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Home/ Questions/Q 990335
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Editorial Team
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Editorial Team
Asked: May 16, 20262026-05-16T05:56:03+00:00 2026-05-16T05:56:03+00:00

Looking through the source code of a binary tree,I find the following function: //definition

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Looking through the source code of a binary tree,I find the following function:

//definition of BTR,in case you'd want to know
template< class Type>
struct BTR 
{
    // The item saved to that specifiec position into the tree
    Type  value;    

    // Points to the left leaf
    BTR<Type>*  left;

    // Points to the right leaf
    BTR<Type>*  right;  
};

//why typename?
template< class Type>
BTR<Type>* CreateEx(typename const std::vector<Type>::iterator start,typename const std::vector<Type>::iterator end) 
{
    //definition
}

Now,what’s confusing me about this function, is its parameters.
Why does it need the keyword typename?
Because if I remove both the typenames,my compiler starts complaining and says I should put a ‘)’ before identifier ‘start’.
And if I changed the parameters so that the function took two vectors instead of two iterators and removed the typenames,my compiler stops complaining(although of course,the function doesn’t work any more).

// perfectly acceptable!
template< class Type>
BTR<Type>* CreateEx( const std::vector<Type> start, const std::vector<Type> end)

So it seems I need the keyword because the function wants two iterators.
But why is this keyword necessary in a situation like this?

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  1. Editorial Team
    Editorial Team
    2026-05-16T05:56:04+00:00Added an answer on May 16, 2026 at 5:56 am

    Because the compiler doesn’t know if std::vector< Type >::iterator is a type or a member of std::vector< Type >, and thus needs a little help in the form of typename.

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