How to copy few variables to other variable?

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I'm suspecting you need this kind of function:

template<typename T>
std::enable_if_t<std::is_integral_v<T>, std::array<uint8_t, sizeof(T)>>
littleEndianBytes(T value)
    static_assert(sizeof(uint8_t) == 1);
    using result_type = std::array<uint8_t, sizeof(T)>;
    result_type result;
    for(auto& x : result) {
        x = value & 0xFF;
        value >>= 8;

    return result;

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In the particular case you've shown, you could move the given arguments into the target using bit-shifting (as suggested in the comments) and logical ORing, which would give code like this:

m_n = (b23 << 16) | (b1 << 8) | b0;

But this is very specific to the case you have given. If your other variables have different types and/or you want to copy things differently, you would have to adapt the code to suit each purpose.

Another way (using the same example), but which is more easily adaptable to different target types, would be something like this:

uint8_t bytes[4] = { b0, b1, uint8_t(b23 & 0xFF), uint8_t(b23 >> 8) };
memcpy(&m_n, bytes, 4);

where you first initialize a byte array to the arguments given (could easily be increased to 16 bytes) and then use memcpy to move the byte array into the target.

This latter approach could be further 'optimized' by making bytes a member of Foo and setting up its values in an initializer list:

struct Foo
    Foo(uint8_t b0, uint8_t b1, uint16_t b23) : bytes{ b0, b1, uint8_t(b23 & 0xFF), uint8_t(b23 >> 8) }
        memcpy(&m_n, bytes, 4);
    uint8_t bytes[4];
    uint32_t m_n;

Feel free to ask for further clarification and/or explanation.

4. Methods Use Instance Variables: How Objects Behave, In other words, methods use instance variable values. Like, “if dog is But we can declare a method to give a specific type of value back to the caller, such as: int giveSecret() The argument passed to the z parameter was only a copy of x. The output of "a+b" is same whether you are using any of object reference variable. This happens, because the assignment of one object reference variable to another didn't create any memory, they will refer to the same object. In other words, any copy of the object is not created, but the copy of the reference is created.

Possible implementation (needs C++17, or see below):

template<typename T, typename... Ts>
constexpr void combine_as_bits_impl(std::size_t offset, unsigned char* out, 
                                    const T& x, const Ts&... xs) {
    std::memcpy(out + offset, &x, sizeof(T));
    if constexpr (sizeof...(Ts) > 0)
        combine_as_bits_impl(offset + sizeof(T), out, xs...);

template<typename Out, typename... Ts>
constexpr Out combine_as_bits(const Ts&... xs) {
    static_assert((sizeof(Ts) + ...) == sizeof(Out));

    unsigned char buff[sizeof(Out)];
    combine_as_bits_impl(0, buff, xs...);

    Out out;
    std::memcpy(&out, buff, sizeof(Out));
    return out;

Usage examples:

auto s = combine_as_bits<std::uint32_t>(
             std::uint8_t{0x1}, std::uint8_t{0x2}, std::uint16_t{0x3456});
assert(s == 0x34560201);


Foo(std::uint8_t b0, std::uint8_t b1, std::uint16_t b23) : 
    m_n(combine_as_bits<std::uint32_t>(b0, b1, b23)) {}

If if constexpr is not available, simple overloading can be used to terminate recursion:

constexpr void combine_as_bits_impl(std::size_t, unsigned char*) {}

template<typename T, typename... Ts>
constexpr void combine_as_bits_impl(std::size_t offset, unsigned char* out, 
                                    const T& x, const Ts&... xs) {
    std::memcpy(out + offset, &x, sizeof(T));
    combine_as_bits_impl(offset + sizeof(T), out, xs...);

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about_Ref, You can use reference type variables to permit a function to change the When you pass a variable by value, you are passing a copy of the data. Some .NET methods may require you to pass a variable as a reference. Numeric variables. All the traditional mathematical operators (i.e., +, -, /, (, ), and *) work in R in the way that you would expect when performing math on variables. For example, to add two numeric variables called q2a_1 and q2b_1, select Insert > New R > Numeric Variable (top of the screen), paste in the code q2a_1 + q2b_1, and click CALCULATE.