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Convert CPP to C#

Posted on 2014-04-12
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Last Modified: 2014-04-12
How do I convert the following to C#

struct fpoint : pair<double> {
    fpoint() : pair<double>() { }
    fpoint(double x, double y) : pair<double>(x, y) { }
};
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Question by:JElster
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by:jkr
jkr earned 100 total points
ID: 39996739
The C# equivalent could be

public struct fpoint : System.Collections.Generic.KeyValuePair<double,double> {
    fpoint() : base() { }
    fpoint(double x, double y) : base((x, y) { }
}; 

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by:JElster
ID: 39996749
Get a bunch of compile errors,, like missing param.
Probably close... thx
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Expert Comment

by:jkr
ID: 39996758
Soory, had my mind tilted towards C++ a bit too much - does

public struct fpoint : System.Collections.Generic.KeyValuePair<double,double> {
    fpoint() : base() { }
    fpoint(double x, double y) : base() { key = x; value = y; }
}; 
                                            

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work better?
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Author Comment

by:JElster
ID: 39996764
Error      3      Structs cannot contain explicit parameterless constructors      

Error      2      Type 'System.Collections.Generic.KeyValuePair<...>' in interface list is not an interface
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käµfm³d   👽 earned 400 total points
ID: 39996766
I disagree slightly, but I don't think it's of major concern. From what I read, the Tuple class would be closer to the exact match for pair.

e.g.

public class fpoint : Tuple<double, double>
{
    public fpoint() : base(default(double), default(double)) { }
    public fpoint(double x, double y) : base(x, y) { }
}

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The reason I disagree is that in C# you cannot reassign struct members once you have created the instance--you have to create a second instance with the new values. This is because KeyValuePair is a value type. A Tuple is a reference type, so you can reassign its member even after the instance is created. Albeit in either case there is no swap method, so you'd need to roll your own.
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Expert Comment

by:käµfm³d 👽
ID: 39996770
Actually, never mind. The Tuple's members are read-only also. I saw "tuple" mentioned in the documentation I linked to, so I naturally thought it would be the equivalent. I think either data structure would yield the same results.

Of course, you could always write your own class/struct to be an exact match!
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