标记为noexcept的函数中是否有异常？

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假设我有一个标记为 noexcept 的函数，但是有一行代码可以抛出。这行代码将在try块中，异常将被捕获。这是否会导致什么？

Let's say that I have a function marked as noexcept but there's a line of code inside that can throw. That line of code will be in a try block and the exception will be caught. Does that cause anything?

void MyFunc() noexcept { try { throw std::exception("..."); } catch (const std::exception & e) { // I'll deal with it here... } }

推荐答案

是的，这是完全合法的，只要异常不漏出函数。

Yes, this is perfectly legal, as long as the exception doesn't leak out of the function.

一个实现不应该拒绝一个表达式，只是因为执行时它抛出或可能抛出一个异常，函数不允许。

An implementation shall not reject an expression merely because when executed it throws or might throw an exception that the containing function does not allow.

[C ++ 11中的except.spec / 11]

[except.spec/11 in C++11]