天啊，我最讨厌的事之一。

内联更像静态或extern，而不是告诉编译器内联函数的指令。Extern, static, inline是链接指令，几乎只由链接器使用，而不是编译器。

据说，内联提示编译器，你认为函数应该内联。这在1998年可能是正确的，但十年后，编译器不需要这样的提示。更不用说，当涉及到优化代码时，人类通常是错误的，所以大多数编译器会忽略“提示”。

static - the variable/function name cannot be used in other translation units. Linker needs to make sure it doesn't accidentally use a statically defined variable/function from another translation unit. extern - use this variable/function name in this translation unit but don't complain if it isn't defined. The linker will sort it out and make sure all the code that tried to use some extern symbol has its address. inline - this function will be defined in multiple translation units, don't worry about it. The linker needs to make sure all translation units use a single instance of the variable/function.

注意:通常，将模板声明为内联是没有意义的，因为它们已经具有内联的链接语义。但是，模板的显式专门化和实例化需要内联使用。

具体问题解答:

When should I write the keyword 'inline' for a function/method in C++? Only when you want the function to be defined in a header. More exactly only when the function's definition can show up in multiple translation units. It's a good idea to define small (as in one liner) functions in the header file as it gives the compiler more information to work with while optimizing your code. It also increases compilation time. When should I not write the keyword 'inline' for a function/method in C++? Don't add inline just because you think your code will run faster if the compiler inlines it. When will the compiler not know when to make a function/method 'inline'? Generally, the compiler will be able to do this better than you. However, the compiler doesn't have the option to inline code if it doesn't have the function definition. In maximally optimized code usually all private methods are inlined whether you ask for it or not. As an aside to prevent inlining in GCC, use __attribute__(( noinline )), and in Visual Studio, use __declspec(noinline). Does it matter if an application is multithreaded when one writes 'inline' for a function/method? Multithreading doesn't affect inlining in any way.

编译时，GCC默认不内联任何函数 启用优化。我不知道visual studio - deft_code

我检查了Visual Studio 9(15.00.30729.01)通过/FAcs编译并查看汇编代码: 编译器产生了对成员函数的调用，而没有在调试模式下启用优化。即使函数被标记为__forceinline，也不会产生内联运行时代码。

什么时候编译器不知道什么时候使一个函数/方法'内联'?

这取决于所使用的编译器。不要盲目相信现在的编译器比人类更了解如何内联，也不要因为性能原因而使用它，因为它是链接指令而不是优化提示。虽然我同意这些观点在意识形态上是正确的，但遇到现实可能是另一回事。

在阅读了多个线程之后，出于好奇，我尝试了内联对我正在工作的代码的影响，结果是我得到了GCC的可测量加速，而英特尔编译器的速度没有提高。

(更多细节:数学模拟与少数关键函数定义类之外，GCC 4.6.3 (g++ -O3)， ICC 13.1.0 (icpc -O3);将内联添加到临界点导致GCC代码加速6%)。

因此，如果你将GCC 4.6限定为现代编译器，那么如果你编写CPU密集型任务，并且知道瓶颈在哪里，内联指令仍然很重要。

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.

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。

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。