C内联函数和“未定义的外部”错误

Question

我试图用内联函数替换一些宏子程序，因此编译器可以优化它们，所以调试器可以进入它们等等。如果我将它们定义为普通函数工作原理：

void do_something(void) 
{ 
    blah; 
} 

void main(void) 
{ 
    do_something(); 
} 

，但如果我将它们定义为内联：

inline void do_something(void) 
{ 
    blah; 
} 

void main(void) 
{ 
    do_something(); 
} 

它说： “错误：未定义的外部”。那是什么意思？在暗处刺，我试过

static inline void do_something(void) 
{ 
    blah; 
} 

void main(void) 
{ 
    do_something(); 
} 

并没有更多的错误。函数定义和对函数的调用都在同一个.c文件中。

有人可以解释为什么一个工作，另一个不是？

（第二个相关的问题：我应该把inline函数，如果我想在一个以上的.c文件中使用它们？）

Answer 1

首先，编译器并不总是内联标记为inline的函数;例如，如果关闭所有优化，它可能不会内联它们。

当你定义一个内联函数

inline void do_something(void) 
{ 
    blah 
} 

，并使用该功能，即使在同一个文件，调用该函数由链接不是编译器解决，因为它是隐含“外部”。但是这个定义本身并没有提供该功能的外部定义。

如果包括在C文件，而无需inline

void do_something(void); 

的声明，可以看到inline定义，编译器将提供功能的外部定义，错误应该消失。

原因static inline作品是，它使得仅在该compilatioin单元可视功能，因此允许编译器来解决调用函数（和优化），并发出代码为编译单元内的功能。链接器不需要解析它，所以不需要外部定义。

最好的地方，把内联函数是在头文件，并声明他们static inline。这消除了对外部定义的任何需求，因此它解决了链接器问题。但是，这会导致编译器在使用它的每个编译单元中为该函数发出代码，因此可能导致代码膨胀。但由于该函数是内联的，因此它可能很小，所以这通常不是问题。

的另一种选择是定义其作为头extern inline，并且在一个C文件提供和extern声明而不inline改性剂。因此

GCC手册解释它：

By declaring a function inline, you can direct GCC to make calls to that function faster. One way GCC can achieve this is to integrate that function's code into the code for its callers. This makes execution faster by eliminating the function-call overhead; in addition, if any of the actual argument values are constant, their known values may permit simplifications at compile time so that not all of the inline function's code needs to be included. The effect on code size is less predictable; object code may be larger or smaller with function inlining, depending on the particular case. You can also direct GCC to try to integrate all "simple enough" functions into their callers with the option -finline-functions .

GCC implements three different semantics of declaring a function inline. One is available with -std=gnu89 or -fgnu89-inline or when gnu_inline attribute is present on all inline declarations, another when -std=c99 , -std=c1x , -std=gnu99 or -std=gnu1x (without -fgnu89-inline), and the third is used when compiling C++.

To declare a function inline, use the inline keyword in its declaration, like this:

static inline int 
inc (int *a) 
{ 
    return (*a)++; 
} 

If you are writing a header file to be included in ISO C90 programs, write __inline__ instead of inline .

The three types of inlining behave similarly in two important cases: when the inline keyword is used on a static function, like the example above, and when a function is first declared without using the inline keyword and then is defined with inline , like this:

extern int inc (int *a); 
inline int 
inc (int *a) 
{ 
    return (*a)++; 
} 

In both of these common cases, the program behaves the same as if you had not used the inline keyword, except for its speed.

When a function is both inline and static , if all calls to the function are integrated into the caller, and the function's address is never used, then the function's own assembler code is never referenced. In this case, GCC does not actually output assembler code for the function, unless you specify the option -fkeep-inline-functions . Some calls cannot be integrated for various reasons (in particular, calls that precede the function's definition cannot be integrated, and neither can recursive calls within the definition). If there is a nonintegrated call, then the function is compiled to assembler code as usual. The function must also be compiled as usual if the program refers to its address, because that can't be inlined.

Note that certain usages in a function definition can make it unsuitable for inline substitution. Among these usages are: use of varargs, use of alloca, use of variable sized data types , use of computed goto, use of nonlocal goto, and nested functions. Using -Winline will warn when a function marked inline could not be substituted, and will give the reason for the failure.

As required by ISO C++, GCC considers member functions defined within the body of a class to be marked inline even if they are not explicitly declared with the inline keyword. You can override this with -fno-default-inline .

GCC does not inline any functions when not optimizing unless you specify the always_inline attribute for the function, like this:

/* Prototype. */ 
inline void foo (const char) __attribute__((always_inline)); 

The remainder of this section is specific to GNU C90 inlining.

When an inline function is not static , then the compiler must assume that there may be calls from other source files; since a global symbol can be defined only once in any program, the function must not be defined in the other source files, so the calls therein cannot be integrated. Therefore, a non- static inline function is always compiled on its own in the usual fashion.

If you specify both inline and extern in the function definition, then the definition is used only for inlining. In no case is the function compiled on its own, not even if you refer to its address explicitly. Such an address becomes an external reference, as if you had only declared the function, and had not defined it.

This combination of inline and extern has almost the effect of a macro. The way to use it is to put a function definition in a header file with these keywords, and put another copy of the definition (lacking inline and extern) in a library file. The definition in the header file will cause most calls to the function to be inlined. If any uses of the function remain, they will refer to the single copy in the library.

Answer 2

你必须把它们放在一个头文件，如果你想从使用它们多个文件。

而对于链接器错误：函数的默认声明意味着它是“extern”，但由于它被内联，链接器可以找到编译器生成的符号存根，因此出现错误。

Answer 3

对于inline功能与C99工作（他们才来到那里进的语言），你就必须给定义在头文件

inline void do_something(void) 
{ 
    blah 
} 

和一个编译单元（又名.C ）你放置某种“实例化”

void do_something(void); 

没有inline。