在c++中，什么时候我不应该为函数/方法写关键字“内联”?

如果函数在头文件中声明，并在.cpp文件中定义，则不应编写关键字。

什么时候编译器不知道什么时候使一个函数/方法'内联'?

没有这种情况。编译器不能使函数内联。它所能做的就是内联对函数的部分或所有调用。如果它没有函数的代码，它就不能这样做(在这种情况下，链接器需要这样做，如果它能够这样做的话)。

当一个应用程序为一个函数/方法写“内联”时，它是否重要?

不，那完全不重要。

在开发和调试代码时，不要使用内联。这会使调试复杂化。

添加它们的主要原因是为了帮助优化生成的代码。通常，这是以增加的代码空间换取速度，但有时内联可以同时节省代码空间和执行时间。

在算法完成之前将这种思想扩展到性能优化是不成熟的优化。

F.5:如果一个函数非常小并且对时间要求很高，那么就内联声明它

原因:一些优化器在没有程序员提示的情况下很擅长内联，但不要依赖它。测量!在过去40年左右的时间里，我们一直被承诺在没有人类提示的情况下，编译器可以比人类更好地内联。我们还在等待。指定inline(在类定义中编写成员函数时显式或隐式)可以鼓励编译器更好地完成工作。

来源:https://isocpp.github.io/CppCoreGuidelines/CppCoreGuidelines.html Rf-inline

有关示例和异常，请访问源代码(参见上面)。

实际上，几乎从来没有。你所做的只是建议编译器将给定的函数内联(例如，替换对该函数的所有调用/w它的函数体)。当然，这不能保证:编译器可能会忽略该指令。

编译器通常会很好地检测和优化这样的事情。

我想用一个令人信服的例子来解释这篇文章中所有的伟大答案，以消除任何剩余的误解。

给定两个源文件，例如:

inline111.cpp: # include < iostream > 空白栏(); Inline fun() { 返回111; ｝ Int main() { std:: cout < <“inline111:有趣 () = " << 有趣的 () << ", & 有趣= " < < (void *)与娱乐; 酒吧(); ｝ inline222.cpp: # include < iostream > Inline fun() { 返回222; ｝ 空格条(){ std:: cout < <“inline222:有趣 () = " << 有趣的 () << ", & 有趣= " < < (void *)与娱乐; ｝

Case A: Compile: g++ -std=c++11 inline111.cpp inline222.cpp Output: inline111: fun() = 111, &fun = 0x4029a0 inline222: fun() = 111, &fun = 0x4029a0 Discussion: Even thou you ought to have identical definitions of your inline functions, C++ compiler does not flag it if that is not the case (actually, due to separate compilation it has no ways to check it). It is your own duty to ensure this! Linker does not complain about One Definition Rule, as fun() is declared as inline. However, because inline111.cpp is the first translation unit (which actually calls fun()) processed by compiler, the compiler instantiates fun() upon its first call-encounter in inline111.cpp. If compiler decides not to expand fun() upon its call from anywhere else in your program (e.g. from inline222.cpp), the call to fun() will always be linked to its instance produced from inline111.cpp (the call to fun() inside inline222.cpp may also produce an instance in that translation unit, but it will remain unlinked). Indeed, that is evident from the identical &fun = 0x4029a0 print-outs. Finally, despite the inline suggestion to the compiler to actually expand the one-liner fun(), it ignores your suggestion completely, which is clear because fun() = 111 in both of the lines.

Case B: Compile (notice reverse order): g++ -std=c++11 inline222.cpp inline111.cpp Output: inline111: fun() = 222, &fun = 0x402980 inline222: fun() = 222, &fun = 0x402980 Discussion: This case asserts what have been discussed in Case A. Notice an important point, that if you comment out the actual call to fun() in inline222.cpp (e.g. comment out cout-statement in inline222.cpp completely) then, despite the compilation order of your translation units, fun() will be instantiated upon it's first call encounter in inline111.cpp, resulting in print-out for Case B as inline111: fun() = 111, &fun = 0x402980.

Case C: Compile (notice -O2): g++ -std=c++11 -O2 inline222.cpp inline111.cpp or g++ -std=c++11 -O2 inline111.cpp inline222.cpp Output: inline111: fun() = 111, &fun = 0x402900 inline222: fun() = 222, &fun = 0x402900 Discussion: As is described here, -O2 optimization encourages compiler to actually expand the functions that can be inlined (Notice also that -fno-inline is default without optimization options). As is evident from the outprint here, the fun() has actually been inline expanded (according to its definition in that particular translation unit), resulting in two different fun() print-outs. Despite this, there is still only one globally linked instance of fun() (as required by the standard), as is evident from identical &fun print-out.

除非您正在编写一个库或有特殊的原因，否则您可以忘记内联，而是使用链接时间优化。它消除了函数定义必须在头文件中才能考虑跨编译单元进行内联的要求，这正是内联所允许的。

(但请参阅为什么不使用链接时间优化?)