typedef struct {
    unsigned int part[4];
} bigint_t;

我想将此数字转换为十进制字符串表示形式并将其输出到文件中.

现在,我使用bigint_divmod10函数将数字除以10,跟踪余数.我重复调用这个函数,输出余数作为数字,直到数字为零.这很慢.这是最快的方法吗？如果是这样,有没有一种聪明的方法来实现我没有看到的这个功能？我试过看GMP的get_str.c,但我发现它非常难以理解.

编辑：这是我能够为divmod10函数提出的最快的代码：

static unsigned uint128_divmod10(uint128 *value)
{
    unsigned int a = value->word[3];
    unsigned int b = value->word[2];
    unsigned int c = value->word[1];
    unsigned int d = value->word[0];

    unsigned int diva = a / 5;
    unsigned int divb = b / 5;
    unsigned int divc = c / 5;
    unsigned int divd = d / 5;

    value->word[3] = diva;
    value->word[2] = divb;
    value->word[1] = divc;
    value->word[0] = divd;

    unsigned int moda = a - diva*5;
    unsigned int modb = b - divb*5;
    unsigned int modc = c - divc*5;
    unsigned int modd = d - divd*5;

    unsigned int mod = 0;
    mod += moda;
    unsigned int carryb = mod*858993459;
    mod += modb;
    if (mod >= 5) {
        mod -= 5;
        carryb++;
    }
    unsigned int carryc = mod*858993459;
    mod += modc;
    if (mod >= 5) {
        mod -= 5;
        carryc++;
    }
    unsigned int carryd = mod*858993459;
    mod += modd;
    if (mod >= 5) {
        mod -= 5;
        carryd++;
    }

    uint128_add(value,carryd,0);
    uint128_add(value,carryc,1);
    uint128_add(value,carryb,2);

    if (value->word[0] & 1) {
        mod += 5;
    }
    uint128_shift(value,-1);
    return mod;
}

其中add函数定义为：

static void uint128_add(uint128 *value,unsigned int k,unsigned int pos)
{
    unsigned int a = value->word[pos];
    value->word[pos] += k;
    if (value->word[pos] < a) {
        // overflow
        for (int i=pos+1; i<4; i++) {
            value->word[i]++;
            if (value->word[i]) {
                break;
            }
        }
    }
}