Convert serial port data to float in C with union - c

Question

I'm using a C program on Linux to read data from a serial port.
The data to read comes from Code Composer Studio from the line: UART_writePolling(uartHandle, (uint8_t*) &value, sizeof(float));
value is the float I want to read in C, where value = 1.5.
When I read in the data from the serial port, in C, into a buffer and print with printf("%u\n", (int)buffer[i]);
I get value to be:
0
0
4294967232
63
and when I insert buffer[i] into a.array and print with
printf("%d\n", a.array[i]);
I get value to be:
0
0
-64
63
I've also tried using unions:
unsigned int value = 0;
for (int j = 3; j >= 0; j--){
//value <<= 8;
value = value + (int)a.array[i+8+j];
}
printf("value: %u\n", value);
data.u = value;
printf("(float): %f\n", data.f);
which doesn't give the correct answer.
How can I use union to get the correct data as a float?
Do I need to use <<?
EDIT: better idea of the code
//headers
typedef struct {
int *array;
size_t used;
size_t size;
} Array;
void initArray(Array *a, size_t initialSize) {
a->array = (int *)malloc(initialSize * sizeof(int));
a->used = 0;
a->size = initialSize;
}
... //more functions/code to resize array and free the memory later
union Data {
float f;
unsigned int u;
};
int main(){
union Data data;
//open serial port code
char buffer[1]; /* Buffer to store the data received,
reading one at a time */
Array a;
initArray(&a, 5); /* initialise an array to store the read data
that is read into buffer*/
//while loop to read in data for some amount of time/data
int b_read = 0;
b_read = read(fd, &buffer, sizeof(buffer));
for (int i=0; i<b_read; i++){
printf("%u\n", (int)buffer[i]);
// how the first set of values above were printed
insertArray(&a, buffer[i]);
// also adding the values read to buffer into array a
}
//end while
// close the port
for(int i=0; i<no. of elements in array a; i++){
printf("%d\n", a.array[i]);
// how the second set of results were printed
}
//below is an attempt at using union and <<:
unsigned int value = 0;
for (int j = 3; j >= 0; j--){
//value <<= 8;
value = value + (int)a.array[i+8+j]; //index used is particular to my code, where value is in a specific place in the array
}
printf("value: %u\n", value);
data.u = value;
printf("(float): %f\n", data.f);
//these printfs don't give a reasonable answer
// free memory
return 0;
}

Answer 1

Once the bytes are in buffer starting at offset i, you can reinterpret the bytes as a float with:
float f;
memcpy(&f, buffer+i, sizeof f);
To use a union, you could use:
union { uint32_t u; float f; } x;
x.u = value;
float f = x.f;
However, this requires that value contain all 32 bits that represent the float. When you attempted to construct the value with:
//value <<= 8;
value = value + (int)a.array[i+8+j];
There are two issues. First, value <<= 8 is needed. I presume you tried it first and did not get a correct answer, so you commented it out. However, it is required. Second, this code to insert the bytes one-by-one into value is order-dependent. Once the shift is restored, it will insert greater-addressed bytes into less-significant bits of value. Systems generally arrange bytes in objects in one of two orders: More significant bytes in lower addresses or more significant bytes in greater addresses. We do not know which order your system uses, so we do not know whether your code to insert the greater-addressed bytes in less significant bytes is correct.
Note: The above assumes that the bytes are read and written in the same order, or that issues of endianness have already been handled in other code.

Answer 2

You use printf with %u but cast into a int. So maybe it's not surprising to have this behavior since 2^32 = 4294967296, and 4294967296 - 64 (your second printf result) = 4294967232 (your first printf result).
Just cast into "unsigned" if you use "%u" or cast into "int" if you use "%d".