CODEWITHSUNDEEP
clinuxembedded
anshu
anshu

Reputation: 665

Convert a 32 bits to float value

I am working on a DSP processor to implement a BFSK frequency hopping mechanism using C on a Linux system. On the receiver part of the program, I am getting an input of a set of samples which I de-modulate using Goertzel algorithm to determine whether the received bit was a 0 or 1.

Right now, I am able to detect the bits individually. But I have to return the data for processing in the form of a float array. So, I need to pack every set of 32 bits received to form a float value. Right I am doing something like :

uint32_t i,j,curBit,curBlk;
unint32_t *outData; //this is intiallized to address of some pre-defined location in DSP memory
float *output;
for(i=0; i<num_os_bits; i++) //Loop for number of data bits
{ 

//Demodulate the data and set curBit=0x0000 or 0x8000

curBlk=curBlk>>1;
curBlk=curBlk|curBit;

bitsCounter+=1;
    if(i!=0 && (bitsCounter%32)==0) //32-bits processed, save the data in array
    {
        *(outData+j)=curBlk;
        j+=1;
        curBlk=0;
    }
}

output=(float *)outData;

Now, the values of the output array are just the values of outData array with 0s after the decimal point. example: if output[i]=12345 the `outData[i]=12345.0000'.

But while testing the program I am generating the sample test data of bits using an array of float float data[] ={123.12456,45.789,297.0956};

So after the demodulation I am expecting the float array output to have a same values as data array. Is there some other method to convert 32-bits of data to a float. Should I store the received bits to a char array and then convert it to float.

Upvotes: 6

Views: 14211

Answers (2)

Ren&#233; Kolař&#237;k
Ren&#233; Kolař&#237;k

Reputation: 1286

Not sure if I get your point - you sequentialy obtain bits and if you got 32 bit you want to make a float from them?

what about:

union conv32
{
    uint32_t u32; // here_write_bits
    float    f32; // here_read_float
};

Which can be used like this in-line:

float f = ((union conv32){.u32 = my_uint}).f32;

Or with a helper variable:

union conv32 x = {.u32 = my_uint};
float f = x.f32;

Upvotes: 6

Clifford
Clifford

Reputation: 93476

You need to copy the bit pattern verbatim without invoking any implicit conversions. The simplest way to do that is to take the address of the data and reinterpret it as a pointer to the intermediate "carrier" type before dereferencing it.

Consider this:

float source_float = 1234.5678f ;
uint32_t transport_bits = *((uint32_t*)&source_float);
float destination_float = *((float*)&transport_bits);

The intermediate result in transport_bits is target dependent, in my test on x86, it is 0x449a522b, this being the bit representation of single precision float on that platform (floating point representation and endianness dependent).

Either way the result in destination_float is identical to source_float having been transported via the uint32_t transport_bits.

However this does require that the floating point representation of the originator is identical to that of the receiver which may not be a given. It is all a bit non-portable, The fact that your code does not work suggests perhaps that the representation does indeed differ between sender and receiver. Not only must the FP representation be identical but also the endianness. If they differ your code may have to do one or both of reorder the bits and calculate the floating point equivalent by extraction of exponent and mantissa. All so you nee to be sure that the transmission order of the bits is is the same order in which you are reassembling them. There are a number of opportunities to get this wrong.

Upvotes: 5

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