49

我在我的 C++ 程序中使用 GCC 内置类型__int128来处理一些事情，没有什么真正重要的，至少不足以证明仅为此使用 BigInt 库是合理的，但足以防止完全删除它。

当我遇到我的课程的打印部分时，我的问题就出现了，这是一个最小的例子：

#include <iostream>

int main()
{
  __int128 t = 1234567890;

  std::cout << t << std::endl;

  return t;
}

注释掉该std::cout行将使此代码可以很好地编译g++，但是拥有它会导致以下错误消息：

int128.c: In function ‘int main()’:
int128.c:7:13: error: ambiguous overload for ‘operator<<’ (operand types are ‘std::ostream {aka std::basic_ostream<char>}’ and ‘__int128’)
   std::cout << t << std::endl;
             ^
int128.c:7:13: note: candidates are:
In file included from /usr/include/c++/4.9/iostream:39:0,
                 from int128.c:1:
/usr/include/c++/4.9/ostream:108:7: note: std::basic_ostream<_CharT, _Traits>::__ostream_type& std::basic_ostream<_CharT, _Traits>::operator<<(std::basic_ostream<_CharT, _Traits>::__ostream_type& (*)(std::basic_ostream<_CharT, _Traits>::__ostream_type&)) [with _CharT = char; _Traits = std::char_traits<char>; std::basic_ostream<_CharT, _Traits>::__ostream_type = std::basic_ostream<char>] <near match>
       operator<<(__ostream_type& (*__pf)(__ostream_type&))
       ^
/usr/include/c++/4.9/ostream:108:7: note:   no known conversion for argument 1 from ‘__int128’ to ‘std::basic_ostream<char>::__ostream_type& (*)(std::basic_ostream<char>::__ostream_type&) {aka std::basic_ostream<char>& (*)(std::basic_ostream<char>&)}’
/usr/include/c++/4.9/ostream:117:7: note: std::basic_ostream<_CharT, _Traits>::__ostream_type& std::basic_ostream<_CharT, _Traits>::operator<<(std::basic_ostream<_CharT, _Traits>::__ios_type& (*)(std::basic_ostream<_CharT, _Traits>::__ios_type&)) [with _CharT = char; _Traits = std::char_traits<char>; std::basic_ostream<_CharT, _Traits>::__ostream_type = std::basic_ostream<char>; std::basic_ostream<_CharT, _Traits>::__ios_type = std::basic_ios<char>] <near match>
       operator<<(__ios_type& (*__pf)(__ios_type&))
       ^
/usr/include/c++/4.9/ostream:117:7: note:   no known conversion for argument 1 from ‘__int128’ to ‘std::basic_ostream<char>::__ios_type& (*)(std::basic_ostream<char>::__ios_type&) {aka std::basic_ios<char>& (*)(std::basic_ios<char>&)}’
/usr/include/c++/4.9/ostream:127:7: note: std::basic_ostream<_CharT, _Traits>::__ostream_type& std::basic_ostream<_CharT, _Traits>::operator<<(std::ios_base& (*)(std::ios_base&)) [with _CharT = char; _Traits = std::char_traits<char>; std::basic_ostream<_CharT, _Traits>::__ostream_type = std::basic_ostream<char>] <near match>
       operator<<(ios_base& (*__pf) (ios_base&))
       ^
/usr/include/c++/4.9/ostream:127:7: note:   no known conversion for argument 1 from ‘__int128’ to ‘std::ios_base& (*)(std::ios_base&)’
/usr/include/c++/4.9/ostream:166:7: note: std::basic_ostream<_CharT, _Traits>::__ostream_type& std::basic_ostream<_CharT, _Traits>::operator<<(long int) [with _CharT = char; _Traits = std::char_traits<char>; std::basic_ostream<_CharT, _Traits>::__ostream_type = std::basic_ostream<char>]
       operator<<(long __n)
       ^
/usr/include/c++/4.9/ostream:170:7: note: std::basic_ostream<_CharT, _Traits>::__ostream_type& std::basic_ostream<_CharT, _Traits>::operator<<(long unsigned int) [with _CharT = char; _Traits = std::char_traits<char>; std::basic_ostream<_CharT, _Traits>::__ostream_type = std::basic_ostream<char>]
       operator<<(unsigned long __n)
       ^
/usr/include/c++/4.9/ostream:174:7: note: std::basic_ostream<_CharT, _Traits>::__ostream_type& std::basic_ostream<_CharT, _Traits>::operator<<(bool) [with _CharT = char; _Traits = std::char_traits<char>; std::basic_ostream<_CharT, _Traits>::__ostream_type = std::basic_ostream<char>]
       operator<<(bool __n)
       ^
In file included from /usr/include/c++/4.9/ostream:609:0,
                 from /usr/include/c++/4.9/iostream:39,
                 from int128.c:1:
/usr/include/c++/4.9/bits/ostream.tcc:91:5: note: std::basic_ostream<_CharT, _Traits>& std::basic_ostream<_CharT, _Traits>::operator<<(short int) [with _CharT = char; _Traits = std::char_traits<char>]
     basic_ostream<_CharT, _Traits>::
     ^
In file included from /usr/include/c++/4.9/iostream:39:0,
                 from int128.c:1:
/usr/include/c++/4.9/ostream:181:7: note: std::basic_ostream<_CharT, _Traits>::__ostream_type& std::basic_ostream<_CharT, _Traits>::operator<<(short unsigned int) [with _CharT = char; _Traits = std::char_traits<char>; std::basic_ostream<_CharT, _Traits>::__ostream_type = std::basic_ostream<char>]
       operator<<(unsigned short __n)
       ^
In file included from /usr/include/c++/4.9/ostream:609:0,
                 from /usr/include/c++/4.9/iostream:39,
                 from int128.c:1:
