Reputation: 39
When doing a math operation, how can I get the part that overflowed?
For example, assuming 32-bit ints:
unsigned int a = 0Xffffffff;
unsigned int b = 0xffffffff;
unsigned int c = a + b;
In this case, c
is 0xfffffffe
, but the answer should be 0x1fffffffe
. How do I get the 1 that overflowed?
How can I do the same for multiplication? Can I multiply two large numbers together and only get the overflowed part?
How do bigint libraries manage this?
Upvotes: 3
Views: 231
Reputation: 37214
How do I get the 1 that overflowed?
To do it afterwards in a portable way (not forgetting that unsigned int
might only be 16 bits):
uint32_t a = 0Xffffffff;
uint32_t b = 0xffffffff;
uint32_t c_low = a + b;
uint32_t c_high;
if(c_low >= a) {
c_high = 0;
} else {
c_high = 1;
}
To do it beforehand in a portable way (without branches):
uint32_t a = 0Xffffffff;
uint32_t b = 0xffffffff;
uint32_t c_low;
uint32_t c_high;
c_high = (a&b) >> 31;
c_low = (a ^ (c_high<<31)) + b;
How can I do the same for multiplication?
Multiplication doesn't have a carry, it has an "upper half". Specifically; if you multiply an unsigned integer that has N bits with an unsigned integer that has M bits then the result will have N+M bits; and if both numbers had the same size then the result will be twice as big.
Sadly C doesn't support "result type is larger than source/s types", so you need to "pre-promote" the source types, like:
uint32_t a = 0Xffffffff;
uint32_t b = 0xffffffff;
uint64_t temp = (uint64_t)a * (uint64_t)b;
uint32_t c_low = temp;
uint32_t c_high = temp >> 32;
Of course if the compiler doesn't support a larger type then you have to split it into smaller pieces, like:
uint32_t a = 0Xffffffff;
uint32_t b = 0xffffffff;
uint32_t a_low = a & 0xFFFF;
uint32_t a_high = a >> 16;
uint32_t b_low = a & 0xFFFF;
uint32_t b_high = b >> 16;
uint32_t temp_0 = a_low * b_low;
uint32_t temp_16a = a_high * b_low;
uint32_t temp_16b = a_low * b_high;
uint32_t temp_32 = a_high * b_high;
uint32_t c_low = temp_0 + (temp16a << 16) + (temp16b << 16);
uint32_t c_high = (temp16a >> 16) + (temp16b >> 16) + temp_32;
How do bigint libraries manage this?
Mainly; they use inline assembly language because most CPUs support instructions to work on bigger integers efficiently and/or because you can access the carry flag directly. For example; for 80x86; the CPU has adc
/sbb
, shld
/shrd
, mul
(with double width result)/div
(with double width numerator); plus maybe extensions (adcx
and adox
).
In 32-bit 80x86 assembly language, the addition might look like:
xor edx,0
add eax,ebx ;c_low = a + b
adc edx,0 ;c_high = carry
..and the multiplication might look like:
mul ebx ;edx:eax = a * b
Upvotes: 2
Reputation: 215173
Assuming unsigned type operands, you can write:
bool cf = a+b<a;
or
bool cf = a>-1-b;
These work regardless of the existence of a larger type to work with.
Multiplication is harder; without a larger type, there is no way to access the upper half of the result. If you do have one you can use it. For example, if your operands are uint32_t
,
uint32_t upper = ((uint64_t)a * b) >> 32;
uint32_t lower = a*b;
Otherwise, you're stuck dropping to a half-sized type and using long multiplication. For example, with uint64_t a,b;
uint32_t al = a, ah = a>>32;
uint32_t bl = b, bh = b>>32;
And then the upper part of the result is ah*bh
plus the carry out of adding al*bh
, ah*bl
, and the upper bits of al*bl
.
Bigint libraries can avoid the pain of this by just choosing a limb type that's at most half the width of the largest integer type.
Upvotes: 2
Reputation: 67476
@Warning NOT PORTABLE@
Example: https://godbolt.org/z/NcyNzR
#include <stdio.h>
int main(void)
{
unsigned int res;
if(__builtin_uadd_overflow(0Xffffffff, 0Xffffffff, &res))
{
printf("Overflowed\n");
}
printf("Result: 0x%x\n", res);
}
Upvotes: 3