What is the meaning of 14.8.2 paragraphs 3 and 4 in the C++ Standard?
c++, c++11, templates
Solution
Consider:
template <class T> void f(T t) { t = 5; }
`f<int>` is well-formed, but `f<const int>` is not, because it attempts to assign to a `const` variable.
See: Use of 'const' for function parameters
Problem
I'm struggling to understand this rule, specially the sentences in bold below (my emphasis): Consider the comment #2 in the snippet below: what does it mean to say that the function type is `f(int)`, but `t` is `const`? `§14.8.2/3`: After this substitution is performed, the function parameter type adjustments described in 8.3.5 are performed. [ Example: A parameter type of “`void ()(const int, int[5])`” becomes “`void(*)(int,int*)`”. —end example ] [ Note: A top-level qualifier in a function parameter declaration does not affect the function type but still affects the type of the function parameter variable within the function. —end note ] [ Example: ``` template <class T> void f(T t); template <class X> void g(const X x); template <class Z> void h(Z, Z*); int main() { // #1: function type is f(int), t is non const f<int>(1); // #2: function type is f(int), t is const f<const int>(1); // #3: function type is g(int), x is const g<int>(1); // #4: function type is g(int), x is const g<const int>(1); // #5: function type is h(int, const int*) h<const int>(1,0); ``` } —end example ] `§14.8.2/4`: [ Note: `f<int>(1)` and `f<const int>(1)` call distinct functions even though both of the functions called have the same function type. —end note ]