byte move in a 16 bit system

I use code like this:

#include <stdint.h>
// Expand nWords into bytes in place.  Be sure the buffer at pWords is
//   of length nWords*2 words!

#define BYTES_PER_WORD 2

void unpack16(uint16_t* pWords, size_t nWords)
{
	uint16_t *pW = &pWords[nWords];  // just past last word	
	// put byte pointer just past last ch
	unsigned char *pB = (unsigned char*) &pWords[nWords*BYTES_PER_WORD]; 
	
	while (pB != (unsigned char*) pWords) {
		uint16_t W = *--pW;
		// LSB, but we're going backwards, so MSB first:
		*--pB = W >>  8 & 0xFF;
		*--pB = W       & 0xFF;
	}
}

Regards,

Bill

So you have used "char" as the "byte" type?

#include <stddef.h>

void *mem2byte(void *src, unsigned int *dest, size_t len)
{
  size_t ii;

  for(ii = 0; ii < len; ii++)
  {
    dest[ii] = __byte((int *) src, ii);
  }

  return dest;
}

You mean something like this?

There is no byte, or 8-bit type on TI DSP processors.  I use the 'char' type simply to mean, here are 16-bit uints, where only the lowest 8-bits are meaningful.

Regards,

Bill

Bill:
Thank you a lot! A variable of type 'char' also occupies 16 bits, but its high byte is meaningless. I didn't know the difference between type 'char' and 'int16' until you told me. I appreciate it!
Regards,
Tony

Thanks!

 I hoped on something faster. Because that is a lot of overhead to copy bytes at speed.

 

People sometimes do byte-packing to save memory. For example, you could define a data type with two 8-bit bitfields in a 16-bit unsigned int to pack two bytes together. Then copy the 16-bit ints. This is very effective for saving memory, say if you have large ASCII strings that you are transmitting out a serial port.

- David

Perhaps, but compared to most I/O interfaces (like UART) it is blazingly fast.  For USB you shouldn't need to pack/unpack.

I do wish the DMA controller could be set up to do the unpacking, but alas it cannot.

Regards,

Bill