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brutto
Joined: 24 Jun 2015 Posts: 2
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about #byte #bit and xxxx.bits |
Posted: Wed Jun 24, 2015 11:38 am |
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hello,
i have a question, i can do:
Code: |
#BYTE IFS0=0X0084
#BIT IEC0_AD1IE=0X0094.13 |
but could i do any word to write 3 bits for one register? for example, the register "IPC3bits.AD1IP" have 3 bits, would be the solution this?
Code: |
#BIT IPC3_AD1IP0=0x00A4.4
#BIT IPC3_AD1IP1=0x00A4.5
#BIT IPC3_AD1IP2=0x00A4.6
ipc3_ad1ip0=0; //IPC3bits.AD1IP=6
ipc3_ad1ip1=1;
ipc3_ad1ip2=1;
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or is there any other solution to write directly "6" in ipc3.ad1ip register??
thanks in advance... |
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PCM programmer
Joined: 06 Sep 2003 Posts: 21708
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Posted: Wed Jun 24, 2015 11:50 am |
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Yes, declare a structure with bitfields. Then write to the bitfields in your
code. The compiler will handle inserting the correct data into the bits.
Look at the .LST file to see this.
Code: |
#include <16F690.h>
#fuses INTRC_IO,NOWDT
#use delay(clock=4M)
// This structure divides PortB into two bitfields.
struct {
int8 Columns:4; // B0-3: Output pins
int8 Rows:4; // B4-7: Input pins
}PortB;
#byte PortB = getenv("SFR:PORTB")
//=========================
void main()
{
set_tris_b(0);
PortB.Columns = 6;
PortB.Rows = 3;
while(TRUE);
} |
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Ttelmah
Joined: 11 Mar 2010 Posts: 19359
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Posted: Wed Jun 24, 2015 12:06 pm |
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There is another thing you can do, and a comment.
#byte creates a byte variable at a location if asked to. #word creates a 16bit word. To use bit 13, you should be using #word.
However #byte can also do something else. If given the name of an already declared variable, it can _locate_ this variable at the location. So you can declare a struct, using bit fields, like:
Code: |
#struct
{
int8 dummy:4; //bits 0 to 3
int8 ip:3;
int8 dummy2:1;
} ipc3bits ;
#byte ipc3bits=0xA4
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You can then write to the variable ipc3bits.ip as a 3bit integer sitting at the required location in the register!.... |
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brutto
Joined: 24 Jun 2015 Posts: 2
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Posted: Sat Jun 27, 2015 2:53 am |
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thanks guys, i will try with that |
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