3.2 Nontype Function Template Parameters
3.2 非类型函数模板参数
You can also define nontype parameters for function templates. For example, the following function template defines a group of functions for which a certain value can be added:
你也可以为函数模板定义非类型参数。例如，下面的函数模板定义了一组可以增加特定值的函数：
template<int Val, typename T> T addValue (T x) { return x + Val; }
These kinds of functions can be useful if functions or operations are used as parameters. For example, if you use the C++ standard library you can pass an instantiation of this function template to add a value to each element of a collection:
如果需要把函数或操作当作参数的话，那么这类函数就非常有用。例如，借助于C++标准库，你可以传递这个函数模板给集合中的每一个元素，让他们都增加一个整数值：
std::transform (source.begin(), //源的开始位置 source.end(), //源的结束位置 dest.begin(), //目标的开始位置 addValue<5,int>); //操作
The last argument instantiates the function template addValue<>() to add 5 to a passed int value. The resulting function is called for each element in the source collection source, while it is translated into the destination collection dest.
在上面的调用中，最后一个实参将addValue()函数模板实例化成：让int元素增加5。源集合(source)中的每一个元素都会调用实例化后addValue()函数，并把调用结果放入目标集合(dest)中。
Note that you have to specify the argument int for the template parameter T of addValue<>(). Deduction only works for immediate calls and std::transform() need a complete type to deduce the type of its fourth parameter. There is no support to substitute/deduce only some template parameters and the see, what could fit, and deduce the remaining parameters.
注意，必须指定addValue()的模板参数为int类型。推导只适用于立即调用，并且std::transform()需要一个完整的类型才能推导出其第4个参数的类型。不支持先部分替换或推导模板参数的类型，然后再根据具体情况去推导其余的模板参数。
Again, you can also specify that a template parameter is deduced from the previous parameter. For example, to derive the return type from the passed nontype:
同样，你也可以指定一个从上一个参数推导而来的模板参数。例如，要传递从非类型中派生出来的返回类型：
template<auto Val, typename T = decltype(Val)>
T foo();
or to ensure that the passed value has the same type as the passed type:
或者为了确保传递的值与其类型相同：
template<typename T, T Val = T{}>
T bar();
