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scaven92
Joined: 06 Oct 2006
Posts: 44
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| How to receive int32 or float over SPI? |
 Posted: Sun Oct 07, 2007 8:22 pm |
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I have searched for this and cannot find the exact answer I am looking for. As I understand it, when receiving/sending data over SPI (in my case from Slave to the Master) you can only send an 8 bit integer at a time. I need to be able to send an int32, or float, to the Master. For example I will have to send the output from a PID controller to the Master, so I will have signed numbers with decimals (ex: -5.34).
How can I do this? In regard to the int32, would I just shift the 32 bit integer and send 8 bits at a time, 4 times. Once I have received this on my Master, would I then just reconstruct the int32 by shifting them in the other way? Also this data will have to be signed.
If I were to use a float how would I do this? Is there any better way? I am trying to conserve time to maintain I high sampling rate for my application. I know that bit operations are not slow, but the less overhead the better.
Thanks for the help! |
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PCM programmer
Joined: 06 Sep 2003
Posts: 21708
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| Posted: Mon Oct 08, 2007 12:24 am |
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You can use this FAQ article as a guide and write some macros to
break up a float into 4 bytes, and then to re-assemble the 4 bytes back
into a float.
http://www.ccsinfo.com/faq.php?page=write_eeprom_not_byte
Example:
| Code: |
#include <18F452.H>
#fuses XT, NOWDT, BROWNOUT, PUT, NOLVP
#use delay(clock=4000000)
#use rs232(baud=9600, xmit=PIN_C6, rcv=PIN_C7, ERRORS)
#define GetByte(x, offset) *((int8 *)&x +offset)
#define MakeDWord(x, b3, b2, b1, b0) \
*(int8 *)&x = b0; \
*((int8 *)&x +1) = b1; \
*((int8 *)&x +2) = b2; \
*((int8 *)&x +3) = b3
//======================================
void main()
{
float value, result;
int8 b0, b1, b2, b3;
value = 123.4567;
printf("value = %7.4f \r", value);
// Break the float into 4 bytes.
b0 = GetByte(value, 0);
b1 = GetByte(value, 1);
b2 = GetByte(value, 2);
b3 = GetByte(value, 3);
// Re-assemble the float from 4 bytes.
MakeDWord(result, b3, b2, b1, b0);
printf("result = %7.4f \n\r", result);
while(1);
}
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Actually, if you want minimum overhead, you should probably do it
with loops as shown in the FAQ. The code above uses un-rolled loops. |
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Guest
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| Exactly what i needed to see! |
| Posted: Mon Oct 08, 2007 12:53 am |
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Wow, that is exactly the information I needed. I appreciate the help greatly. You source code will be a perfect thing to work off of, as soon as I can assemble a circuit to test it. Thank you very much!
I searched the forum, and I should have searched the CCS FAQ also. Thanks for the heads up I will need to look through that more often. |
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