Question

这是一个基本问题，但是我想将3个数字相乘，这些数字在寄存器$ t2，$ t3和$ t4中。假设，假设每个寄存器只能保持8位，当我们将这三个数相乘时，结果可能大于8位。如果它大于8位，如果每个寄存器只能保存8位，结果如何存储在寄存器$ t0中？

Answer 1

Usually CPUs have some kind of an instruction that multiplies two registers and gives a result twice as large, spread across multiple registers. This is necessary to implement extended precision arithmetic.

For example on amd64, to multiply two 64-bit numbers and get a 128-bit result you would do:

mov rax, qword ptr [t2]
mul qword ptr [t3]
; the result is now in rdx : rax
mov qword ptr [result], rax
mov qword ptr [result + 8], rdx

To multipliy three 64-bit numbers and get a 192-bit result you would have to implement extended precision multiplication by multiplying the relevant parts as above and summing them together.

Answer 2

在该特定情况下，算术单元将设置overflow flag

Answer 3

Nohow. This is well known arithmetic overflow.

Sometimes you can change the order of operators to avoid such overflow. F.e. instead if (a + b > INT_MAX) ... you can write if (a > INT_MAX - b) ... and so on.

Sometimes you should use int64 instead of int32, or double instead of float.

And sometimes you should use big numbers library like GMP.

寄存器中存储的数字有多大

3 个答案: