Variable Storage

[robcarter 0]

Hi

In an application i have to implement a 40 bit shift register, does anyone know if i define 5 char variables will they always be placed in consecutive memory locations?

e.g using

volatile unsigned char data1,data2,data3,data4,data5;

or using

volatile unsigned long date1;

volatile unsigned char data2;

 

I am using volatile because they are accessed from an interrupt.

 

or would an array be better?

 

thanks

Edited December 3, 2014 by robcarter

[Reynard 0]

Hi Rob,

 

An array or structure will keep them together otherwise the linker may put them individually anywhere it can find space.

 

Cheers

 

Reynard

[robcarter 0]

Thanks, if I use a structure can i access it from assembly for example using:

movf _structure_name.element_name,w

[trossin 0]

Why do the variables have to be together? The following code pieces can shift a 40 bit variable left or right by one bit and shift in the input Input. The 40 bit variable in Verilog speak is {High[31:0],Low[7:0]}.

 

unsigned long High;

unsigned char Low,Carry;

unsigned char Input;

 

// Shift Left

Carry = (Low & 0x80) ? 1 : 0;

High = (High<<1) | Carry;

Low = (Low<<1) | Input;

 

// Shift Right

 

Carry = (High & 1) ? 0x80 : 0;

High = (High>>1) | (Input ? 0x80000000 : 0);

Low = (Low>>1) | Carry;

Edited December 5, 2014 by trossin

[robcarter 0]

i theory they don't need to be together, but I wanted to implement the shifting in assembler because the 36 shifts (I don't need all 40 bits) are output serially on an interupt and didn't want the extra overhead of bank switching code being inserted. I think the best way is probably to assign fixed addresses at 0x7? so they an be accessed from any bank.