javascript - How to understand endianness in these function: 'writeFloatLE', 'writeFloatBE', 'writeDoubleLE', 'writeDoubleBE' -

i know endianness, such 0xdead(using writeintle) can transform ad de.

however, in these demos:

var buf = new buffer(4); buf.writefloatbe(0xcafebabe, 0);  console.log(buf);  buf.writefloatle(0xcafebabe, 0); console.log(buf);  // <buffer 4f 4a fe bb> // <buffer bb fe 4a 4f> 

and

var buf = new buffer(8); buf.writedoublebe(0xdeadbeefcafebabe, 0);  console.log(buf);  buf.writedoublele(0xdeadbeefcafebabe, 0);  console.log(buf);  // <buffer 43 eb d5 b7 dd f9 5f d7> // <buffer d7 5f f9 dd b7 d5 eb 43> 

i don't know how transform 0xcafebabe 4f 4a fe bb or bb fe 4a 4f. me understanding steps, pls.

this has nothing (or not much) endianness. you're passing number written hex litteral writedouble or writefloat, turn 32 or 64-bit floating-point representation. is, transformation observe unsigned integer floating-point.

floating-point means number represented m×10^e m , e signed integers (well, technically sign stored in separate sign bit).

in case of 64-bit float, 53 bits used digits m , rest (11) of bits encode exponent e. javascript has single number type, , that's 64-bit floating-point. therefore, precision 53 bits (52 unsigned), means cannot store 64-bit integer @ full precision, way.

when parsing code, javascript interpreter convert number literals (such hex literals) internal 64-bit float representation. that number passed writewhatever @ runtime, not original integer. expect if it's writedouble there won't conversion (just byte reordering maybe).