如何在 C 中反转 16 位十六进制值？

Question

I'm trying to reverse my value of Pressure data and store this reversed result in the integer uint16_t Pcounts. I have found a macro that can be used and it seems to flip some bits but not others. Here is my code:

#define REV(X) ((X << 24) | (((X>>16)<<24)>>16) | (((X<<16)>>24)<<16) | (X>>24))

uint16_t PressureData;
uint16_t Pcounts;
Pcounts = REV(PressureData);

If PressureData = 0xAABB I would expect Pcounts to be 0xBBAA. However, the value comes out to be 0xBB00.

PressureData is the real-time value of pressure detected from a sensor.

Answer 1

If PressureData = 0xAABB I would expect Pcounts to be 0xBBAA.

Let us assume with abcd , OP wants cdab (byte swap) and not dcba (nibble swap):

The below reverses bytes for a 32-bit value. @Some programmer dude

#define REV(X) ((X << 24) | (((X>>16)<<24)>>16) | (((X<<16)>>24)<<16) | (X>>24))

To reverse bytes for a 16-bit unsigned value. @Eric (Notice () about X .)

#define REV16_A(X) (((X) << 8) | ((X)>>8))

Better code would not rely on sign-ness, but force the issue.

#define REV16_B(X) (((uint16_t)(X) << 8) | ((uint16_t)(X)>>8))

Yet I see little reason for a macro here and simply use a helper function. Make it inline if desired.

uint16_t rev16(uint16_t x) {
  return (x << 8) | (x >>8);
}

---

Further, I expect the desire to flip the bytes is an endian issue. So if the code is ported, flipping the bytes might / might not be desired. In that case, it is likely the hardware is presenting the bytes in a certain endian which may/may not match the native endian of code. Consider a helper function to handle that.

uint16_t endian_hw_to_native16(uint16_t x) {
  // endian sensitive code ...
}

Answer 2

The problem is you are overriding some bits.

Try using a function instead a macro:

swapped = ((input>>8)&0x0f) | // move byte 1 to byte 0
           (input<<8)&0xf0)); // byte 0 to byte 1