what is the result (type) of ternary operation?

Question

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

Answer 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).

Answer 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;

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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