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Editorial Team
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Editorial Team
Asked: May 12, 20262026-05-12T07:07:14+00:00 2026-05-12T07:07:14+00:00

I have method that transforms some input value by the user passing it a

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I have method that transforms some input value by the user passing it a Func delegate wich returns the new value (very over simplified for what I am trying to achieve)

    public L Coerce<L>(string value, Func<string, L> coercer)
    {
        return coercer(value);
    }

    Coerce<int>("123", v => int.Parse(v));

This is fine however I also want to be able to write methods that override the behaviour for a specific type eg…

    public int Coerce<int>(string value)
    {
        return Coerce<int>(value, v => int.Parse(v));
    }

So basically calling

    Coerce<int>("123"); // equivalent Coerce<int>("123", v => int.Parse(v));

will save me having to re-write the int.Parse for every Coerce. Of course this should then extend to handle

    public decimal Coerce<decimal>(string value)
    {
        return Coerce<decimal>(value, v => int.Parse(v));
    }

Etc etc.

Can this be done neatly?

James

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-12T07:07:14+00:00Added an answer on May 12, 2026 at 7:07 am

    Well, if you really don’t want to do

    Convert.ToInt32(value)
    

    Then this will do what you are asking:

    public T Coerce<T>(string value) where T : IConvertible
    {
        return (T)(((IConvertible)value).ToType(typeof(T),
           CultureInfo.InvariantCulture));
    }
    

    Hence:

    int x = Coerce<int>("123");
    

    or

    byte b = Coerce<byte>("123");
    

    This will give you a compile-time error if you try to coerce to a non-convertible type, for example:

    var x = Coerce<MyClass>("123"); //compile-time error
    

    In which case you force the caller to use your Coerce(string value, Func<string,T> coercer) overload.

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