How to get float in bytes?
c, floating-point
Solution
You can do it like this:
char data[sizeof(float)];
float f = 0.6f;
memcpy(data, &f, sizeof f); // send data
float g;
memcpy(&g, data, sizeof g); // receive data
In order for this to work, both machines need to use the same floating point representations.
As was rightly pointed out in the comments, you don't necessarily need to do the extra `memcpy`; instead, you can treat `f` directly as an array of characters (of any signedness). You still have to do `memcpy` on the receiving side, though, since you may not treat an arbitrary array of characters as a float! Example:
unsigned char const * const p = (unsigned char const *)&f;
for (size_t i = 0; i != sizeof f; ++i)
{
printf("Byte %zu is %02X\n", i, p[i]);
send_over_network(p[i]);
}
Problem
I am using the HIDAPI to send some data to a USB device. This data can be sent only as byte array and I need to send some `float` numbers inside this data array. I know floats have `4` bytes. So I thought this might work: ``` float f = 0.6; char data[4]; data[0] = (int) f >> 24; data[1] = (int) f >> 16; data[2] = (int) f >> 8; data[3] = (int) f; ``` And later all I had to do is: ``` g = (float)((data[0] << 24) | (data[1] << 16) | (data[2] << 8) | (data[3]) ); ``` But testing this shows me that the lines like `data[0] = (int) f >> 24;` returns always `0`. What is wrong with my code and how may I do this correctly (i.e. break a `float` inner data in `4 char` bytes and rebuild the same `float` later)? EDIT: I was able to accomplish this with the following codes: ``` float f = 0.1; unsigned char *pc; pc = (unsigned char*)&f; // 0.6 in float pc[0] = 0x9A; pc[1] = 0x99; pc[2] = 0x19; pc[3] = 0x3F; std::cout << f << std::endl; // will print 0.6 ``` and ``` *(unsigned int*)&f = (0x3F << 24) | (0x19 << 16) | (0x99 << 8) | (0x9A << 0); ``` I know `memcpy()` is a "cleaner" way of doing it, but this way I think the performance is somewhat better.