I can’t figure out why

std::enable_if<is_allowed_type<T>::value>::type*=nullptr

does not work, while

std::enable_if<is_allowed_type<T>::value>::type* =nullptr

works well.

The difference is just a blank space. Does the *= is treated as an operator?

Here is the code snippet:

#include <stdint.h>
#include <type_traits>

template<class T>
class is_allowed_type:public std::false_type{};

template<>
class is_allowed_type<uint8_t>:public std::true_type{};

template<>
class is_allowed_type<uint16_t>:public std::true_type{};

template<class T>
typename std::enable_if<is_allowed_type<T>::value>::type foo(T val){}

template<class T, typename = typename std::enable_if<is_allowed_type<T>::value>::type>
void goo(T val){}

template<class T>
void koo(T val, typename std::enable_if<is_allowed_type<T>::value>::type*=nullptr){}

template<class T>
void hoo(T val, typename std::enable_if<is_allowed_type<T>::value>::type* = nullptr){}

int main() {
    unsigned char val{'0'};
    uint16_t num{0};
    foo(val);  //works;
    foo(num);  //works;
    //foo(6);    //not compile as expected
    //foo(3.14); //not compile as expected 
    return 0;
}

Here is what the compiler complains:

<source>:20:73: error: expected ')'
   20 | void koo(T val, typename std::enable_if<is_allowed_type<T>::value>::type*=nullptr){}
      |                                                                         ^
<source>:20:9: note: to match this '('
   20 | void koo(T val, typename std::enable_if<is_allowed_type<T>::value>::type*=nullptr){}
      |         ^
1 error generated.