c++ - 返回时不允许隐式转换

Question

#include <optional>

bool f() {
  std::optional<int> opt;
  return opt;
}

不编译: 'return': cannot convert from 'std::optional<int>' to 'bool'
咨询引用我本来想找到一个解释，但我读了它应该没问题。

Implicit conversions are performed whenever an expression of some type T1 is used in context that does not accept that type, but accepts some other type T2; in particular:

• when the expression is used as the argument when calling a function that is declared with T2 as parameter;
• when the expression is used as an operand with an operator that expects T2;
• when initializing a new object of type T2, including return statement in a function returning T2;
• when the expression is used in a switch statement (T2 is integral type);
• when the expression is used in an if statement or a loop (T2 is bool).

Answer 1

std::optional没有任何隐式转换为 bool 的工具. (允许隐式转换为 bool 通常被认为是一个坏主意，因为 bool 是一个整数类型，所以像 int i = opt 这样的东西会编译并做完全错误的事情。)
std::optional确实有一个“上下文转换”到 bool，其定义看起来类似于强制转换运算符:explicit operator bool() .这不能用于隐式转换；它仅适用于某些特定情况，其中预期的“上下文”是 bool 值，例如 if 语句的条件。

您要的是opt.has_value() .