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我有一个嵌入式 16 位 CPU。在这台机器上，整数是 16 位宽的，它支持 32 位宽的长整数。我需要做一些需要以 64 位存储的乘法运算（例如，将 32 位数乘以 16 位数）。我怎样才能在给定的约束下做到这一点？我没有数学库来做到这一点。

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2 回答 2

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C 中的一个建议。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容易

// 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;
}
于 2013-08-25T06:25:27.727 回答
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您可能想查看Hacker's Delight（这是一本书和一个网站）。他们有来自The Art of Computer Programming Vol.2的 Knuth 的有符号多字乘法和无符号多字乘法的 C 实现。

于 2013-12-11T06:31:45.080 回答