Question

I read this code in a library which is used to display a bitmap (.bmp) to an LCD.
I do really hard in understanding what is happening at the following lines, and how it does happen.

Maybe someone can explain this to me.

uint16_t s, w, h;
uint8_t* buffer;   // does get malloc'd

s = *((uint16_t*)&buffer[0]);
w = *((uint16_t*)&buffer[18]);
h = *((uint16_t*)&buffer[22]);

I guess it’s not that hard for a real C programmer, but I am still learning, so I thought I just ask 🙂
As far as I understand this, it sticks somehow together two uint8_tvariables to an uint16_t.

Thanks in advance for your help here!

Answer 1

In the code you’ve provided, buffer (which is an array of bytes) is read, and values are extracted into s, w and h.

The (uint16_t*)&buffer[n] syntax means that you’re extracting the address of the n^th byte of buffer, and casting it into a uint16_t*. The casting tells the compiler to look at this address as if points at a uint16_t, i.e. a pair of uint8_ts.
The additional * in the code dereferences the pointer, i.e. extracts the value from this address. Since the address now points at a uint16_t, a uint16_t value is extracted.

As a result:

1. s gets the value of the first uint16_t, i.e. bytes 0 and 1.
2. w gets the value of the tenth uint16_t, i.e. bytes 18 and 19.
3. h gets the value of the twelveth uint16_t, i.e. bytes 22 and 23.