c++11: trailing return type in functions（函数返回类型后置）

In C++03, the return type of a function template cannot be generalized if the return type relies on those of the template arguments. Here is an example, mul(T a, T b) is a function template that calculates the product of a and b. Arguments a and b are of an arbitrary type T, which is not decided until the template instantiation. Thus, we cannot declare a return type for mul to generalize all the cases for a*b. If we must define such a function template, we have to introduce another template parameter as follows:In C++03, the return type of a func