C++ pass variable type to function

Question

How can I pass a variable type to a function? like this:

void foo(type){     cout << sizeof(type); } 

Answer 1

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:

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 T replaced with int. Look up function templates.

Answer 2

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:

template <typename T> void foo(T) {   cout << sizeof(T) } 

That allows you to call the function as foo(x), where x is a variable of type T. The parameterless version would have to be called as foo<T>().