Detecting endianness programmatically in a C++ pro

Question

Is there a programmatic way to detect whether or not you are on a big-endian or little-endian architecture? I need to be able to write code that will execute on an Intel or PPC system and use exactly the same code (i.e. no conditional compilation).

Answer 1

Unless you're using a framework that has been ported to PPC and Intel processors, you will have to do conditional compiles, since PPC and Intel platforms have completely different hardware architectures, pipelines, busses, etc. This renders the assembly code completely different between the two.

As for finding endianness, do the following:

short temp = 0x1234;
char* tempChar = (char*)&temp;

You will either get tempChar to be 0x12 or 0x34, from which you will know the endianness.

Answer 2

As pointed out by Coriiander, most (if not all) of those codes here will be optimized away at compilation time, so the generated binaries won't check "endianness" at run time.

It has been observed that a given executable shouldn't run in two different byte orders, but I have no idea if that is always the case, and it seems like a hack to me checking at compilation time. So I coded this function:

#include <stdint.h>

int* _BE = 0;

int is_big_endian() {
    if (_BE == 0) {
        uint16_t* teste = (uint16_t*)malloc(4);
        *teste = (*teste & 0x01FE) | 0x0100;
        uint8_t teste2 = ((uint8_t*) teste)[0];
        free(teste);
        _BE = (int*)malloc(sizeof(int));
        *_BE = (0x01 == teste2);
    }
    return *_BE;
}

MinGW wasn't able to optimize this code, even though it does optimize the other codes here away. I believe that is because I leave the "random" value that was alocated on the smaller byte memory as it was (at least 7 of its bits), so the compiler can't know what that random value is and it doesn't optimize the function away.

I've also coded the function so that the check is only performed once, and the return value is stored for next tests.

Answer 3

As stated above, use union tricks.

There are few problems with the ones advised above though, most notably that unaligned memory access is notoriously slow for most architectures, and some compilers won't even recognize such constant predicates at all, unless word aligned.

Because mere endian test is boring, here goes (template) function which will flip the input/output of arbitrary integer according to your spec, regardless of host architecture.

#include <stdint.h>

#define BIG_ENDIAN 1
#define LITTLE_ENDIAN 0

template <typename T>
T endian(T w, uint32_t endian)
{
    // this gets optimized out into if (endian == host_endian) return w;
    union { uint64_t quad; uint32_t islittle; } t;
    t.quad = 1;
    if (t.islittle ^ endian) return w;
    T r = 0;

    // decent compilers will unroll this (gcc)
    // or even convert straight into single bswap (clang)
    for (int i = 0; i < sizeof(r); i++) {
        r <<= 8;
        r |= w & 0xff;
        w >>= 8;
    }
    return r;
};

Usage:

To convert from given endian to host, use:

host = endian(source, endian_of_source)

To convert from host endian to given endian, use:

output = endian(hostsource, endian_you_want_to_output)

The resulting code is as fast as writing hand assembly on clang, on gcc it's tad slower (unrolled &,<<,>>,| for every byte) but still decent.

Answer 4

The C++ way has been to use boost, where preprocessor checks and casts are compartmentalized away inside very thoroughly-tested libraries.

The Predef Library (boost/predef.h) recognizes four different kinds of endianness.

The Endian Library was planned to be submitted to the C++ standard, and supports a wide variety of operations on endian-sensitive data.

As stated in answers above, Endianness will be a part of c++20.

Answer 5

int i=1;
char *c=(char*)&i;
bool littleendian=c;

Answer 6

compile time, non-macro, C++11 constexpr solution:

union {
  uint16_t s;
  unsigned char c[2];
} constexpr static  d {1};

constexpr bool is_little_endian() {
  return d.c[0] == 1;
}