Compiler Error C7510

'type-name': use of dependent template name must be prefixed with 'template'
'type-name': use of dependent type name must be prefixed with 'typename'

Remarks

In /permissive- mode, the compiler requires the template keyword to precede a template name when it comes after a dependent nested-name-specifier. Similar rules hold for types qualified by typename.

Compiler behavior has changed starting in Visual Studio 2017 version 15.8 under /permissive- mode. The compiler requires the template or typename keyword to precede a template or type name when it comes after a dependent nested-name-specifier. For more information, see Name resolution for dependent types and Templates and name resolution.

Examples

The following code under /permissive- mode now raises C7510:

template<typename T> struct Base
{
    template<class U> void example() {}
};

template<typename T>
struct X : Base<T>
{
    void example()
    {
        Base<T>::example<int>(); // C7510: 'example': use of dependent
            // template name must be prefixed with 'template'
            // note: see reference to class template instantiation
            // 'X<T>' being compiled
    }
};

To fix the error, add the template keyword to the Base<T>::example<int>(); statement, as shown in the following example:

template<typename T> struct Base
{
    template<class U> void example() {}
};

template<typename T>
struct X : Base<T>
{
    void example()
    {
        // Add template keyword here:
        Base<T>::template example<int>();
    }
};

In Visual Studio 2019 under /std:c++20 or later, function template bodies that have if constexpr statements have extra parsing-related checks enabled. For example, in Visual Studio 2017 the following code produces C7510 only if the /permissive- option is set. In Visual Studio 2019 the same code raises errors even when the /permissive option is set:

// C7510.cpp
// compile using: cl /EHsc /W4 /permissive /std:c++latest C7510.cpp
#include <iostream>

template <typename T>
int f()
{
    T::Type a; // error C7510: 'Type': use of dependent type name must be prefixed with 'typename'
    // To avoid the error, add the 'typename' keyword. Use this declaration instead:
    // typename T::Type a;

    if constexpr (a.val)
    {
        return 1;
    }
    else
    {
        return 2;
    }
}

struct X
{
    using Type = X;
    constexpr static int val = 1;
};

int main()
{
    std::cout << f<X>() << "\n";
}

See also

/permissive- (Standards conformance)
Name resolution for dependent types
Templates and name resolution
typename