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Home/ Questions/Q 6738719
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Editorial Team
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Editorial Team
Asked: May 26, 20262026-05-26T11:24:07+00:00 2026-05-26T11:24:07+00:00

Are there extensions for C++ like there are in C#? For example in C#

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Are there extensions for C++ like there are in C#?

For example in C# you can do:

public static uint SwapEndian(this uint value)
{
    var tmp = BitConverter.GetBytes(value);
    Array.Reverse(tmp);
    return BitConverter.ToUInt32(tmp, 0);
}

someuint.SwapEndian();

Is there anything like that in C++?

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-26T11:24:08+00:00Added an answer on May 26, 2026 at 11:24 am

    Extension methods (and also “static classes”) exist in C#/Java languages solely because the designers decided that (the Java way of) OOP is The One True Way and that everything must be a method from a class:

    This is not C++ way of doing things. In C++ you have namespaces, free functions and Koenig lookup to extend the behavior of a class:

    namespace foo
    {
        struct bar { ... };
    
        void act_on_bar(const bar& b) { ... };
    }
    
    ...
    
    foo::bar b;
    act_on_bar(b); // No need to qualify because of Koenig lookup
    

    I usually consider extension methods harmful. If you attach too much behavior to a class, you are proabably failing to capture the reason why the class exists. Also (like “partial classes”), they tend to make the code related to a class non local. Which is bad.

    As to your problem, in C++ you simply do:

    template <typename T>
    T swap_endian(T x)
    {
        union { T value; char bytes[sizeof(T)]; } u;
        u.value = x;
    
        for (size_t i = 0; i < sizeof(T)/2; i++) 
            swap(u.bytes[i], u.bytes[sizeof(T) - i - 1]);
    
        return u.value;
    }
    

    Usage:

    swap_endian<std::uint32_t>(42);
    

    or, if the type can be deduced:

    std::uint64_t x = 42;
    std::uint64_t y = swap_endian(x);
    
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