endianness

So, I was wondering, why some architectures use little-endian and others big-endian. I remember I read somewhere that it has to do with performance, however, I don't understand how can endianness influence it. Also I know that: The little-endian system has the property that the same value can be read from memory at different lengths without using different addresses. Which seems a nice feature, but, even so, many systems use big-endian, which probably means big-endian has some advantages too (If so, which?). I'm sure there's more to it, most probably digging down to hardware level. Would love

How to convert the 32 bit integer to network byte order. What is the right way to do that? [1024].pack("N") OR [1,0,2,4].pack("N") Thanks To start, look at the output of each: >> [1024].pack("N") => "\000\000\004\000" >> [1,0,2,4].pack("N") => "\000\000\000\001" Note what the second is missing: >> [1,0,2,4].pack("NNNN") => "\000\000\000\001\000\000\000\000\000\000\000\002\000\000\000\004" 来源： https://stackoverflow.com/questions/12571715/how-to-convert-32-bit-integer-to-network-byte-order

I read on internet that standard byte order for networks is big endian, also known as network byte order. Before transferring data on network, data is first converted to network byte order (big endian). But can any one please let me know who will take care of this conversion. Whether the code developer do really worry about this endianness? If yes, can you please let me know the examples where we need to take care (in case of C, C++). In C and C++, you will have to worry about endianness in low level network code. Typically the serialization and deserialization code will call a function or

to convert a byte array from another machine which is big-endian, we can use: long long convert(unsigned char data[]) { long long res; res = 0; for( int i=0;i < DATA_SIZE; ++i) res = (res << 8) + data[i]; return res; } if another machine is little-endian, we can use long long convert(unsigned char data[]) { long long res; res = 0; for( int i=DATA_SIZE-1;i >=0 ; --i) res = (res << 8) + data[i]; return res; } why do we need the above functions? shouldn't we use hton at sender and ntoh when receiving? Is it because hton/nton is to convert integer while this convert() is for char array? The hton /

I was helping someone with their homework and ran into this strange issue. The problem is to write a function that reverses the order of bytes of a signed integer(That's how the function was specified anyway), and this is the solution I came up with: int reverse(int x) { int reversed = 0; reversed = (x & (0xFF << 24)) >> 24; reversed |= (x & (0xFF << 16)) >> 8; reversed |= (x & (0xFF << 8)) << 8; reversed |= (x & 0xFF) << 24; return reversed; } If you pass 0xFF000000 to this function, the first assignment will result in 0xFFFFFFFF . I don't really understand what is going on, but I know it has