C++ size_type does not name a type

Question

While trying to create a function to calculate the length of a vector, I encountered error: 'size_type' does not name a type for line 2. Is size_type not already a type? What exactly does it mean to name a type?

template<class vecType>
size_type len(vector<vecType> inVector) {
    size_type vecSize = inVector.size();
    return vecSize;
}

FYI : Using gcc/g++ 4.9.2

"name" is a verb in this case. It means `size_type` is not a name of a data type. It means you have not told the compiler what `size_type` is. — , Mar 17 '17 at 02:01
`size_type` is usually attached to another class, such as `vector::size_type` — user4581301, Mar 17 '17 at 02:02
Related: http://stackoverflow.com/questions/4849632/vectorintsize-type-in-c — user4581301, Mar 17 '17 at 02:04

score 2 · Accepted Answer · edited May 23 '17 at 12:17

Let's take a look at a regular function, not a function template.

int add_two(int in)
{
   return (in + 2);
}

The int before the name of the function is the return type. If the compiler cannot determine that it represents a type, it will report that as an error. If you had a typo and wrote

imt add_two(int in)
{
   return (in + 2);
}

the compiler will complain that imt is not a type.

A function template must also have a return type. You have:

template<class vecType>
size_type len(vector<vecType> inVector) {
    size_type vecSize = inVector.size();
    return vecSize;
}

Here, you have size_type before the function name. If the compiler cannot determine that size_type is indeed a type, it will complain. If size_type is a known type, the compiler will proceeds with that type being the return type of the function.

You can use:

template<class vecType>
typename std::vector<vecType>::size_type len(vector<vecType> inVector) {
   ...
}

to let the compiler know that std::vector<vecType> has a type called size_type and the return value of that function is that type. You will need to use typename std::vector<vecType>::size_type since it is a dependent type name. See Where and why do I have to put the "template" and "typename" keywords? for more on the subject.

Also, you have to fix the declaration of the variable vecSize in the function.

template<class vecType>
typename std::vector<vecType>::size_type len(vector<vecType> inVector) {
   typename std::vector<vecType>::size_type vecSize = inVector.size();
   return vecSize;
}

If you are able to use a C++11 compiler, you can simplify the variable declaration to

   auto vecSize = inVector.size();

score -1 · Answer 2 · answered Mar 17 '17 at 02:06

-1

Try

vector<vecType>::size_type vecSize = inVector.size ();

answered Mar 17 '17 at 02:06

user481779

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C++ size_type does not name a type

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