Q: C++ determine the type of auto and pass to function template I have a function :
template<typename T> f(T x) { do something with x; }

I want to pass this auto pointer into the function
auto x = ...
f<???>(x)

Is there anyway for me to do so ?

A: Just call it like
auto x = ...
f(x)

templated functions automatically deduce the type depending on the arguments you pass it. In fact that's the preferred way to call a templated function.  
If you really want to explicitly give it the type (I don't recommend doing that) you can use decltype for it:
auto x = ...
f<decltype(x)>(x)


Here a minimal proof: http://coliru.stacked-crooked.com/a/d01070d90c0b9803

A: compiler should be smart enough to figure out the type, so 
auto x = ...
f(x);

or you can use decltype
auto x = ...
f<decltype(x)>(x);

