我正在尝试将作为4个无符号整数数组存储的128位无符号整数转换为C中的十进制字符串表示：
unsigned int src[] = { 0x12345678, 0x90abcdef, 0xfedcba90, 0x8765421 };
printf("%s", some_func(src)); // gives "53072739890371098123344"
(上述的输入和输出示例是完全虚构的；我不知道那个输入会产生什么结果。)
如果我要使用十六进制、二进制或八进制，这将是一个简单的问题，只需要使用掩码和位移来获取最低有效字符即可。但是，我认为我需要进行十进制除法。不幸的是，我不记得如何在多个整数之间执行除法，并且我正在使用的系统不支持大于32位的数据类型，因此无法使用128位类型。使用不同的编程语言也不行，而且我宁愿避免为了这一个操作而使用大数库。
不需要进行除法运算：
#include <string.h>
#include <stdio.h>
typedef unsigned long uint32;
/* N[0] - contains least significant bits, N[3] - most significant */
char* Bin128ToDec(const uint32 N[4])
{
// log10(x) = log2(x) / log2(10) ~= log2(x) / 3.322
static char s[128 / 3 + 1 + 1];
uint32 n[4];
char* p = s;
int i;
memset(s, '0', sizeof(s) - 1);
s[sizeof(s) - 1] = '\0';
memcpy(n, N, sizeof(n));
for (i = 0; i < 128; i++)
{
int j, carry;
carry = (n[3] >= 0x80000000);
// Shift n[] left, doubling it
n[3] = ((n[3] << 1) & 0xFFFFFFFF) + (n[2] >= 0x80000000);
n[2] = ((n[2] << 1) & 0xFFFFFFFF) + (n[1] >= 0x80000000);
n[1] = ((n[1] << 1) & 0xFFFFFFFF) + (n[0] >= 0x80000000);
n[0] = ((n[0] << 1) & 0xFFFFFFFF);
// Add s[] to itself in decimal, doubling it
for (j = sizeof(s) - 2; j >= 0; j--)
{
s[j] += s[j] - '0' + carry;
carry = (s[j] > '9');
if (carry)
{
s[j] -= 10;
}
}
}
while ((p[0] == '0') && (p < &s[sizeof(s) - 2]))
{
p++;
}
return p;
}
int main(void)
{
static const uint32 testData[][4] =
{
{ 0, 0, 0, 0 },
{ 1048576, 0, 0, 0 },
{ 0xFFFFFFFF, 0, 0, 0 },
{ 0, 1, 0, 0 },
{ 0x12345678, 0x90abcdef, 0xfedcba90, 0x8765421 }
};
printf("%s\n", Bin128ToDec(testData[0]));
printf("%s\n", Bin128ToDec(testData[1]));
printf("%s\n", Bin128ToDec(testData[2]));
printf("%s\n", Bin128ToDec(testData[3]));
printf("%s\n", Bin128ToDec(testData[4]));
return 0;
}
输出：
0
1048576
4294967295
4294967296
11248221411398543556294285637029484152
unsigned int不能保证至少有32位长。unsigned long可以。无论系统是否为64位，都与C标准无关且超出其范围。 - Alexey Frunzeuint32 而不是 unsigned。这使得解决方案可以在使用 16 位 int/unsigned 的嵌入式设计中移植。 - chux - Reinstate Monicauint32，而是使用stdint.h中的标准uint32_t。 - user694733这是一个基于2^32的简单除法，按照倒序打印十进制位数，并使用64位算术。其复杂度为 O(n)，其中n是表示中的十进制数字位数：
#include <stdio.h>
unsigned int a [] = { 0x12345678, 0x12345678, 0x12345678, 0x12345678 };
/* 24197857161011715162171839636988778104 */
int
main ()
{
unsigned long long d, r;
do
{
r = a [0];
d = r / 10;
r = ((r - d * 10) << 32) + a [1];
a [0] = d;
d = r / 10;
r = ((r - d * 10) << 32) + a [2];
a [1] = d;
d = r / 10;
r = ((r - d * 10) << 32) + a [3];
a [2] = d;
d = r / 10;
r = r - d * 10;
a [3] = d;
printf ("%d\n", (unsigned int) r);
}
while (a[0] || a[1] || a[2] || a[3]);
return 0;
}
编辑：修正了循环，如果数组a仅包含零，则显示0。此外，该数组从左到右读取，a [0] 是最高有效位，a [3] 是最低有效位。
x >= y 的函数和另一个计算 x -= y 的函数（当这种情况发生时）。
然后，您可以开始计算可以减去多少次10^38（这将是最高位），然后计算可以减去多少次10^37 ... 直到可以减去1的次数。#include <stdio.h>
typedef unsigned ui128[4];
int ge128(ui128 a, ui128 b)
{
int i = 3;
while (i >= 0 && a[i] == b[i])
--i;
return i < 0 ? 1 : a[i] >= b[i];
}
void sub128(ui128 a, ui128 b)
{
int i = 0;
int borrow = 0;
while (i < 4)
{
int next_borrow = (borrow && a[i] <= b[i]) || (!borrow && a[i] < b[i]);
a[i] -= b[i] + borrow;
borrow = next_borrow;
i += 1;
}
}
ui128 deci128[] = {{1u,0u,0u,0u},
{10u,0u,0u,0u},
{100u,0u,0u,0u},
{1000u,0u,0u,0u},
{10000u,0u,0u,0u},
{100000u,0u,0u,0u},
{1000000u,0u,0u,0u},
{10000000u,0u,0u,0u},
{100000000u,0u,0u,0u},
{1000000000u,0u,0u,0u},
{1410065408u,2u,0u,0u},
{1215752192u,23u,0u,0u},
{3567587328u,232u,0u,0u},
{1316134912u,2328u,0u,0u},
{276447232u,23283u,0u,0u},
{2764472320u,232830u,0u,0u},
{1874919424u,2328306u,0u,0u},
