Why does the C/C++ bitwise XOR operator care about sign?

Question

I've been struggling with some low-level messaging for quite some time and it turned out to be an issue with the checksum calculation. I thought the bitwise XOR operator didn't care about sign, so I was using a QByteArray to store the bytes, and using the at method, that returns a char, to calculate the checksum. The messages were properly acknowledged sometimes, but not always.

Looks like the guys on the other end were using uint8_t to store the bytes, and the checksum was failing in some situations. I solved it by casting the char to uint8_t, but I'm seriously baffled about that.

Why does the bitwise XOR operator care about sign? I thought it worked on a bit-level, regardless of what they represented. Here's a piece of code I've used to try to understand it.

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

int main ()
{    
    uint8_t a = 0b10010101;
    char b    = 0b10010101;

    uint32_t checksum;

    checksum = 55;
    checksum ^= a;
    std::cout << std::bitset<32>(checksum) << std::endl;

    checksum = 55;
    checksum ^= b;
    std::cout << std::bitset<32>(checksum) << std::endl;    
}

Even tho both integers hold the same bits, the operation yields different results in each case.

It's `cstdio` in C++, not `stdio.h` – KamilCuk Sep 17 '18 at 10:45 — KamilCuk, Sep 17 '18 at 10:45
I do not see any or operation here. All I see are XORs – bartop Sep 17 '18 at 10:45 — bartop, Sep 17 '18 at 10:45
OR is |. ^ is XOR – Swift - Friday Pie Sep 17 '18 at 10:50 — Swift - Friday Pie, Sep 17 '18 at 10:50
You're right about the OR / XOR. – José Tomás Tocino Sep 17 '18 at 11:26 — José Tomás Tocino, Sep 17 '18 at 11:26

score 9 · Accepted Answer · answered Sep 17 '18 at 10:46

9

First of all you have to remember that it's implementation-defined if char is signed or unsigned.

Then you have to remember that values of smaller types like char or uint8_t are promoted to int (or possibly unsigned int) when used in expressions. This promotion brings with it sign extension of signed types.

So if char is signed, the value 0b10010101 will be promoted to 0b11111111111111111111111110010101 (using two's complement for negative numbers). Which is very different from the unsigned value 0b00000000000000000000000010010101.

answered Sep 17 '18 at 10:46

Some programmer dude

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While integral promotions exist, it's fairly obvious that an "in-place" operator like `^=` cannot promote its left-hand operand. The _usual arithmetic conversions_ apply to _binary_ operators: `a = b ^c;`. It's wrong (incorrect generalization) to say that promotion happens is _expressions_, because `a^=b` is also an expression. – MSalters Sep 17 '18 at 10:52
but `a` and `b` are expressions there too.. and operator can't be implemented to work "in-place" with different type of operands. it's more equivalent to `uint32_t operator^=(int32_t a, int32_t b) { uint32_t c = a ^ b; return c; }` with factual operands being `char` and `uint8_t`. I guess, it's just matter of perspective – Swift - Friday Pie Sep 17 '18 at 10:59

Why does the C/C++ bitwise XOR operator care about sign?

1 Answers1