C ++模板和内联 [英] C++ template and inline

问题描述

当我写一个简单的（非模板）类时，如果函数实现是就地提供的，它会被自动处理为 inline p>

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，这个规则怎么样？

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?
};

此外，如何应用 inline 到非嵌套模板函数，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; }

在第一种情况和第二种情况下会发生什么？

What would happen in the first and in the second case here?

谢谢。

推荐答案

我知道的模板函数自动内联。然而，现实是大多数现代编译器经常忽略内联限定符。编译器的优化启发式方法很可能比人类程序员更好地选择内联函数。

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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