构造函数初始化列表中的执行顺序是否可确定?我知道成员中的成员顺序是这些成员初始化的顺序,但如果我有这样的场景:
class X()
{
X_Implementation* impl_;
};
and then providing that allocator is available:
X::X():impl_(Allocate(sizeof(X_Implementation)))//HERE I'M ALLOCATING <--1
,impl_(Construct<X_Implementation>(impl_))//AND HERE I'M CONSTRUCTING <--2
{
}
但为了使这个可靠,这个顺序必须是从左到右.是否保留了高级书籍std ::或者没有?如果不是,我总是可以将第二条线移动到身体中.
解决方法:
根据ISO / IEC 14882:2003(E)第12.6.2节:
Initialization shall proceed in the following order:
- First, and only for the constructor of the most derived class as described below, virtual base classes shall
be initialized in the order they appear on a depth-first left-to-right traversal of the directed acyclic graph
of base classes, where “left-to-right” is the order of appearance of the base class names in the derived
class base-specifier-list.- Then, direct base classes shall be initialized in declaration order as they appear in the base-specifier-list
(regardless of the order of the mem-initializers).- Then, nonstatic data members shall be initialized in the order they were declared in the class definition
(again regardless of the order of the mem-initializers).- Finally, the body of the constructor is executed.
因此,按照该顺序,您将获得订单.同样根据标准,订单被如此规定,使得对象可以以完全相反的顺序未初始化.
标签:c,operator-precedence 来源: https://codeday.me/bug/20190918/1810907.html
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