{1569325056u,23283064u,0u,0u},
{2808348672u,232830643u,0u,0u},
{2313682944u,2328306436u,0u,0u},
{1661992960u,1808227885u,5u,0u},
{3735027712u,902409669u,54u,0u},
{2990538752u,434162106u,542u,0u},
{4135583744u,46653770u,5421u,0u},
{2701131776u,466537709u,54210u,0u},
{1241513984u,370409800u,542101u,0u},
{3825205248u,3704098002u,5421010u,0u},
{3892314112u,2681241660u,54210108u,0u},
{268435456u,1042612833u,542101086u,0u},
{2684354560u,1836193738u,1126043566u,1u},
{1073741824u,1182068202u,2670501072u,12u},
{2147483648u,3230747430u,935206946u,126u},
{0u,2242703233u,762134875u,1262u},
{0u,952195850u,3326381459u,12621u},
{0u,932023908u,3199043520u,126217u},
{0u,730304488u,1925664130u,1262177u},
{0u,3008077584u,2076772117u,12621774u},
{0u,16004768u,3587851993u,126217744u},
{0u,160047680u,1518781562u,1262177448u}};
void print128(ui128 x)
{
int i = 38;
int z = 0;
while (i >= 0)
{
int c = 0;
while (ge128(x, deci128[i]))
{
c++; sub128(x, deci128[i]);
}
if (i==0 || z || c > 0)
{
z = 1; putchar('0' + c);
}
--i;
}
}
int main(int argc, const char *argv[])
{
ui128 test = { 0x12345678, 0x90abcdef, 0xfedcba90, 0x8765421 };
print128(test);
return 0;
}
11248221411398543556294285637029484152
Python也认为这是正确的值（当然，这并不意味着代码是正确的!!! ;-)）
同样的事情，但使用32位整数算术：
#include <stdio.h>
unsigned short a [] = {
0x0876, 0x5421,
0xfedc, 0xba90,
0x90ab, 0xcdef,
0x1234, 0x5678
};
int
main ()
{
unsigned int d, r;
do
{
r = a [0];
d = r / 10;
r = ((r - d * 10) << 16) + a [1];
a [0] = d;
d = r / 10;
r = ((r - d * 10) << 16) + a [2];
a [1] = d;
d = r / 10;
r = ((r - d * 10) << 16) + a [3];
a [2] = d;
d = r / 10;
r = ((r - d * 10) << 16) + a [4];
a [3] = d;
d = r / 10;
r = ((r - d * 10) << 16) + a [5];
a [4] = d;
d = r / 10;
r = ((r - d * 10) << 16) + a [6];
a [5] = d;
d = r / 10;
r = ((r - d * 10) << 16) + a [7];
a [6] = d;
d = r / 10;
r = r - d * 10;
a [7] = d;
printf ("%d\n", r);
}
while (a[0] || a[1] || a[2] || a[3] || a [4] || a [5] || a[6] || a[7]);
return 0;
}
假设您拥有快速的32位乘法和除法，可以通过实现bigint除法/模10000来每次计算4个数字，并使用（s）printf输出数字组的结果。
这种方法也很容易扩展到更高（甚至可变）精度...
#include <stdio.h>
typedef unsigned long bigint[4];
void print_bigint(bigint src)
{
unsigned long int x[8]; // expanded version (16 bit per element)
int result[12]; // 4 digits per element
int done = 0; // did we finish?
int i = 0; // digit group counter
/* expand to 16-bit per element */
x[0] = src[0] & 65535;
x[1] = src[0] >> 16;
x[2] = src[1] & 65535;
x[3] = src[1] >> 16;
x[4] = src[2] & 65535;
x[5] = src[2] >> 16;
x[6] = src[3] & 65535;
x[7] = src[3] >> 16;
while (!done)
{
done = 1;
{
unsigned long carry = 0;
int j;
for (j=7; j>=0; j--)
{
unsigned long d = (carry << 16) + x[j];
x[j] = d / 10000;
carry = d - x[j] * 10000;
if (x[j]) done = 0;
}
result[i++] = carry;
}
}
printf ("%i", result[--i]);
while (i > 0)
{
printf("%04i", result[--i]);
}
}
int main(int argc, const char *argv[])
{
bigint tests[] = { { 0, 0, 0, 0 },
{ 0xFFFFFFFFUL, 0, 0, 0 },
{ 0, 1, 0, 0 },
{ 0x12345678UL, 0x90abcdefUL, 0xfedcba90UL, 0x8765421UL } };
{
int i;
for (i=0; i<4; i++)
{
print_bigint(tests[i]);
printf("\n");
}
}
return 0;
}
@Alexey Frunze的方法很简单，但非常慢。你应该使用@chill上面提到的32位整数方法。另一种不需要任何乘除法的简单方法是double dabble。这可能比chill的算法慢，但比Alexey的快得多。运行后，您将获得十进制数字的打包BCD。
uint265_t 和 uint128_t。
https://github.com/calccrypto/uint256_t
不，我与那个项目没有关联，但我曾经为这样的目的使用过它，但我想它对其他人也可能有用。
实际上您不需要实现长除法，而是需要实现乘以二的幂和加法。您有四个uint_32。首先将它们中的每一个转换为字符串。分别将它们乘以 (2^32)^3、(2^32)^2、(2^32)^1和(2^32)^0，然后将它们相加。您不需要进行基础转换，只需处理将这四个部分放在一起即可。显然需要确保这些字符串可以处理高达UINT_32_MAX*(2^32)^3的数字。