Though the short answer "it's not an lvalue" is correct, that's perhaps just begging the question. Why isn't it an lvalue? Or, as we say in C#, a variable.
The reason is because you cannot have your cake and eat it too. Work it out logically:
First off, the meaning of a ++ operator in C#, whether postfix or prefix, is "take the value of this variable, increment the value, assign the new value to the variable, and produce a value as a result". The value produced as the result is either the original value or the incremented value, depending on whether it was a postfix or a prefix. But either way, you produce a value.
Second, the value of a variable is always the current contents of that variable. (Modulo certain bizarre threading scenarios that would take us far afield.)
I hope you agree that these are perfectly sensible rules.
Now it should be clear why the result of i++ cannot be a variable, but in case it isn't, let me make it clear:
Suppose i is 10. The meaning of i++ should be "get the value of i — 10 — increment it — 11 — store it — i is now 11 — and give the original value as the result — 10". So when you say print(i++) it should print 10, and 11 should be stored in i.
Now suppose the meaning of i++ is to return the variable, not the value. You say print(i++) and what happens? You get the value of i — 10 — increment it — 11 — store it — i is now 11 — and give the variable back as a result. What's the current value of the variable? 11! Which is exactly what you DON'T want to print.
In short, if i++ returned a variable then it would be doing exactly the opposite of the intended meaning of the operator! Your proposal is logically inconsistent, which is why no language does it that way.
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