CODEWITHSUNDEEP
ccastingbuffershift
sam
sam

Reputation: 63

casting and shifting to from uint16_t to int16_t

I have a noob question about casting and shifting. I'm trying to store two elements of uint8_t from an array as a single, signed element in an array of type int16_t. However my result is not correct, and I' don't know why.

I have this code:

uint8_t buffer[BUFFER_SIZE];
int16_t mp3_stereo_buffer[BUFFER_SIZE];

for (i = 0; i < BUFFER_SIZE; i += 2) {
    mp3_stereo_buffer[i] = ((uint16_t)buffer[i] << 8) | ((uint16_t)buffer[i + 1]);
}

Upvotes: 1

Views: 4962

Answers (2)

Koorosh Hajiani
Koorosh Hajiani

Reputation: 94

#define LITTLE_ENDIAN
int main()
{
uint8_t buffer[4]={0x15,0xff,0x63,0xee};
int16_t mp3_stereo_buffer[2];

for (int i = 0; i < 2; i += 1) {
    mp3_stereo_buffer[i] = ((int16_t*)buffer)[i];

#ifdef LITTLE_ENDIAN
    mp3_stereo_buffer[i]=(mp3_stereo_buffer[i]>>8 &0x00ff) | (mp3_stereo_buffer[i]<<8 & 0xff00);              
#endif

    printf("%x\n",mp3_stereo_buffer[i]&0xffff);
 }
return 0;
}

Upvotes: 0

rustyx
rustyx

Reputation: 85341

First, you can pack N uint8_t into N/2 uint16_t elements.

uint8_t buffer[BUFFER_SIZE];
uint16_t mp3_stereo_buffer[BUFFER_SIZE / 2];

Then you need to know if your data is little-endian or big-endian.

For little-endian:

for (i = 0; i < BUFFER_SIZE / 2; i++) {
  mp3_stereo_buffer[i] = (uint16_t) (buffer[i*2] | (buffer[i*2+1] << 8));
}

For big-endian:

for (i = 0; i < BUFFER_SIZE / 2; i++) {
  mp3_stereo_buffer[i] = (uint16_t) ((buffer[i*2] << 8) | buffer[i*2+1]);
}

p.s. If your data is in fact signed, then you can change the type and the casting to int16_t but beware that the way signed numbers are represented is not guaranteed to be portable.

Upvotes: 1

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