Casting float to unsigned short

Question

Hi everyone, I've got a strange bug with my system.

I have a float struct member which I divide with a float resolution value. Then I cast it to an unsigned short so that I can put it in a two bytes message.

Like this:

float calc, num = -150.35, res = 0.703125;
unsigned short val;

val = (unsigned short)(num / res);
msg_1 = val >> 8;
msg_2 = val;

The problem here, with a positive num variable, I cast it to the val successfully; with a negative one, val is 0 no matter what is the num's value. I couldn't find out why.

But when I changed the type of val to signed short, I managed to calculate the result successfully for both of the positive and the negative values. But I'm not sure if this is healthy for messages. I usually use unsigned short and unsigned char without knowing why.

Answer 1

I guess I found my answer:

https://stackoverflow.com/questions/10541200/is-the-behaviour-of-casting-a-negative-double-to-unsigned-int-defined-in-the-c-s

Answer 2

There is a reason why "unsigned short" is named "unsigned"
whenever a data type is unsigned, it means only positive values can be stored in it as there is no space reserved to store the "sign"( positive/negative ) of the number.

Answer 3

Powehi,
Your question topic was about type casting from `float` to `unsigned int`. The way you can print an `unsigned int` using a format specifier intended for `int` isn't an actual example to the topic (type casting).

And for signed integers, as you probably know already, the number of usable bits to represent a value is subtracted by one, as the MSB is acting as the sign bit (signed 7bits, unsigned 8bits).

And please don't forget, there's a risk of data loss when cast from a higher ranked type to a lower ranked type, `float` usually takes 4 bytes (higher ranked), `unsigned short` usually takes 2 bytes (lower ranked).

Answer 4

Arsenic false. You can store all the bits in the unsigned data anyway.

u can try :

int x = -5;
unsigned int y = x;
printf("%d",y);

Answer 5

Powehi
Wrong, you are using %d for printing an `unsigned int` variable when you should've used %u, the correct format specifier used for printing/reading `unsigned int`.

Answer 6

Ipang I obviously know. It's just pointing out that the data can be stored as-is. It's just ignoring the sign bit or not.

The system is coded with C99. Maybe relevant?

Answer 7

I mean, let's talk about bytes.

1111 1111 -> if unsigned; 255
1111 1111 -> if signed; -1

Whether signed or unsigned, there are 8 bits. And you can converse these. Assigning -1 to an unsigned char, means assigning 255.

Answer 8

Thanks Ipang, I am aware of the data-loss. Just can't figure out the way the system calculates 0. Doesn't make any sense + with other IDE's the casting is OK, the result it successfully calculated.

Answer 9

Yeah that makes sense 👍