C++ 将变量类型传递给函数
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C++ pass variable type to function
提问by user3325976
How can I pass a variable type to a function? like this:
如何将变量类型传递给函数？像这样：
void foo(type){
cout << sizeof(type);
}
回答by Joseph Mansfield
You can't pass types like that because types are not objects. They do not exist at run time. Instead, you want a template, which allows you to instantiate functions with different types at compile time:
你不能传递这样的类型，因为类型不是对象。它们在运行时不存在。相反，您需要一个模板，它允许您在编译时实例化不同类型的函数：
template <typename T>
void foo() {
cout << sizeof(T);
}
You could call this function with, for example, foo<int>(). It would instantiate a version of the function with Treplaced with int. Look up function templates.
例如，您可以使用foo<int>(). 它将实例化函数的一个版本，T替换为int. 查找函数模板。
回答by microtherion
As Joseph Mansfield pointed out, a function template will do what you want. In some situations, it may make sense to add a parameter to the function so you don't have to explicitly specify the template argument:
正如 Joseph Mansfield 指出的那样，函数模板可以满足您的需求。在某些情况下，向函数添加参数可能是有意义的，这样您就不必显式指定模板参数：
template <typename T>
void foo(T) {
cout << sizeof(T)
}
That allows you to call the function as foo(x), where xis a variable of type T. The parameterless version would have to be called as foo<T>().
这允许您将函数调用为foo(x)，其中x是 T 类型的变量。无参数版本必须调用为foo<T>()。
