How would I set the value of the 4 and 5th bits to the value of i?

 

You are wanting to change just bits 4 and 5 of the SFR to equal i every time it increments? So your bit sequence would look like:
11000100
11010100
11100100

Is that correct?

 

Make a hole, slip in the bits:

SFR = (SFR & 0b11001111) | (i << 4);

 

Make a hole, slip in the bits:

SFR = (SFR & 0b11001111) | (i << 4);

Ah so I clear those bits first then set them?

 

Take i and shift it left by 4 bits
(i << 4)

Take SFR and mask bits 4 and 5
(SFR & 0xCF)

then OR the shifted bits
(SFR & 0xCF) | (i << 4)

Edit: ErnieM beat me to it!

 

The smallest and fastest compiled code may be from using bit tests;

sfr.F4 = 0;
if(i.F0) sfr.F4 = 1;
sfr.F5 = 0;
if(i.F1) sfr.F5 = 1;

(If your C compiler is one of those poor ones that doesn't allow bit testing/setting then you may not have that option.)

 

The smallest and fastest compiled code may be from using bit tests;

sfr.F4 = 0;
if(i.F0) sfr.F4 = 1;
sfr.F5 = 0;
if(i.F1) sfr.F5 = 1;

(If your C compiler is one of those poor ones that doesn't allow bit testing/setting then you may not have that option.)

This is basically what I do now.

 

If the compiler optimises i<<4 as the swapf instruction then that would be pretty fast too.