如何在 16 位机器上进行 64 位乘法? [英] How to do 64 bit multiply on 16 bit machine?
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问题描述
我有一个嵌入式 16 位 CPU.在这台机器上,整数是 16 位宽,它支持 32 位宽的长整数.我需要做一些需要以 64 位存储的乘法(例如,将 32 位数字乘以 16 位数字).我怎样才能在给定的约束下做到这一点?我没有数学库来做这件事.
I have an embedded 16 bit CPU. On this machine ints are 16 bit wide and it supports longs that are 32 bits wide. I need to do some multiplications that will need to be stored in 64 bits (e.g. multiply a 32 bit number by a 16 bit number). How can I do that with the given constraints? I do not have a math library to do this.
推荐答案
C 中的建议.请注意,使用内联汇编器可能更容易实现此代码,因为 C 中的进位检测似乎并不那么容易
A suggestion in C. Note that this code probably will be easier to implement with inline assembler as carry detection in C doesn't seem that easy
// Change the typedefs to what your compiler expects
typedef unsigned __int16 uint16 ;
typedef unsigned __int32 uint32 ;
// The 64bit int
typedef struct uint64__
{
uint32 lower ;
uint32 upper ;
} uint64;
typedef int boolt;
typedef struct addresult32__
{
uint32 result ;
boolt carry ;
} addresult32;
// Adds with carry. There doesn't seem to be
// a real good way to do this is in C so I
// cheated here. Typically in assembler one
// would detect the carry after the add operation
addresult32 add32(uint32 left, uint32 right)
{
unsigned __int64 res;
addresult32 result ;
res = left;
res += right;
result.result = res & 0xFFFFFFFF;
result.carry = (res - result.result) != 0;
return result;
}
// Multiplies two 32bit ints
uint64 multiply32(uint32 left, uint32 right)
{
uint32 lleft, uleft, lright, uright, a, b, c, d;
addresult32 sr1, sr2;
uint64 result;
// Make 16 bit integers but keep them in 32 bit integer
// to retain the higher bits
lleft = left & 0xFFFF;
lright = right & 0xFFFF;
uleft = (left >> 16) ;
uright = (right >> 16) ;
a = lleft * lright;
b = lleft * uright;
c = uleft * lright;
d = uleft * uright;
sr1 = add32(a, (b << 16));
sr2 = add32(sr1.result, (c << 16));
result.lower = sr2.result;
result.upper = d + (b >> 16) + (c >> 16);
if (sr1.carry)
{
++result.upper;
}
if (sr2.carry)
{
++result.upper;
}
return result;
}
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