Why can't C++ deduce template type from assignment?

Question

int x = fromString(\"test\") :could not deduce template argument for \'ValueType\'

int x = fromString(\"test\") : works

Answer 1

The return type of a function is dependent on overload resolution, not the other way around.

There is a trick that works though: operator= usually exists only for equal LHS/RHS argument types, except when an explicit operator= is defined (whether as standalone or as a member does not matter).

Thus, overload resolution will find operator=(int &, int), and see if the return value from your function is convertible to int. If you return a temporary that has an operator int, this is an acceptable resolution (even if the operator int is in the generic form of a template operator T).

Thus:

template
U convert_impl(T const &t);

template
struct convert_result {
    convert_result(T const &t) : t(t) { }
    template operator U(void) const { return convert_impl(t); }
    T const &t;
};

template
convert_result convert(T const &t) { return t; }