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C++ Variadics

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130 Discussions

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  • trinhmaibabk2012
     + 0 comments
    template <bool...digits>
int reversed_binary_value(){
    bool arr[] = {digits...};
    int sum = 0;
    int a = 0;
    for(bool i : arr){
        sum = sum | ((1&i) << a++);
    }
    return sum;
}

  • Ahmed_Lotfy
     + 0 comments

    This is a stupid example, doesn't make any sense

  • harun0450
     + 0 comments
    template <bool... digits> 
int reversed_binary_value(){
    std::vector<int> parameterPack = {digits...};
    //std::reverse(parameterPack.begin(), parameterPack.end());
    int a = 0;
    int sum = 0;
    for (bool i : parameterPack)
    {
        sum = sum + std::pow(2,a)*i;
        a++;
    }
    return sum;
}

  • sezzaldiin0
     + 0 comments
    template <bool a> int reversed_binary_value() { return a; }

template <bool a, bool b, bool... d> int reversed_binary_value() {
  return (reversed_binary_value<b, d...>() << 1) + a;
}

  • pedroandre
     + 0 comments

    Decided to play with MetaProgramming,

    template struct constexpr_pow { constexpr static uint64_t value() { return base * constexpr_pow::value(); } };

    template struct constexpr_pow { constexpr static uint64_t value() { return 1; } }; static_assert(constexpr_pow<2, 0>::value() == 1, "Failed pow(2,0)"); static_assert(constexpr_pow<2, 1>::value() == 2, "Failed pow(2,1)"); template constexpr uint64_t reversed_binary_value_impl() { return 0; } // 0, 0, [1...] template constexpr uint64_t reversed_binary_value_impl() { return digit * constexpr_pow<2, n>::value() + reversed_binary_value_impl(); } template constexpr uint64_t reversed_binary_value() { return reversed_binary_value_impl<0, digits...>(); } static_assert(reversed_binary_value<0>() == 0, "Failed 0b0"); static_assert(reversed_binary_value<1>() == 1, "Failed 0b1");