1)如今，几乎从来没有。如果内联一个函数是个好主意，编译器会在没有你帮助的情况下完成它。

2)。看到# 1。

(经过编辑，反映出你把问题分成了两个问题……)

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.

一个用例可能发生在继承上。例如，如果以下所有情况都为真:

你有某个类的基类 基类需要是抽象的 基类除了析构函数之外没有纯虚方法 您不希望为基类创建CPP文件，因为这是徒劳的

然后你必须定义析构函数;否则，你会有一些未定义的引用链接错误。此外，你不仅要定义，还要用inline关键字定义析构函数;否则，您将有多个定义链接错误。

这可能发生在一些只包含静态方法或编写基异常类的辅助类上。

让我们举个例子:

Base.h:

class Base {
public:
    Base(SomeElementType someElement) noexcept : _someElement(std::move(someElement)) {}

    virtual ~Base() = 0;

protected:
    SomeElementType _someElement;
}

inline Base::~Base() = default;

Derived1.h:

#include "Base.h"

class Derived1 : public Base {
public:
    Derived1(SomeElementType someElement) noexcept : Base(std::move(someElement)) {}

    void DoSomething1() const;
}

Derived1.cpp:

#include "Derived1.h"

void Derived1::DoSomething1() const {
    // use _someElement 
}

Derived2.h:

#include "Base.h"

class Derived2 : public Base {
public:
    Derived2(SomeElementType someElement) noexcept : Base(std::move(someElement)) {}

    void DoSomething2() const;
}

Derived2.cpp:

#include "Derived2.h"

void Derived2::DoSomething2() const {
    // use _someElement 
}

通常，抽象类有一些纯虚方法，而不是构造函数或析构函数。因此，你不必分离基类的虚析构函数的声明和定义，你可以只写virtual ~ base () = default;关于类声明。然而，在我们的案例中，情况并非如此。

据我所知，MSVC允许你在类声明上写这样的东西:virtual ~Base() = 0{}。所以你不需要用内联关键字分离声明和定义。但它将只与MSVC编译器工作。

现实世界的例子:

BaseException.h:

#pragma once

#include <string>

class BaseException : public std::exception {
public:
    BaseException(std::string message) noexcept : message(std::move(message)) {}
    virtual char const* what() const noexcept { return message.c_str(); }

    virtual ~BaseException() = 0;

private:
    std::string message;
};

inline BaseException::~BaseException() = default;

SomeException.h:

#pragma once

#include "BaseException.h"

class SomeException : public BaseException {
public:
    SomeException(std::string message) noexcept : BaseException(std::move(message)) {}
};

SomeOtherException.h:

#pragma once

#include "BaseException.h"

class SomeOtherException : public BaseException {
public:
    SomeOtherException(std::string message) noexcept : BaseException(std::move(message)) {}
};

main.cpp:

#include <SomeException.h>
#include <SomeOtherException.h>

#include <iostream>

using namespace std;

static int DoSomething(int argc) {
    try {
        switch (argc) {
        case 0:
            throw SomeException("some");
        case 1:
            throw SomeOtherException("some other");
        default:
            return 0;
        }
    }
    catch (const exception& ex) {
        cout << ex.what() << endl;
        return 1;
    }
}

int main(int argc, char**) {
    return DoSomething(argc);
}

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。

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?)