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.

Inline关键字请求编译器用函数体替换函数调用，它首先计算表达式，然后传递。它减少了函数调用开销，因为不需要存储返回地址，函数参数也不需要堆栈内存。

使用时间:

提高绩效 减少呼叫开销。 因为它只是对编译器的请求，所以某些函数不会被内联 *大功能 有太多条件参数的函数 递归代码和带有循环的代码等等。

你要把它放在最开始，在返回类型之前。但是大多数编译器忽略了它。如果它是定义的，并且它的代码块较小，大多数编译器都认为它是内联的。

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.

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.

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。