为什么 64 位 Windows 上的本机 long 原语的大小只有 4 个字节？

Question

Will someone please tell me how this makes any sense, and how to make it stop? Seriously, am I crazy or is the 64-bit Windows long type only 4 bytes? How does that make any sense? I thought the native long primitive size was supposed to be the same as the native register size.

[32-bit Linux]

me@u32:~$ ./sizes32
sizeof(char):      1
sizeof(short):     2
sizeof(int):       4
sizeof(long):      4
sizeof(long long): 8

[64-bit Linux]

me@u64:~$ ./sizes64
sizeof(char):      1
sizeof(short):     2
sizeof(int):       4
sizeof(long):      8
sizeof(long long): 8

[32-bit Windows]

C:\Users\me\Downloads>sizes32.exe
sizeof(char):      1
sizeof(short):     2
sizeof(int):       4
sizeof(long):      4
sizeof(long long): 8

[64-bit Windows]

C:\Users\me\Downloads>sizes64.exe
sizeof(char):      1
sizeof(short):     2
sizeof(int):       4
sizeof(long):      4
sizeof(long long): 8

Answer 1

Backward compatibility!

Windows came from a 16-bit platform where sizeof(long) == 4 and it makes extensive use of custom types like LONG , DWORD ... in its API. Microsoft takes a very serious stance on backward compatibility (sometimes evenmodifying its code to make stupid old code work ) and changing that would make a lot of issues

Over on Channel 9, member Beer28 wrote, "I can't imagine there are too many problems with programs that have type widths changed." I got a good chuckle out of that and made a note to write up an entry on the Win64 data model.

The Win64 team selected the LLP64 data model, in which all integral types remain 32-bit values and only pointers expand to 64-bit values. Why?

In addition to the reasons give on that web page, another reason is that doing so avoids breaking persistence formats. For example, part of the header data for a bitmap file is defined by the following structure:

 typedef struct tagBITMAPINFOHEADER { DWORD biSize; LONG biWidth; LONG biHeight; WORD biPlanes; WORD biBitCount; DWORD biCompression; DWORD biSizeImage; LONG biXPelsPerMeter; LONG biYPelsPerMeter; DWORD biClrUsed; DWORD biClrImportant; } BITMAPINFOHEADER, FAR *LPBITMAPINFOHEADER, *PBITMAPINFOHEADER;

If a LONG expanded from a 32-bit value to a 64-bit value, it would not be possible for a 64-bit program to use this structure to parse a bitmap file.

Why did the Win64 team choose the LLP64 model?

Answer 2

long has to be at least 32-bits, at least as big as int and no bigger than long long . That's it. Period.

Answer 3

You got already plenty of valid responses.

Just for the records, here the precise definition in the C++ standard:

3.9.1/2: There are five standard signed integer types : “signed char”, “short int”, “int”, “long int”, and “long long int”. In this list, each type provides at least as much storage as those preceding it in the list. (...) Plain ints have the natural size suggested by the architecture of the execution environment (44).

The last sentence suggests that int is of the size corresponding to the register. Unfortunately, rather than telling the full story of the origin of the universe, its footnote just says: " (44) that is, large enough to contain any value in the range of INT_MIN and INT_MAX, as defined in the header <climits> "