Hex to Uint32 Little Endian in Delphi 7

Question

There is a hex byte sequence: 04 9A 01 00 The HxD hex editor decodes them as "104964" (UInt 32 Little Endian). I need to implement a similar conversion in Delphi 7, but in the opposite direction: so that the number 104964 turns into 04 9A 01 00 (i.e. with this flipping of Hex bytes, because the usual calculator thinks that this is "19A04") .

Answer 1

If you want to do that unbound to the CPU architecture you're in then you create each byte by simple arithmetics:

var
  iNumber: Cardinal;  // Input value: 32 bit unsigned
  aHex: Array[1.. 4] of Byte;  // Output of raw bytes
begin
  iNumber:= 104964;

  aHex[1]:= iNumber and $FF;  // First 8 of 32 bit
  aHex[2]:= (iNumber shr 8) and $FF;  // Next=higher 8 bit...
  aHex[3]:= (iNumber shr 16) and $FF;
  aHex[4]:= (iNumber shr 24) and $FF;  // ...last=highest 8 bit

  Caption:= IntToHex( aHex[1], 2 )+ ' '
          + IntToHex( aHex[2], 2 )+ ' '
          + IntToHex( aHex[3], 2 )+ ' '
          + IntToHex( aHex[4], 2 );  // Converting to text
end;

However, if you know that you're already on a LE architecture (Windows) then you can directly access your variable per byte, because in memory it is already in Little Endian:

var
  iNumber: Cardinal;  // Input value
  aHex: Array[1.. 4] of Byte absolute iNumber;  // Input as raw bytes
begin
  iNumber:= 104964;

  Caption:= IntToHex( aHex[1], 2 )+ ' '
          + IntToHex( aHex[2], 2 )+ ' '
          + IntToHex( aHex[3], 2 )+ ' '
          + IntToHex( aHex[4], 2 );  // Converting into text
end;