what is the result (type) of ternary operation?

问题

Does the ternary operation return a copy or reference?

I checked the following code

vector<int> v0 = { 1, 2 };
vector<int> v1 = { 3 };

vector<int>& v = true ? v0 : v1;
v.clear(); // v0 will be cleared also

I think the ternary operation returns a copy of v0. And then pass it to v. Thus v and v0 has different storage of data. Testing doesn't show it.

Thanks, Kerrek SB! I add a "should-not-compiled" example (Thanks WhiZTiM!) to show the point.

vector<int>& v = true ? v0 : vector<int>{3};
v.clear(); // v0 will not be cleared

回答1:

The type of a conditional expression is the common type of the operands.

But I think you aren't actually interested in that. What matters is what the value category of a conditional expression is.

If both operands are, or can be converted to, lvalues of the common type, then the conditional expression is an lvalue; otherwise it is an rvalue (potentially requiring lvalue-to-rvalue conversion of one of the operands).

回答2:

The rules are found here: Relevant to your expression:

E1 ? E2 : E3

4) If E2 and E3 are glvalues of the same type and the same value category, then the result has the same type and value category, and is a bit-field if at least one of E2 and E3 is a bit-field.

In your case:

true ? v0 : v1;

v0 and v1 are lvalues (broadly glvalue). So the return will be an lvalue v0. Hence, your expression will be equivalent to:

vector<int>& v = v0;

---

As to your Edit:

vector<int>& v = true ? v0 : vector<int>{3};
v.clear(); // v0 will not be cleared

Should not compile, because the value category of the result will be an rvalue, and you cannot bind a non-const reference to an rvalue

来源：https://stackoverflow.com/questions/42395118/what-is-the-result-type-of-ternary-operation