I'm having trouble reading in a 16bit .wav file. I have read in the header information, however, the conversion does not seem to work.
For example, in Matlab if I read in wave file I get the following type of data:
-0.0064, -0.0047, -0.0051, -0.0036, -0.0046, -0.0059, -0.0051
However, in my C++ program the following is returned:
0.960938, -0.00390625, -0.949219, -0.00390625, -0.996094, -0.00390625
I need the data to be represented the same way. Now, for 8 bit
.wav files I did the following:
uint8_t c;
for(unsigned i=0; (i < size); i++)
{
c = (unsigned)(unsigned char)(data[i]);
double t = (c-128)/128.0;
rawSignal.push_back(t);
}
This worked, however, when I did this for 16bit:
uint16_t c;
for(unsigned i=0; (i < size); i++)
{
c = (signed)(signed char)(data[i]);
double t = (c-256)/256.0;
rawSignal.push_back(t);
}
Does not work and shows the output (above).
I'm following the standards found Here
Where data
is a char
array and rawSignal
is a std::vector<double>
I'm probably just handing the conversion wrong but cannot seem to find out where. Anyone have any suggestions?
Thanks
EDIT:
This is what is now displaying (In a graph):
This is what it should be displaying:
There are a few problems here:
Assuming a little-endian architecture, your code would look more like this (very close to Carl's answer):
for (int i = 0; i < size; i += 2)
{
int c = (data[i + 1] << 8) | data[i];
double t = c/32768.0;
rawSignal.push_back(t);
}
for a big-endian architecture:
for (int i = 0; i < size; i += 2)
{
int c = (data[i] << 8) | data[i+1];
double t = c/32768.0;
rawSignal.push_back(t);
}
That code is untested, so please LMK if it doesn't work.