[英]c++ exception safety in constructor
What about following code 怎么样下面的代码
MyClass a(new Foo(), new Bar());
if "new Foo()" is successful, but "new Bar()" throws, will Foo leak? 如果“new Foo()”成功,但“new Bar()”抛出,Foo会泄漏吗?
Is taking 正在服用
std::unique_ptr<Foo>
or 要么
std::shared_ptr<Foo>
as parameters, enough to prevent the leak? 作为参数,足以防止泄漏?
if "new Foo()" is successful, but "new Bar()" throws, does Foo will leak? 如果“new Foo()”成功,但“new Bar()”抛出,Foo会泄漏吗?
Yes. 是。
Is taking [...] as parameters, enough to prevent the leak? 以[...]为参数,足以防止泄漏?
Not necessarily. 不必要。 It depends on how you pass the parameters. 这取决于您如何传递参数。 For instance, even supposed your class constructor looks like this: 例如,即使你认为你的类构造函数如下所示:
MyClass::MyClass(std::unique_ptr<Foo> foo, std::unique_ptr<Bar> bar)
The following may still cause a leak: 以下可能仍会导致泄漏:
MyClass a(std::unique_ptr<Foo>(new Foo()), std::unique_ptr<Bar>(new Bar())
That is because the compiler may is allowed to evaluate the above expressions in the following order: 这是因为可以允许编译器按以下顺序评估上述表达式:
new Foo()
评估表达式new Foo()
new Bar()
评估表达式new Bar()
std::unique_ptr<Foo>
temporary from the result of 1. 从结果1构造std::unique_ptr<Foo>
临时。 std::unique_ptr<Bar>
temporary from the result of 2. 从结果2构造临时std::unique_ptr<Bar>
。 If 2) throws an exception, you've lost your Foo
. 如果2)抛出异常,你就失去了你的Foo
。
However, it is possible to make this safe by using std::make_unique<>()
(C++14 only) or std::make_shared<>()
, like so: 但是,通过使用std::make_unique<>()
(仅限C ++ 14)或std::make_shared<>()
,可以使其安全,如下所示:
MyClass a(std::make_unique<Foo>(), std::make_unique<Bar>());
Now no leak could possibly happen, because std::make_unique<>()
(and std::make_shared<>()
) immediately associate the object they create to the corresponding smart pointer, without these two operations (dynamic allocation and construction of the smart pointer) being interleaved with any other operation. 现在不会发生泄漏,因为std::make_unique<>()
(和std::make_shared<>()
)会立即将它们创建的对象关联到相应的智能指针,而不需要这两个操作(动态分配和构造智能指针)与任何其他操作交错。
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