C++ 模板和内联
当我编写一个简单的(非模板)类时，如果函数实现是就地"提供的，它会自动被视为inline.
When I'm writing a simple (non-template) class, if the function implementation is provided "right in place", it's automatically treated as inline.
class A {
void InlinedFunction() { int a = 0; }
// ^^^^ the same as 'inline void InlinedFunction'
}
当谈论基于模板的类时，这条规则怎么样?
What about this rule when talking about template-based classes?
template <typename T> class B {
void DontKnowFunction() { T a = 0; }
// Will this function be treated as inline when the compiler
// instantiates the template?
};
另外，inline 规则是如何应用于非嵌套模板函数的，比如
Also, how is the inline rule applied to non-nested template functions, like
template <typename T> void B::DontKnowFunction() { T a = 0; }
template <typename T> inline void B::DontKnowFunction() { T a = 0; }
这里第一种和第二种情况会发生什么?
What would happen in the first and in the second case here?
谢谢.
推荐答案
据我所知，模板化函数是自动内联的.然而，现实是大多数现代编译器经常忽略内联限定符.在选择要内联的函数方面，编译器的优化启发式方法很可能比人类程序员做得更好.
Templated functions as far as I know are automatically inline. However, the reality is that most modern compilers regularly ignore the inline qualifier. The compiler's optimizing heuristics will most likely do a far better job of choosing which functions to inline than a human programmer.
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