C++ 模板和内联

C++ template and inline(C++ 模板和内联)

问题描述

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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