为什么uint128_t没有uint128_t的移位超载？

Question

boost::multiprecision::uint128_t doesn't seem to have an overload for the left/right shift operator when both operands are uint128_t s:

uint128_t operator<<(uint128_t, uint128_t); // Missing
uint128_t operator>>(uint128_t, uint128_t); // Also Missing

Some sample code:

#include <boost/multiprecision/cpp_int.hpp>

int main() {
    using uint128_t = boost::multiprecision::uint128_t;
    uint128_t number = 100;
    uint128_t ten = 10;

    auto leftShift = number << ten;  // fail
    auto rightShift = number >> ten; // fail
    return 0;
}

Here is a demo .

The error message is pretty long, so here is the (modified) first message for the left shift operator:

prog.cpp:8:26: error: no match for 'operator<<' (operand types 
                      are 'uint128_t {aka boost_template_typedef_for_uint128_t}' 
                      and 'uint128_t {aka boost_template_typedef_for_uint128_t}')
  auto leftShift = number << ten;  // fail

^{You can look at the demo for the full error messages.}

Why is that the case? I don't see a reason why they wouldn't be implemented, as unsigned a = 100u << 2u; works perfectly fine for the same types.

Answer 1

It's because shifting so many bits to the left is rarely useful. It kind-of defines floating point representations (use cpp_dec_float or similar).

Here's a workaround:

Live On Coliru

#include <boost/multiprecision/integer.hpp>
#include <boost/multiprecision/cpp_int.hpp>
#include <iostream>

int main() {
    boost::multiprecision::uint128_t v("1"), u("20");

    v = v << u.convert_to<size_t>();
    std::cout << v;
}

Prints

1048576