使用typeid检查模板类型 [英] Using typeid to check for template type
问题描述
我想知道如果进行以下操作是安全的:
I would like to know if doing the following is safe:
template<class T>
void Parameters::add(Parameter<T> p)
{
std::string sprobe("");
int iprobe = 0;
double dprobe = 0.;
if (typeid(T) == typeid(sprobe))
this->mstrings[p.name()] = p;
if (typeid(T) == typeid(iprobe))
this->mints[p.name()] = p;
if (typeid(T) == typeid(dprobe))
this->mdoubles[p.name()] = p;
}
我有一个用于存储参数的类。它有3个boost :: unordered_map成员变量用于存储类型int,double和std :: string的参数;
I have a Class for storing parameters. It has 3 boost::unordered_map member variables for storing parameters of type int, double and std::string;
我创建了一个模板类Parameter。
I created a template class Parameter.
我理解如果我的参数不是我预期的3种类型之一,这将失败。但它不是一个问题,因为我知道参数只能是这些类型。
I understand that if my Parameter is not one of the 3 types that I anticipated this will fail. But it is not a problem since I know that the Parameters can only be of these types.
感谢您的帮助
推荐答案
代码不会编译,而不是因为 typeid
。问题是,即使使用正确的如果
-clauses,您的方法的代码需要编译 - 所有。这与代码的一部分是否被执行(=被评估)无关。这导致的问题是,如果 T
是 int
,您仍然需要能够编译另一个例如,此行:
The code won't compile, but not because of typeid
. The problem is that even with the correct if
-clauses, the code of your method needs to be compiled - all of it. That is independent of whether or not a part of the code is executed (=evaluated) or not. This leads to the problem that if T
is int
, you still need to be able to compile the code for the other cases, e.g., this line:
this->mstrings[p.name()] = p;
mstrings
的类型很可能不兼容将参数
p
,因此您将收到编译错误。
The type of mstrings
is very likely incompatible with passing Parameter<int>
as p
, hence you will get a compile error.
解决方案是使用重载,其中每个方法只能编译一个案例,但不能编译其他案例,例如 int
:
The solution is to use overloading where each method must only compile one case, but not the others, example for int
:
void Parameters::add(Parameter<int> p)
{
this->mints[p.name()] = p;
}
,对于其他情况也是如此。
and likewise for the other cases.
最后一点:即使你使用 typeid
,你也不需要探针。您可以直接使用 typeid(int)
。
Final note: Even if you use typeid
, you don't need the probes. You can simply use typeid(int)
directly.
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