[C++] Cast int16_t memory to float

Question

Hi Stackoverflow,

I have a function from an external source that returns an array of 2 uint16_t elements (which I cast to int).

I have already been able to cast these to one "big" int ((i1 << 16) + i2)

Now I need to be able to cast this to float, keeping the point value as is in memory.

Can anyone suggest a way or point me in the right direction?

Thanks in advance!

Answer 1

Perhaps something like this would work?

uint32_t a = ....; // Contains your big int.
float b = *(float*)&a;

Of course that would require you to know int has the same size as float...

Answer 2

I suggest to keep very clear that you are messing around and use a memcpy:

float a;
int b;

memcpy(&a,&b,min(sizeof(a),sizeof(b)));

someone might encounter your code when you're long gone, in which case this will show there's something special happening intentionally.

Answer 3

It depends on what you mean:

uint16_t *ip = external_source();
float f = *(float *)ip;

It could also mean:

uint16_t *ip = external_source();
uint32_t  iv = ((uint32_t)ip[0] << 16) | ip[1];
float      f = iv;

Or you could have some other interpretation in mind. The first assumes that the bits in the array are in the correct sequence to be interpreted as a float. But you need to explain what conversion you require more clearly.

Answer 4

I'd use a union:

union fltconv {
    float f;
    uint32_t i;
} u;
u.i = ((uint32_t) i1 << 16) | i2;
float f = u.f;

It's more explicit about what you're doing. Bear in mind that this is very nonportable.