/usr/include/c++/4.9/bits/ostream.tcc:105:5: note: std::basic_ostream<_CharT, _Traits>& std::basic_ostream<_CharT, _Traits>::operator<<(int) [with _CharT = char; _Traits = std::char_traits<char>]
     basic_ostream<_CharT, _Traits>::
     ^
In file included from /usr/include/c++/4.9/iostream:39:0,
                 from int128.c:1:
/usr/include/c++/4.9/ostream:192:7: note: std::basic_ostream<_CharT, _Traits>::__ostream_type& std::basic_ostream<_CharT, _Traits>::operator<<(unsigned int) [with _CharT = char; _Traits = std::char_traits<char>; std::basic_ostream<_CharT, _Traits>::__ostream_type = std::basic_ostream<char>]
       operator<<(unsigned int __n)
       ^
/usr/include/c++/4.9/ostream:201:7: note: std::basic_ostream<_CharT, _Traits>::__ostream_type& std::basic_ostream<_CharT, _Traits>::operator<<(long long int) [with _CharT = char; _Traits = std::char_traits<char>; std::basic_ostream<_CharT, _Traits>::__ostream_type = std::basic_ostream<char>]
       operator<<(long long __n)
       ^
/usr/include/c++/4.9/ostream:205:7: note: std::basic_ostream<_CharT, _Traits>::__ostream_type& std::basic_ostream<_CharT, _Traits>::operator<<(long long unsigned int) [with _CharT = char; _Traits = std::char_traits<char>; std::basic_ostream<_CharT, _Traits>::__ostream_type = std::basic_ostream<char>]
       operator<<(unsigned long long __n)
       ^
/usr/include/c++/4.9/ostream:220:7: note: std::basic_ostream<_CharT, _Traits>::__ostream_type& std::basic_ostream<_CharT, _Traits>::operator<<(double) [with _CharT = char; _Traits = std::char_traits<char>; std::basic_ostream<_CharT, _Traits>::__ostream_type = std::basic_ostream<char>]
       operator<<(double __f)
       ^
/usr/include/c++/4.9/ostream:224:7: note: std::basic_ostream<_CharT, _Traits>::__ostream_type& std::basic_ostream<_CharT, _Traits>::operator<<(float) [with _CharT = char; _Traits = std::char_traits<char>; std::basic_ostream<_CharT, _Traits>::__ostream_type = std::basic_ostream<char>]
       operator<<(float __f)
       ^
/usr/include/c++/4.9/ostream:232:7: note: std::basic_ostream<_CharT, _Traits>::__ostream_type& std::basic_ostream<_CharT, _Traits>::operator<<(long double) [with _CharT = char; _Traits = std::char_traits<char>; std::basic_ostream<_CharT, _Traits>::__ostream_type = std::basic_ostream<char>]
       operator<<(long double __f)
       ^
/usr/include/c++/4.9/ostream:245:7: note: std::basic_ostream<_CharT, _Traits>::__ostream_type& std::basic_ostream<_CharT, _Traits>::operator<<(const void*) [with _CharT = char; _Traits = std::char_traits<char>; std::basic_ostream<_CharT, _Traits>::__ostream_type = std::basic_ostream<char>] <near match>
       operator<<(const void* __p)
       ^
/usr/include/c++/4.9/ostream:245:7: note:   no known conversion for argument 1 from ‘__int128’ to ‘const void*’
In file included from /usr/include/c++/4.9/ostream:609:0,
                 from /usr/include/c++/4.9/iostream:39,
                 from int128.c:1:
/usr/include/c++/4.9/bits/ostream.tcc:119:5: note: std::basic_ostream<_CharT, _Traits>& std::basic_ostream<_CharT, _Traits>::operator<<(std::basic_ostream<_CharT, _Traits>::__streambuf_type*) [with _CharT = char; _Traits = std::char_traits<char>; std::basic_ostream<_CharT, _Traits>::__streambuf_type = std::basic_streambuf<char>] <near match>
     basic_ostream<_CharT, _Traits>::
     ^
/usr/include/c++/4.9/bits/ostream.tcc:119:5: note:   no known conversion for argument 1 from ‘__int128’ to ‘std::basic_ostream<char>::__streambuf_type* {aka std::basic_streambuf<char>*}’
In file included from /usr/include/c++/4.9/iostream:39:0,
                 from int128.c:1:
/usr/include/c++/4.9/ostream:493:5: note: std::basic_ostream<char, _Traits>& std::operator<<(std::basic_ostream<char, _Traits>&, unsigned char) [with _Traits = std::char_traits<char>]
     operator<<(basic_ostream<char, _Traits>& __out, unsigned char __c)
     ^
/usr/include/c++/4.9/ostream:488:5: note: std::basic_ostream<char, _Traits>& std::operator<<(std::basic_ostream<char, _Traits>&, signed char) [with _Traits = std::char_traits<char>]
     operator<<(basic_ostream<char, _Traits>& __out, signed char __c)
     ^
/usr/include/c++/4.9/ostream:482:5: note: std::basic_ostream<char, _Traits>& std::operator<<(std::basic_ostream<char, _Traits>&, char) [with _Traits = std::char_traits<char>]
     operator<<(basic_ostream<char, _Traits>& __out, char __c)
     ^
/usr/include/c++/4.9/ostream:476:5: note: std::basic_ostream<_CharT, _Traits>& std::operator<<(std::basic_ostream<_CharT, _Traits>&, char) [with _CharT = char; _Traits = std::char_traits<char>]
     operator<<(basic_ostream<_CharT, _Traits>& __out, char __c)
     ^

