069_link_variadic_template.cpp Test void void func (int p) func int p A function typename... Args Args void void func (int p, Args &... args) func int p Args &... args A variadic template function overload typename... Args Args void void func (int p, Args &&... args) func int p Args &&... args A variadic template function overload typename... Args Args void void func (int p, const Args &... args) func int p const Args &... args A variadic template function overload typename... Args Args void void func (int p, const Args &&... args) func int p const Args &&... args A variadic template function overload typename... Args Args void void func (int p, Args *... args) func int p Args *... args A variadic template function overload typename... Args Args void void func (int p, Args **... args) func int p Args **... args A variadic template function overload typename... Args Args void void func (int p, const Args *... args) func int p const Args *... args A variadic template function overload typename... Args Args void void func (int p, const Args **... args) func int p const Args **... args A variadic template function overload typename... Args Args void void func (int p, Args... args) func int p Args... args A variadic template function overload At the time of writing, the part between <> is totally ignored: func<Args...>(Args... args) is interpreted as func(Args... args). Beware that a function parameter with either a & or * operator, e.g. 'const Args&... args', requires \link and \ref to specify such parameter as verbatim, i.e. 'const Args&... args'. At the time of writing, the form func(const Args&...) will fail, unless the function parameter was declared just as 'const Args&...'. variadic template method Links to the variadic template function overloads: First overloadSecond overloadThird overloadFourth overloadFifth overloadSixth overloadSeventh overloadEighth overloadNinth overload The following are interpreted the same: without template argumentwith template argument See the test class.