No - this same code cannot be used.
12 or 16 bytes is normal because the total number of bytes for an integer at least needs to be divisible by 4 for efficient hardware implementation. You can access an individual byte, but the hardware can access a "row of bytes" in the same time that it can access a single byte within a row. A "row" is either 32 (4 bytes) or 64 bits (8 bytes).

I presume this for loop is within Perl code after a read of a binary file? You should consider pack() and unpack() to get the data into a Perl binary value (big endian). The right kind of pack() or unpack() is dependent upon the architecture (endianness).

It would be helpful to show some code and a short explanation of why you are accessing the individual bytes of what apparently is a multi-byte integer?
Update: too many bytes for an int.. I guess this is a float? It is quite common for binary values to written with big endian values to disk, etc.

In reply to Re: [OT] Endianness and extended precision (80-bit) long doubles by Marshall
in thread [OT] Endianness and extended precision (80-bit) long doubles by syphilis

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