是的，我知道，很多行解释__int128只是没有妥善处理......

有没有一种简单的方法可以像任何其他数字类型一样__int128被打印？iostream

编辑：对于那些仍然混淆 C 和 C++ 的人，是的，我读到了这个问题：如何使用 gcc 打印 __uint128_t 数字？但是，正如我现在所问的那样，这是针对 C 而不是针对 C++ 的。

4

6 回答 6

40

如果您不需要任何花哨的格式化选项，编写自己的<<运算符是微不足道的。形式上，我怀疑写一个 for__int128_t会被认为是未定义的行为，但实际上，我认为它会起作用，直到库开始为它提供实际支持（此时，你会退休你的转换运算符）。

无论如何，应该像下面这样工作：

std::ostream&
operator<<( std::ostream& dest, __int128_t value )
{
    std::ostream::sentry s( dest );
    if ( s ) {
        __uint128_t tmp = value < 0 ? -value : value;
        char buffer[ 128 ];
        char* d = std::end( buffer );
        do
        {
            -- d;
            *d = "0123456789"[ tmp % 10 ];
            tmp /= 10;
        } while ( tmp != 0 );
        if ( value < 0 ) {
            -- d;
            *d = '-';
        }
        int len = std::end( buffer ) - d;
        if ( dest.rdbuf()->sputn( d, len ) != len ) {
            dest.setstate( std::ios_base::badbit );
        }
    }
    return dest;
}

