Converting an byte array into 2 unsigned long in C and Delphi

Question

The following C code:

typedef unsigned char byte;

byte temp[8] = {...};

unsigned long iONE = 0;
unsigned long iTWO = 0;

memcpy(((byte *)&iONE)+sizeof(unsigned long)-4, temp, 4);
memcpy(((byte *)&iTWO)+sizeof(unsigned long)-4, temp+4, 4);

What is the significance of +sizeof(unsigned long)-4?

Could the memcpy statements be written as:

memcpy((byte *)&iONE, temp, 4);
memcpy((byte *)&iTWO, temp+4, 4);

And is the Delphi conversion valid:

var
  temp: array[0..7] of Byte;
  iONE: Cardinal;
  iTWO: Cardinal;
begin
  temp := ...;
  iONE := 0;
  iTWO := 0;

  CopyMemory(iOne, temp[0], SizeOf(Cardinal));
  CopyMemory(iTwo, temp[4], SizeOf(Cardinal));
end;

Ask yourself what this does on a platform where sizeof(unsigned long) is greater than 4 octets (think 8, then do the math). You may also consider the ramifications of running this on a big-vs-little endian system. — WhozCraig
– WhozCraig, Commented Jan 25, 2015 at 0:57
Let's get serious. This is horrible, non portable code, and it's not possible to tell without context, what "the significance" of those operations should be. The code seems to make assumptions about endianess and sizeof(unsigned long), that at least I am not able to decipher in purpose. — πάντα ῥεῖ
– πάντα ῥεῖ, Commented Jan 25, 2015 at 1:03
It is little endian system (Windows). And the unsigned long is 4 bytes. Not sure if I answered the question. — Mike
– Mike, Commented Jan 25, 2015 at 20:46

David Heffernan · Accepted Answer · 2015-01-25 08:45:08Z

1

(byte *)&iONE)+sizeof(unsigned long)-4

is trying read 4 bytes from iONE but also account for the possibility that unsigned long is larger than 4 bytes. The code does so by picking off the last 4 bytes of iONE. Since the code always reads from the same end of the variable its behaviour varies between little and big endian machines. Without any context it is hard to say what this code is trying to achieve but my guess is that it was originally running on a big endian system with 8 byte unsigned long. Perhaps you know more.

In Delphi you'd probably write it like this.

var
  temp: array[0..7] of Byte;
  iONE: Cardinal;
  iTWO: Cardinal;
begin
  temp := ...;
  iONE := PCardinal(@temp[0])^;
  iTWO := PCardinal(@temp[4])^;;
end;

However, it would help to know more about the original code. Personally if I was faced with this task I'd write the code from scratch rather than make a literal port. The original code already looks poor and once it has been translated it will be even worse. Work out what the code does and write a new program using the idiom of the target language.

Two minor points on your code:

It is pointless to initialise a variable that is immediately over-written. That just confuses the future reader.
CopyMemory is a Windows API function and as such not portable. System.Move is the function to use for raw memory copies.

answered Jan 25, 2015 at 8:45

David Heffernan

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6 Comments

Mike Over a year ago

Your comment very useful. Thank you for the succinct Delphi code.

Mike Over a year ago

I deleted the post to avoid straying. Will post as new question at the right time. Thank you for your comment.

David Heffernan Over a year ago

What about this question?

Mike Over a year ago

Thank you David. The C code is likely not original. The intended unsigned long is 4 bytes (Windows). The temp[8] is big endian, and holds IBM doubles that needs to be converted into IEEE double (little endian). iONE and iTWO are bytes swapped before further manipulation.

David Heffernan Over a year ago

I mean, I think this question has been answered

|

Iharob Al Asimi · Accepted Answer · 2015-01-25 00:53:07Z

0

memcpy(((byte *)&iONE)+sizeof(unsigned long)-4, temp, 4);

is copying 4 bytes from temp into iONE's address at offset sizeof(unsigned long)-4, and

memcpy((byte *)&iONE, temp, 4);

will copy 4 bytes from temp at the beginning of the address where iONE is stored.

So sizeof(unsigned long)-4 is just an offset in bytes, at which to copy the 4 bytes from temp.

answered Jan 25, 2015 at 0:53

Iharob Al Asimi

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2 Comments

πάντα ῥεῖ Over a year ago

While this is technically correct, that doesn't tell anything about the significance :-P ...

Mike Over a year ago

Your comment helpful. Thank you.

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