how to easily make u12 into u16 with sign extension?

sweethomela8 · May 3, 2011

Hi, whats the easiest way to transform a 12 bit unsigned variable into a 16 bit unsigned variable where the additional 4 bits equal to the 12th bit of the original variable? (sign extension)

ie. 0xF0A -> 0xFF0A
ie. 0xA12 -> 0xFA12
ie. 0x612 -> 0x0612

thanks.

In the C programming language.

andre_luis · May 3, 2011

Check if operation bellow works :

Code:

Value = Value | ( 15 * 256 * 256 * 256 ) ; // Shift left 3 nibles

You can declarte it like constant (#define .... ).
I just posted this way to clarify.

+++

alexan_e · May 3, 2011

You can do it using typecasting, when you shift a signed number the sign is preserved (arithmetic shift)

Code C - [expand]
1
2
3
4
volatile unsigned int x;
 
x=0xF0A;
x=((signed int)x<<4)>>4;  // shift left 4 bits so that bit11 goes to bit15 as sign and then shift right 4 and the "sign" will be preserved

Alex

horace1 · May 3, 2011

the following is OK if >> is arithmetic shift

Code:

x=((signed int)x<<4)>>4;

it will not work if >> is logical (the C standard does not specify if >> is arithmetic or logical shift it is implementation dependent)
try

Code:

    y = y | ((y & 0x800) << 1) | ((y & 0x800) << 2) | ((y & 0x800) << 3) | ((y & 0x800) << 4);

or

Code:

   if(y & 0x800) y |= 0xf000;

how to easily make u12 into u16 with sign extension?

sweethomela8

Member level 4

andre_luis

Super Moderator

alexan_e

Administrator

horace1

Advanced Member level 5

Similar threads

Code C - [expand]
1 2 3 4	volatile unsigned int x; x=0xF0A; x=((signed int)x<<4)>>4; // shift left 4 bits so that bit11 goes to bit15 as sign and then shift right 4 and the "sign" will be preserved

how to easily make u12 into u16 with sign extension?

sweethomela8

Member level 4

andre_luis

Super Moderator

alexan_e

Administrator

horace1

Advanced Member level 5

Similar threads

Privacy & Transparency

Privacy & Transparency