[英]Check empty vector in initializer list
I would like initialize base class by a first item form a vector which contains base classes, eg: 我想通过第一项初始化基类,形成一个包含基类的向量,例如:
struct Base { ... };
struct Derived : public Base
{
Derived(const Base& baseClass)
: Base(baseClass)
{
}
Derived(const std::vector<Base*>& listOfBaseClass)
: Base(*(listOfBaseClass[0]))
{
}
};
This solution fails when listOfBaseClass
vector will be empty. 当
listOfBaseClass
向量为空时,此解决方案失败。 How can prevent this situation. 怎么能防止这种情况。 This is simplified example, so calling base class constructor like this:
这是简化的示例,因此调用基类构造函数如下:
std::vector<Base> baseClasses;
if(!baseClasses.empty())
{
Derived myDerived(baseClasses[0]);
}
it is not solution for my problem. 这不是我的问题的解决方案。 Thanks in advance.
提前致谢。
Throw an exception if the list is empty: 如果列表为空,则抛出异常:
: Base(listOfBaseClass.empty()
? throw std::invalid_argument("Missing base class")
: *(listOfBaseClass[0])
)
Non-exception variant by request: 根据要求的非例外变体:
: Base(( (listOfBaseClass.empty() && exit(0)),
*(listOfBaseClass[0])
))
Derived(const std::vector<Base*>& listOfBaseClass)
: Base(*(listOfBaseClass->at(0)))
{
}
"at" provides exception safety. “at”提供异常安全。 It throws out_of_range if n is out of bounds.
如果n超出范围,它会抛出out_of_range。
http://www.cplusplus.com/reference/vector/vector/at/ http://www.cplusplus.com/reference/vector/vector/at/
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