Why does std::string_view create a dangling view in a ternary expression?

Question

Consider a method that returns a std::string_view either from a method that returns a const std::string& or from an empty string. To my surprise, w

Answer 1

Because that's how the conditional operator works.

You're invoking ?: on two operands, one of which is an lvalue of type std::string const and the other is an lvalue of type char const[1]. The language rule for the conditional operator is... really complicated. The relevant rule is:

Otherwise, if the second and third operand have different types and either has (possibly cv-qualified) class type, or if both are glvalues of the same value category and the same type except for cv-qualification, an attempt is made to form an implicit conversion sequence from each of those operands to the type of the other. [ Note: Properties such as access, whether an operand is a bit-field, or whether a conversion function is deleted are ignored for that determination. — end note ] Attempts are made to form an implicit conversion sequence from an operand expression E1 of type T1 to a target type related to the type T2 of the operand expression E2 as follows:

• If E2 is an lvalue, the target type is “lvalue reference to T2”, subject to the constraint that in the conversion the reference must bind directly ([dcl.init.ref]) to a glvalue.
• If E2 is an xvalue, [...]

• If E2 is a prvalue or if neither of the conversion sequences above can be formed and at least one of the operands has (possibly cv-qualified) class type:

  • if T1 and T2 are the same class type [...]
  • otherwise, if T2 is a base class of T1, [...]
  • otherwise, the target type is the type that E2 would have after applying the lvalue-to-rvalue, array-to-pointer, and function-to-pointer standard conversions.

Using this process, it is determined whether an implicit conversion sequence can be formed from the second operand to the target type determined for the third operand, and vice versa. If both sequences can be formed, or one can be formed but it is the ambiguous conversion sequence, the program is ill-formed. If no conversion sequence can be formed, the operands are left unchanged and further checking is performed as described below. Otherwise, if exactly one conversion sequence can be formed, that conversion is applied to the chosen operand and the converted operand is used in place of the original operand for the remainder of this subclause. [ Note: The conversion might be ill-formed even if an implicit conversion sequence could be formed. — end note ]

Can't convert std::string const to either char const(&)[1] or char const*, but you can convert char const[1] to std::string const (the inner nested bullet)... so that's what you get. A prvalue of type std::string const. Which is to say, you're either copying one string or constructing a new one... either way, you're returning a string_view to a temporary which goes out of scope immediately.

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What you want is either what you had:

std::string_view myMethod(bool bla) {
    return bla ? std::string_view(otherMethod()) : "";
}

or:

std::string_view myMethod(bool bla) {
    return bla ? otherMethod() : ""sv;
}

The result of that conditional operator is a string_view, with both conversions being safe.