Assembling a float from an array of bytes

Question

I would like to convert an array of bytes, received from serial, to a float.

Let's consider this exemple. I'm sending 0.12 byte-wise from an Android application. The conversion from float to array of bytes in the Android side is handled by this function

public static byte [] float2Bytes(float value)
 {
 return ByteBuffer.allocate(4)./*order(ByteOrder.LITTLE_ENDIAN).*/putFloat(value).array();
 }

When bytes arrive, they are stored in an array and the assembled.

hiBytew1 = bufferBytes[9];
//Serial.println(hiBytew1);
loBytew1 = bufferBytes[10];
//Serial.println(loBytew1);
hiBytew2 = bufferBytes[11];
//Serial.println(hiBytew2);
loBytew2 = bufferBytes[12];
//Serial.println(loBytew2);
float conAck = assemble(hiBytew1, loBytew1, hiBytew2,loBytew2);

How can I achieve this kind of conversion assuming that bytes are received in the correct order?

Answer 1

Just cast the pointer.

float conAck = *((float*)(bufferBytes + 9));

• Edgar Bonet: This needs parentheses around "bufferBytes + 9".
• Ignacio Vazquez-Abrams: Right, otherwise it will advance sizeof(float) * 9 bytes instead.
• UserK: Grazie Ignacio, so it will automatically assemble the sequence in a float variable? Nice
• Ignacio Vazquez-Abrams: @UserK: It will tell the compiler that the bytes it will be looking at are a float.
• xyz: @userk there is no assembly. Floats are already stored as bytes in memory. You are asking android to put the 4 most meaningful bytes in the buffer. Then you are trekking arduino to consider the bytes together as a float. If you are to store the same number in memory it week have the same representation. In a nutshell, this is an O(1) operation.