Bob Plantz wrote:
> On Tue, 2008-04-15 at 13:16 -0400, John Fine wrote:
>> When x and y are both unsigned int (x*y) is also an unsigned int.
>> (long int)(x*y) means first compute the unsigned in (x*y) then promote
>> it to long int.
>> To get the answer you want, you need to promote one of the arguments of
>> the multiply before multiplying. I don't know whether the optimizer
>> will figure out to do the 32 bit multiply you want and store the 64 bit
>> result or whether it would do a 64 bit multiply.
>
> Thank you, John, for your remarks.
>
> I knew about multiplying first, then the promotion, but it slipped my
> mind. So I tried:
> unsigned int x;
> unsigned long int y;
>
> printf("Enter two integers: ");
> scanf("%u %lu", &x, &y);
>
> y = x * y;
>
> In this case 32-bit * 64-bit -> 96-bit result. But the compiler does:
> movl -4(%rbp), %eax # load x
> mov %eax, %edx # promote to 64-bit (zeroes high
> 32 bits)
> movq -16(%rbp), %rax # load y (64-bit value)
> imulq %rdx, %rax # truncate high-order 32 bits of
> result
> movq %rax, -16(%rbp)
>
> I've also tried using 64-bit for all the ints. Basically, if the result
> is wider than the widest int in the multiplication, there is overflow.
>
> My thought is that this is a place where assembly language is needed if
> the high-order bits need to be preserved. At least one needs to check
> the CF and OF. (They get set to one if there is signed multiply
> overflow.)
__uint128_t prod = (__uint128_t) x * y;
in asm:
asm ("mulq %3" : "=a"(z1), "=d"(z2) : "a"(x), "r"(y));
Andrew.
