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OlegUP
Asked:2022-03-25 17:19:59 +0000 UTC2022-03-25 17:19:59 +0000 UTC

C++：编译器不在模板中内联函数

  • 772

将函数替换为模板时，编译器由于某种原因看不到它。虽然他看到了上面的两条线。

#include <iostream>
#include <iomanip>
#include <string>
#include <endian.h>

using namespace std;

template <typename RESULT_T, typename letoh_func, typename T> 
static RESULT_T compose_high_low(T high, T low)
{
    RESULT_T res = letoh_func(high);
    res <<= sizeof(T) * 8;
    res |= letoh_func(low);
    return res;
}

int main()
{
    uint16_t k = le16toh(1);
    uint16_t high = 1, low = 2;
    uint32_t n = compose_high_low<uint32_t, le16toh>(high, low); // error: 'le16toh' was not declared in this scope
    cout << hex << n << endl;
    return 0;
}

在这里运行：http: //cpp.sh/7w7afl

c++

1 个回答

  1. Best Answer

    le16toh通常是一个宏。并且要将函数指针作为模板参数传递，您必须使用非类型参数。一般来说，使用boost::endian.

    #include <boost/endian/conversion.hpp>
#include <ios>
#include <iostream>
#include <cstdint>

template <typename x_Result, typename x_Value>
static x_Result compose_high_low(x_Value const high, x_Value const low) noexcept
{
    x_Result res{::boost::endian::little_to_native(high)};
    res <<= sizeof(x_Value) * CHAR_BIT;
    res |= ::boost::endian::little_to_native(low);
    return res;
}

int main()
{
    ::std::cout << ::std::hex
    << compose_high_low<::std::uint32_t>(::std::uint16_t{1}, ::std::uint16_t{2})
    << ::std::endl;
    return 0;
}
    • 5

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