wrong multiplication result for __uint128_t result

Question

I want to multiple two 64 bits numbers and store their results into two 64 bits (total 128) bits variables

for this I wrote following simple function

void multiply2(uint64_t a, uint64_t b, uint64_t& hi, uint64_t& lo) {
__uint128_t product = static_cast<__uint128_t>(a) * static_cast<__uint128_t>(b);
hi = static_cast<uint64_t>(product >> 64);
lo = static_cast<uint64_t>(product);}}

and then I test is using following

int main(int argc, char* argv[]){

uint64_t a = 4539155444;
uint64_t b = 4539155444;
uint64_t hi, lo;
multiply1(a, b, hi, lo);
std::cout << a << " * " << b << " = " << hi << lo << std::endl;
return 0;}

The output I am getting is

4539155444 * 4539155444 = 12157188071085285520

But expected output is 4539155444 * 4539155444 = 20603932144794837136

I am confused what am I doing wrong here?

You can't just print out two decimal numbers next to each other and get a valid result (it would work in hex), you'll need to convert back to 128bit for printing, your answer is `1` and `2157188071085285520` which are the correct 64-bit numbers — Alan Birtles, Apr 28 '23 at 07:20
It does not work like this. You can only print it this way using hexadecimal notation, but not a decimal number — 0___________, Apr 28 '23 at 07:20
The split happens between two binary digits, not between two decimal digits. — BoP, Apr 28 '23 at 07:21

0___________ · Accepted Answer · 2023-04-28T14:12:12.937

1

You cant simply take two parts of the number and print the next to each other and get the correct decimal number. You need to write your own function to print it.

void printi128(__int128 val)
{
    if(val < 10 && val > -10) printf("%d", val);
    else 
    {
        printi128(val / 10);
        printf("%d", abs(val % 10));
    }
}

https://godbolt.org/z/bTYsEbrsM

edited Apr 28 '23 at 14:12

answered Apr 28 '23 at 07:24

0___________

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1

`printf("%d", val)` has undefined behavior. It should be `printf("%d", static_cast(val))` – Aykhan Hagverdili Apr 28 '23 at 07:48
No need to use recursion here either, a simple loop would be better – Alan Birtles Apr 28 '23 at 08:08
@0___________ My comments were about defects in code that you originally posted. I see that you fixed the algorithm. – Öö Tiib Apr 28 '23 at 13:29
@AyxanHaqverdili it is not UB. Read the code – 0___________ Apr 28 '23 at 13:42
@ÖöTiib it was the original intention. – 0___________ Apr 28 '23 at 13:43
@0___________ You added the cast after my comment? The edit history is public. – Aykhan Hagverdili Apr 28 '23 at 14:11
@AyxanHaqverdili removed the cast – 0___________ Apr 28 '23 at 14:12
@0___________ It is undefined without the cast. `%d` specifier is for `int`. It is not for `__int128` regardless of the runtime magnitude of the value you're passing. – Aykhan Hagverdili Apr 28 '23 at 14:13
@AyxanHaqverdili no it is not because of the implicit conversion when you pass to variadic function – 0___________ Apr 28 '23 at 14:14
@0___________ Implicit promotion to `int` does not apply to `__int128`. Just like how it doesn't apply to `long long`, otherwise there would be no way to print `long long`. The compiler itself will tell you that if you turn on the warnings: https://godbolt.org/z/favWzcEcb – Aykhan Hagverdili Apr 28 '23 at 14:24

wrong multiplication result for __uint128_t result

1 Answers1