Why does this output differently in C# and C++?

Question

In C#

var buffer = new byte[] {71, 20, 0, 0, 9, 0, 0, 0};

var g = (ulong) ((uint) (buffer[0] | buffer[1] << 8 | buffer[2] << 16 | buffer[3] << 24) |
                    (long) (buffer[4] | buffer[5] << 8 | buffer[6] << 16 | buffer[7] << 24) << 32);

In C++

#define byte unsigned char
#define uint unsigned int
#define ulong unsigned long long

byte buffer[8] = {71, 20, 0, 0, 9, 0, 0, 0};

ulong g = (ulong) ((uint) (buffer[0] | buffer[1] << 8 | buffer[2] << 16 | buffer[3] << 24) |
                    (long) (buffer[4] | buffer[5] << 8 | buffer[6] << 16 | buffer[7] << 24) << 32);

C# outputs 38654710855, C++ outputs 5199.

Why? I have been scratching my head on this for hours...

Edit: C# has the correct output.

Thanks for the help everyone :) Jack Aidley's answer was the first so I will mark that as the accepted answer. The other answers were also correct, but I can't accept multiple answers :\

Shifting a 32-bit integer by 32 is undefined behaviour in C++. I don't know how big your long is, though. — Kerrek SB
– Kerrek SB, Commented Feb 2, 2013 at 12:46

score 6 · Accepted Answer · 2013-02-02 15:36:44Z

6

The C++ is not working because you're casting to long which is typically 32-bits in most current C++ implementation but whose exact length is left to the implementor. You want long long.

Also, please read Bikeshedder's more complete answer below. He's quite correct that fixed size typedefs are a more reliable way of doing this.

edited Feb 2, 2013 at 15:36

answered Feb 2, 2013 at 12:47

anon

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2 Comments

Some programmer dude Over a year ago

If I'm not mistaken, long is still 32 bits in MSVC++ while 64 bits in gcc and clang (on 64 bit platforms). So I wouldn't say it's typical. :)

anon Over a year ago

I'd guess, since he's used long long in his type define he is working on a compiler where long is 32 bits. I'd also note that long is guaranteed to be 32 bits or more, while long long is guaranteed to be 64 bits or more, so long long is probably a better choice anyway.

Ivaylo Strandjev · Accepted Answer · 2013-02-02 12:48:28Z

4

The problem is that long type in C++ is still 4 byte or 32 bit(on most compilers) and thus your calculation overflows it. In C# however long is equivelent to C++'s long long and is 64 bit and so the result of the expression fits into the type.

answered Feb 2, 2013 at 12:48

Ivaylo Strandjev

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Comments

Community · Accepted Answer · 2020-06-20 09:12:55Z

2

Your unsigned long is not 64 bits long. You can easily check this using sizeof(unsigned long) which should return 4 (=32 bits) instead of 8 (=64 bits).

Don't use int/short/long if you expect them to be of a specific size. The standard does only say that short <= int <= long <= long long and defines a minimum size. They can actually be all the same size. long is guaranteed to be at least 32 bits and long long is guaranteed to be at least 64 bits. (See Page 22 of the C++ Standard) Still I would highly recommend against this and stick to stdint if you really want to work with a specific size.

Use <cstdint> (C++11) or <cstdint.h> (C++98) and the defined types uint8_t, uint16_t, uint32_t, uint64_t instead.

Corrected C++ code

#include <stdint.h>
#include <iostream>

int main(int argc, char *argv[]) {
    uint8_t buffer[8] = {71, 20, 0, 0, 9, 0, 0, 0};
    uint64_t g = (uint64_t) ((uint32_t) (buffer[0] | buffer[1] << 8 | buffer[2] << 16 | buffer[3] << 24) |
                              (int64_t) (buffer[4] | buffer[5] << 8 | buffer[6] << 16 | buffer[7] << 24) << 32);
    std::cout << g << std::endl;
    return 0;
}

Demo with output: http://codepad.org/e8GOuvMp

edited Jun 20, 2020 at 9:12

CommunityBot

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answered Feb 2, 2013 at 12:54

bikeshedder

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2 Comments

Benjamin Lindley Over a year ago

No, you were right originally. The standard just says the minimum sizes (actually it gives minimum ranges, the size is implied by this). But they can all exceed that. They could all be the same size, if they are all 64 bits (or more).

bikeshedder Over a year ago

@BenjaminLindley the standard defines the value ranges. According to this a long long must be at least 64 bits in size. It can be bigger though. The short <= int <= long <= long long still holds true though. I crossed out a bit too much.

Dirk · Accepted Answer · 2013-02-02 12:49:30Z

1

There is a subtle error in your castings.

long in C# is a 64-bit integer. long in C++ is usually a 32-bit integer.

Thus your (long) (buffer[4] | buffer[5] << 8 | buffer[6] << 16 | buffer[7] << 24) << 32) has a different meaning when you execute it in C# or C++.

answered Feb 2, 2013 at 12:49

Dirk

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