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考虑一个从返回 astd::string_view的方法const std::string&或从空字符串返回 a 的方法。令我惊讶的是，以这种方式编写方法会导致悬空字符串视图：

const std::string& otherMethod();

std::string_view myMethod(bool bla) {
    return bla ? otherMethod() : ""; // Dangling view!
}

https://godbolt.org/z/1Hu_p2

似乎编译器首先将std::string结果的临时副本otherMethod()放在堆栈上，然后返回此临时副本的视图，而不是仅返回引用的视图。首先我想到了一个编译器错误，但是 G++ 和 clang 都这样做了。

修复很简单：包装otherMethod成显式构造string_view解决了这个问题：

std::string_view myMethod(bool bla) {
    return bla ? std::string_view(otherMethod()) : ""; // Works as intended!
}

https://godbolt.org/z/Q-sEkr

为什么会这样？为什么原始代码会在没有警告的情况下创建隐式副本？

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1 回答 1

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因为这就是条件运算符的工作方式。

您正在调用?:两个操作数，其中一个是 type 的左值，std::string const另一个是 type 的左值char const[1]。条件运算符的语言规则......真的很复杂。相关规则是：

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. [ <em>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. [ <em>Note: The conversion might be ill-formed even if an implicit conversion sequence could be formed. — <em>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.

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.

于 2019-06-17T15:22:47.240 回答