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c# postfix increment operator overload
Sal-versij
on Jun 17, 2020
public static Counter operator ++(Counter c) {
return new Counter(c.v + 1);
}
// In C# the increment operator must not mutate its argument. Rather it must only compute the incremented value and return it, without producing any side effects. The side effect of mutating the variable will be handled by the compiler.
Counter c1 = new Counter(1);
// Call the object that c1 refers to right now W. W.v is 1.
Counter c2 = c1++;
// This has the semantics of:
temp = c1
c1 = operator++(c1) // create object X, set X.v to 2
c2 = temp
// So c1 now refers to X, and c2 refers to W. W.v is 1 and X.v is 2.
Counter c3 = ++c1;
// This has the semantics of
temp = operator++(c1) // Create object Y, set Y.v to 3
c1 = temp
c3 = temp
// So c1 and c3 now both refer to object Y, and Y.v is 3.
c3++;
// This has the semantics of
c3 = operator++(c3) // Create object Z, set Z.v to 4All those coders who are working on the C# based application and are stuck on c# postfix increment operator overload can get a collection of related answers to their query. Programmers need to enter their query on c# postfix increment operator overload related to C# code and they'll get their ambiguities clear immediately. On our webpage, there are tutorials about c# postfix increment operator overload for the programmers working on C# code while coding their module. Coders are also allowed to rectify already present answers of c# postfix increment operator overload while working on the C# language code. Developers can add up suggestions if they deem fit any other answer relating to "c# postfix increment operator overload". Visit this developer's friendly online web community, CodeProZone, and get your queries like c# postfix increment operator overload resolved professionally and stay updated to the latest C# updates.
