C++ – 64 bit ntohl() in C++

64-bitcendiannesslinux

The man pages for htonl() seem to suggest that you can only use it for up to 32 bit values. (In reality, ntohl() is defined for unsigned long, which on my platform is 32 bits. I suppose if the unsigned long were 8 bytes, it would work for 64 bit ints).

My problem is that I need to convert 64 bit integers (in my case, this is an unsigned long long) from big endian to little endian. Right now, I need to do that specific conversion. But it would be even nicer if the function (like ntohl()) would NOT convert my 64 bit value if the target platform WAS big endian. (I'd rather avoid adding my own preprocessor magic to do this).

What can I use? I would like something that is standard if it exists, but I am open to implementation suggestions. I have seen this type of conversion done in the past using unions. I suppose I could have a union with an unsigned long long and a char[8]. Then swap the bytes around accordingly. (Obviously would break on platforms that were big endian).

Best Answer

Documentation: man htobe64 on Linux (glibc >= 2.9) or FreeBSD.

Unfortunately OpenBSD, FreeBSD and glibc (Linux) did not quite work together smoothly to create one (non-kernel-API) libc standard for this, during an attempt in 2009.

Currently, this short bit of preprocessor code:

#if defined(__linux__)
#  include <endian.h>
#elif defined(__FreeBSD__) || defined(__NetBSD__)
#  include <sys/endian.h>
#elif defined(__OpenBSD__)
#  include <sys/types.h>
#  define be16toh(x) betoh16(x)
#  define be32toh(x) betoh32(x)
#  define be64toh(x) betoh64(x)
#endif

(tested on Linux and OpenBSD) should hide the differences. It gives you the Linux/FreeBSD-style macros on those 4 platforms.

Use example:

  #include <stdint.h>    // For 'uint64_t'

  uint64_t  host_int = 123;
  uint64_t  big_endian;

  big_endian = htobe64( host_int );
  host_int = be64toh( big_endian );

It's the most "standard C library"-ish approach available at the moment.

Setting a bit

Use the bitwise OR operator (|) to set a bit.

number |= 1UL << n;

That will set the nth bit of number. n should be zero, if you want to set the 1st bit and so on upto n-1, if you want to set the nth bit.

Use 1ULL if number is wider than unsigned long; promotion of 1UL << n doesn't happen until after evaluating 1UL << n where it's undefined behaviour to shift by more than the width of a long. The same applies to all the rest of the examples.

Clearing a bit

Use the bitwise AND operator (&) to clear a bit.

number &= ~(1UL << n);

That will clear the nth bit of number. You must invert the bit string with the bitwise NOT operator (~), then AND it.

Toggling a bit

The XOR operator (^) can be used to toggle a bit.

number ^= 1UL << n;

That will toggle the nth bit of number.

Checking a bit

You didn't ask for this, but I might as well add it.

To check a bit, shift the number n to the right, then bitwise AND it:

bit = (number >> n) & 1U;

That will put the value of the nth bit of number into the variable bit.

Changing the nth bit to x

Setting the nth bit to either 1 or 0 can be achieved with the following on a 2's complement C++ implementation:

number ^= (-x ^ number) & (1UL << n);

Bit n will be set if x is 1, and cleared if x is 0. If x has some other value, you get garbage. x = !!x will booleanize it to 0 or 1.

To make this independent of 2's complement negation behaviour (where -1 has all bits set, unlike on a 1's complement or sign/magnitude C++ implementation), use unsigned negation.

number ^= (-(unsigned long)x ^ number) & (1UL << n);

unsigned long newbit = !!x;    // Also booleanize to force 0 or 1
number ^= (-newbit ^ number) & (1UL << n);

It's generally a good idea to use unsigned types for portable bit manipulation.

number = (number & ~(1UL << n)) | (x << n);

(number & ~(1UL << n)) will clear the nth bit and (x << n) will set the nth bit to x.

It's also generally a good idea to not to copy/paste code in general and so many people use preprocessor macros (like the community wiki answer further down) or some sort of encapsulation.

Why does an 8-bit field have endianness

There are two kind of order. One is byte order, one is bitfield order. There is no standard order about the bitfield order in C language. It depends on the compiler. Typically, the order of bitfields are reversed between big and little endian.