endianness

I've got the following C++ code : template <int isBigEndian, typename val> struct EndiannessConv { inline static val fromLittleEndianToHost( val v ) { union { val outVal __attribute__ ((used)); uint8_t bytes[ sizeof( val ) ] __attribute__ ((used)); } ; outVal = v; std::reverse( &bytes[0], &bytes[ sizeof(val) ] ); return outVal; } inline static void convertArray( val v[], uint32_t size ) { // TODO : find a way to map the array for (uint32_t i = 0; i < size; i++) for (uint32_t i = 0; i < size; i++) v[i] = fromLittleEndianToHost( v[i] ); } }; Which work and has been tested (without the used

I am inquiring about how to tell when one element in an array has finished and another is beginning in an endian architecture. I have 2 arrays where the size of long is 8 and the size of char is 1 long x[2] = {0x012345,0xFEDC}; char c[12] = {'a','b','c','d','e','f','g','h','0','1','2','3'}; And I was wondering how these values would be stored in the different Endian architectures if we consider x starting at memory address 0x100 and c starting at memory address 0x200 . I thought that the Big Endian address would be {01,23,45,FE,DC} where the first element of the set is at memory address 0x100

I need to get 64-bit little-endian integer as byte-array with upper 32 bits zeroed and lower 32 bits containing some integer number, let say it's 51. Now I was doing it in this way: byte[] header = ByteBuffer .allocate(8) .order(ByteOrder.LITTLE_ENDIAN) .putInt(51) .array(); But I'm not sure is it the right way. Am I doing it right? What about trying the following: private static byte[] encodeHeader(long size) { if (size < 0 || size >= (1L << Integer.SIZE)) { throw new IllegalArgumentException("size negative or larger than 32 bits: " + size); } byte[] header = ByteBuffer .allocate(Long.BYTES)

I'm confused when I was reading an article about Big/Little Endian. Code goes below: #include <iostream> using namespace std; int i = 12345678; int main() { char *p = (char*)&i; //line-1 if(*p == 78) //line-2 cout << "little endian" << endl; if(*p == 12) cout << "big endian" << endl; } Question: In line-1, can I do the conversion using static_cast<char*>(&i) ? In line-2, according to the code, if it's little-endian, then 78 is stored in the lowest byte, else 12 is stored in the lowest byte. But what I think is that, i = 12345678; will be stored in memory in binary. If it's little-endian, then

The examples in the libpq documentation show how to get the the integer value by converting it to the host-endian representation. I am curious what must be done to get the double precision value using libpq (without libpqtyppes)? I have tried reinterpret_cast with no luck. Also why text and byte data doesn't need endian conversions? The DB runs locally on Windows 7, I am using Visual C++ 2013. pptr is a double vaule I am trying to retrieve. #include <iostream> #include <memory> #include <vector> #include <libpq-fe.h> #include <Winsock2.h> static void show_binary_results(PGresult *res) { int i,