cstm32
I was looking for solution how convert float value to binary in STM32 and I found this:
float foo = 1.23;
uint32_t bar;
bar = *((uint32_t *)&foo);
My question is how does it works? I know it's some kind of pointer conversion, but I don't understand how precisely does it works.
Looking at the hardware of the basic STM unit I'd say there was one ADC and that means no simultaneous sampling. The 16 inputs are likely to be available due to it having an addressable multiplxer which can route one of the 16 to the single ADC input: -
EDIT However, looking at the STM34F4 device it says this (and thanks to Chris Stratton for pointing this out): -
Because the device can do simultaneous sample and hold then this will fit the bill.
1) You need to intialize frame as array and it's length is the maximum frame length you may recieve. Now you are only making a pointer which point to one place then you write to bytes after that in RAM and it may cause errors. If you didn't see errors you only get lucky!!
2) Numbers is numbers 1 is the same as (0b00000001) in binary as the same as (0x01) in hex. you only need to compare as you want. For example to compare the start byte you could do one of the followings:
if(frame[0] == 0x7E){
// Here you are comparing with Hexadicimal
}
if(frame[0] == 126){
// Here you are comparing with Decimal
}
if(frame[0] == 0b01111110){
// Here you are comparing with Binary
}
Note that any conditions of the above have the same machine code in AVR. You are only represent the number to yourself.
Also i have some notes on that code:
1) Why are you add statickeyword to ch , *frame , index it has no meaning at all here.
2) For frame it is better to be unsigend char or uint8_t as char alone is signed. intialize as array like uint8_t frame[10];
Best Answer
Literally nothing happens. The
floatis already in a binary representation; the cast merely tells the compiler that it should treat the 32 bits in the variable as a 32 bit unsigned integer instead of a single precision floating point value.