I tried to make an Array template class but when I try to build the compiler fails to link the constructor and the method and I get :  undefined reference to `Array::Array()  undefined reference to `Array::getSize()  Here is the header file:  #pragma once  template&lt;typename type, int length&gt;  struct Array{ public:     Array();     int getSize();  private:     type data[length];     int m_length; };   The Array.cpp file:  #include "Array.h"  template&lt;typename t, int l&gt; Array&lt;t, l&gt;::Array() {     m_length = l; }  template&lt;typename type, int length&gt; Array&lt;type, length&gt;::getSize() {     return m_length; }   And the main function:  #define LOG(x) cout&lt;&lt;x&lt;&lt;endl int main() {     Array&lt;int, 10&gt; array;     LOG(array.getSize()); }   If someone has any idea about why I am getting this, I would really appreciate.
Quote from :     The only portable way of using templates at the moment is to implement them in header files by using inline functions.   Why is this?  (Clarification: header files are not the only portable solution. But they are the most convenient portable solution.)
How to Map a subset of list elements to a function? If a nonempty set of real numbers is open and closed, is it $\mathbb{R}$? Why/Why not? In other words, are $\emptyset$ and $\mathbb{R}$ the only open and closed sets in $\mathbb{R}$? Why/Why not?  I tried by assuming a set is equal to its interior points and contains its limit points.  A bounded set will not do since stuff like $[1,4]$ and $\{5\}$ will not work, though that is not really proof. Help please?  Anyway, it must then be unbounded.  If $a$ is a real number then $(a,\infty)$, $(-\infty,a)$, $[a,\infty)$ and $(-\infty,a]$ don't seem to cut it so it must be $\mathbb{R}$. Why use the 'ref' keyword when passing an object? If I am passing an object to a method, why should I use the ref keyword? Isn't this the default behaviour anyway?  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Now, both of the remaining selected faces have the nice color. This is the result I'd rather see instead of image 1. 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Am I limited to just 1, or can I make more than one character at once?