Counting bits in a long long not working as expected

Question

uint64_t bitsToInt64(char *bs) {
    uint64_t r, i;
    r = 0;
    for(i = 0; i < 64; i++)
        if(bs[i] == 1) 
            r |= 1LL << i;          
    return r;
}

int countBits64(uint64_t i) {
    uint64_t x, t;
    t = 0LL;
    for(x = 0LL; x < 64LL; x++) {
        if(i & (1LL << x))
            t += 1;
    }
    return t;
}

char bits [] = { 
    0, 0, 0, 0, 1, 0, 0, 0,
    0, 0, 0, 0, 0, 0, 0, 0,
    0, 0, 0, 0, 0, 0, 0, 0,
    0, 0, 0, 0, 0, 0, 0, 0,
    0, 0, 0, 0, 0, 0, 0, 0,
    0, 0, 0, 0, 0, 0, 0, 0,
    0, 0, 0, 0, 0, 0, 0, 0,
    1, 1, 1, 1, 1, 1, 1, 1,};

uint64_t mask = bitsToInt64(bits);
int nbits = countBits64(mask);
printf("%d", nbits);

The above is printing "1". What am I doing wrong?

What is the output if you do `printf("%ld", mask);` before the last line? — Chriszuma, Jun 30 '11 at 19:32
I don't think this is the only problem, but you should use `1ULL` instead of `1LL`, since a `1 << 63` won't fit into a signed long long. Can you print `mask` so we can determine which function the bug is in? — ughoavgfhw, Jun 30 '11 at 19:33
printf("%llu", mask); prints 10. But the bitsToInt64() must be working before i am using uint64_t as bitboards for chess pieces and they are all getting/setting correctly... — , Jun 30 '11 at 19:40

DarkDust · Answer 1 · 2011-06-30T20:02:37.753

2

Your shifting of 1LL, which is signed, yields undefined behavior. Shifting a 64bit signed integer by 63 bits is thus allowed to make the compiler do funny things (like making daemons fly out of your nose).

The solution is to use 1ULL << x instead in this case.

See also this excellent article from an Chris Lattner of LLVM that explains why things like this can result in strange behavior.

edited Jun 30 '11 at 20:02

answered Jun 30 '11 at 19:48

DarkDust

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Joe · Accepted Answer · 2011-06-30T19:57:03.957

1

With the code below I was able to get the following output

(0xFF00000000000010) 9

Make sure you use the correct size values. The literal constants need ULL so that they match uint64_t and the for loop variables only need to be an int, as well as the return value for countBits64.

uint64_t bitsToInt64(char *bs) {
    uint64_t r;
    int i;
    r = 0;
    for(i = 0; i < 64; i++)
        if(bs[i] == 1) 
            r |= 1ULL << i;          
    return r;
}

int countBits64(uint64_t i) {
    int x, t = 0;
    for(x = 0; x < 64; x++) {
        if(i & (1ULL << x))
            t += 1;
    }
    return t;
}

char bits [] = { 
    0, 0, 0, 0, 1, 0, 0, 0,
    0, 0, 0, 0, 0, 0, 0, 0,
    0, 0, 0, 0, 0, 0, 0, 0,
    0, 0, 0, 0, 0, 0, 0, 0,
    0, 0, 0, 0, 0, 0, 0, 0,
    0, 0, 0, 0, 0, 0, 0, 0,
    0, 0, 0, 0, 0, 0, 0, 0,
    1, 1, 1, 1, 1, 1, 1, 1,};

uint64_t mask = bitsToInt64(bits);
int nbits = countBits64(mask);
printf("(0x%016llX) %d",mask, nbits);

edited Jun 30 '11 at 19:57

answered Jun 30 '11 at 19:45

Joe

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Just copied and pasted it again I am still getting what I posted. I am running x64 intel mac and gcc, uint64_t is typedef'd as unsigned long long. I am also able to get the same results with your original code. What compiler, OS, architecture are you on? – Joe Jun 30 '11 at 20:03
Just updated my last comment, wonder if it some compiler flag or optimization that is causing that for you. – Joe Jun 30 '11 at 20:07
(0x0000000000000010) 1 with the 0x%016llX but thats still not right. I am doing plan old `gcc input -o output` with no flags – Jun 30 '11 at 20:08
Here is something fun to do, make them all 1's then if your output looks like 0x00000000FFFFFFFF, your long long is not quite as long as you think. – Joe Jun 30 '11 at 20:13
OMFG it was because i wasnt importing the a header file in one of the files that uses these functions. No warnings or anything, just a half working function... unbelievable. thanks for the help – Jun 30 '11 at 20:31
Glad you figured out, were you just missing ``? I would like to update the answer so someone can find the solution easier. – Joe Jun 30 '11 at 20:39

Counting bits in a long long not working as expected

2 Answers2