# high-Low-Bit Shifting

Hi,

I have 3 bytes, a,b and c.

a= byte 1: contains high 8 bits of a
b= byte 2: contains high 8 bits of b
c= byte 3: contains 4 bits low(for byte 1) and 4bits low (for byte 2)

i want to take the first 4 bits of c and insert them in the low of a .So result should be

a= {lo 4 bits  of c}{high 8 bits of a} = 12 bits;

thanks
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Commented:
Hi XPUSR,

lo 4 bits of c
c & 0x15

a anyway has 8 bits
a= {lo 4 bits  of c}{high 8 bits of a}

((c & 0x15 ) <<8) | a

to get

Cheers!
Sunny:o)

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Commented:
Sunny, I confess that I don't know which language this is, but doesn't 0x usually precede a hexadecimal value?

So you meant 15 decimal or 0x0F

?

Just checking ;-)
Commented:
>but doesn't 0x usually precede a hexadecimal value?

>So you meant 15 decimal or 0x0F
yes ... you are right :o)

May be I was in too much hurry when I typed the response

modified response

Comment from sunnycoder
Date: 12/11/2003 06:11PM IST

Hi XPUSR,

lo 4 bits of c
c & 0x0F

a anyway has 8 bits
a= {lo 4 bits  of c}{high 8 bits of a}

((c & 0x0F ) <<8) | a

thanks jimmack
Commented:
LOL ... looks like I copied too much from the post

I am becoming absent minded :o(
Commented:
;-)
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