How to check for the type of a template parameter?

Use is_same:

#include <type_traits>

template <typename T>
void foo()
{
    if (std::is_same<T, animal>::value) { /* ... */ }  // optimizable...
}

Usually, that's a totally unworkable design, though, and you really want to specialize:

template <typename T> void foo() { /* generic implementation  */ }

template <> void foo<animal>()   { /* specific for T = animal */ }

Note also that it's unusual to have function templates with explicit (non-deduced) arguments. It's not unheard of, but often there are better approaches.

You can specialize your templates based on what's passed into their parameters like this:

template <> void foo<animal> {

}

Note that this creates an entirely new function based on the type that's passed as T. This is usually preferable as it reduces clutter and is essentially the reason we have templates in the first place.

I think todays, it is better to use, but only with C++17.

#include <type_traits>

template <typename T>
void foo() {
    if constexpr (std::is_same_v<T, animal>) {
        // use type specific operations... 
    } 
}

If you use some type specific operations in if expression body without constexpr, this code will not compile.

std::is_same() is only available since C++11. For pre-C++11 you can use typeid():

template <typename T>
void foo()
{
    if (typeid(T) == typeid(animal)) { /* ... */ }
}