How to tell if two 8-bit chars are gray codes in c++?

Question

The problem is to tell if two 8-bit chars are gray codes(differ only in 1 bit) in C++? I found an elegant C++ solution:

bool isGray(char a, char b) {
    int m = a ^ b;
    return m != 0 && (m & (m - 1) & 0xff) == 0;
}

I was confused that what does the "& 0xff" do?

Unless I'm crazy, just `return m & (m-1);` will work. (Though the underflow might technically be UB.) — David Schwartz, Mar 18 '15 at 04:57
You'd need `unsigned char` IIRC. And that UB is more than "technically UB", modern optimizers will use it to prove m!=0. That proof can have rather unexpected effects on your code. — MSalters, Mar 18 '15 at 08:22

score 0 · Accepted Answer · answered Mar 18 '15 at 04:54

0

& 0xff extracts the 8 lowest bits from the resulting value, ignoring any higher ones.

answered Mar 18 '15 at 04:54

David Schwartz

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Can we use "char m = a ^ b"? – user3692521 Mar 18 '15 at 04:55
Sure, but how will that help? It will just get converted back to an integer in the expression in the next line. – David Schwartz Mar 18 '15 at 04:56
Yes you answered my question thank you! just wondering why the author of this piece of code chose int over char? – user3692521 Mar 18 '15 at 04:58
looks like char is only 8 bits and don't even have to "& 0xff" – user3692521 Mar 18 '15 at 04:58
You mean m get converted back when "(m - 1)"? – user3692521 Mar 18 '15 at 05:00
It won't be a char in the next expression. – David Schwartz Mar 18 '15 at 05:00
Sorry I am still confused where does m get converted back if "char m = a ^ b"? – user3692521 Mar 18 '15 at 05:02
@user3692521 In `m & (m - 1)`. – David Schwartz Mar 18 '15 at 18:38

score 0 · Answer 2 · answered Mar 18 '15 at 08:30

It's wrong. The mistaken idea is that char is 8 bit.

It's also pointless. The presumed problem is that m can have more bits than char (true) so that "unnecessary" bits are masked off.

But m is sign-extended. That means the sign bit is copied to the higher bits. Now, when we're comparing x==0 we're checking whether all bits are zero, and with x & 0xff we're comparing if the lower 8 bits are zero. If the 8th bit of x is copied to all higher positions (by sign extension), then the two conditions are the same regardless of whether the copied bit was 0 or 1.

How to tell if two 8-bit chars are gray codes in c++?

2 Answers2