请注意，这只是一个快速的临时修复，直到 g++ 库支持该类型。它依靠 2 的补码，在溢出时环绕， for __int128_t，但如果不是这种情况，我会感到非常惊讶（正式地说，这是未定义的行为）。如果没有，您需要修复 tmp. 当然，它不处理任何格式选项。您可以根据需要添加。（处理填充和 adjustfield正确的处理可能很重要。）

于 2014-08-04T09:07:38.637 回答
22

我建议不要operator<<为__int128_t. 原因是，每当您看到cout << x某种整数类型时，您会期望各种操纵器都喜欢std::hex或std::setw应该也可以工作。重载运算符时最重要的准则是：“像整数一样做”。

作为替代方案，我建议使用decimal_string(__int128_t)可以cout << decimal_string(x);在代码中使用的函数。对于字符串转换，您可以使用任何 C 相关问答中的算法。这清楚地表明您有 128 位整数的特殊代码。每当标准库升级到 128 位支持时，您都可以删除它（并且很容易grep实现这些功能）。

于 2014-08-04T08:47:44.567 回答
3

库存cout不处理__int128，但您可以使用自己的功能对其进行扩展。

首先，编写如下代码：

std::ostream& operator<<(std::ostream& os, __int128 t) {
    // TODO: Convert t to string
    return os << str;
}

SO上有很多解决方案可以将128位数字转换为字符串，这里不再赘述。

关于评论中的库兼容性：

如果标准库不提供此类处理程序，您只需要滚动您自己的函数。一旦库支持该类型，您应该会在构建时看到冲突，例如 [注意：内置候选运算符<<]，请尝试使用 int64_t。

于 2014-08-04T08:48:25.147 回答
3

一种看似简单的方法

std::ostream& operator<<(std::ostream& o, const __int128& x) {
    if (x == std::numeric_limits<__int128>::min()) return o << "-170141183460469231731687303715884105728";
    if (x < 0) return o << "-" << -x;
    if (x < 10) return o << (char)(x + '0');
    return o << x / 10 << (char)(x % 10 + '0');
}
于 2020-01-26T00:42:27.260 回答
1

到目前为止的答案都很好，但我只是想补充一下 James Kanze 的答案。首先请注意，由于无符号转换，它不适用于 number -0x80000000000000000000000000000000。其次，您可以利用使用 64 位整数进行打印这一事实来优化函数实现，如下所示：

std::ostream& operator<<(std::ostream& os, __int128_t value) {
    if (value < 0) {
        os << '-';
        value = -value;
    }
    // save flags to restore them
    std::ios_base::fmtflags flags(os.flags());
    // set zero fill
    os << std::setfill('0') << std::setw(13);

    // 128-bit number has at most 39 digits,
    // so the below loop will run at most 3 times
    const int64_t modulus = 10000000000000; // 10**13
    do {
        int64_t val = value % modulus;
        value /= modulus;
        if (value == 0) {
            os.flags(flags);
            return os << val;
        }
        os << val;
    } while (1);
}
于 2019-12-05T19:29:08.027 回答
1

如果它不是性能关键，这里有一个简单易读的方法将非负int128 转换为 base-10 字符串（当然可以打印出来）：

std::string toString(__int128 num) {
    std::string str;
    do {
        int digit = num % 10;
        str = std::to_string(digit) + str;
        num = (num - digit) / 10;
    } while (num != 0);
    return str;
}

我们可以通过将数字分成更大的块而不是一次一个来使这个速度提高几倍。但它要求我们检查每个块是否有任何丢失的前导零并将它们重新添加：

std::string toString(__int128 num) {
    auto tenPow18 = 1000000000000000000;
    std::string str;
    do {
        long long digits = num % tenPow18;
        auto digitsStr = std::to_string(digits);
        auto leading0s = (digits != num) ? std::string(18 - digitsStr.length(), '0') : "";
        str = leading0s + digitsStr + str;
        num = (num - digits) / tenPow18;
    } while (num != 0);
    return str;
}

注意：我还在这里为unsigned int128s发布了这个答案的一个版本。

于 2019-05-03T16:26:03.467 回答