C內聯函數和「未定義的外部」錯誤

Question

我試圖用內聯函數替換一些宏子程序，因此編譯器可以優化它們，所以調試器可以進入它們等等。如果我將它們定義爲普通函數工作原理：

void do_something(void) 
{ 
    blah; 
} 

void main(void) 
{ 
    do_something(); 
} 

，但如果我將它們定義爲內聯：

inline void do_something(void) 
{ 
    blah; 
} 

void main(void) 
{ 
    do_something(); 
} 

它說： 「錯誤：未定義的外部」。那是什麼意思？在暗處刺，我試過

static inline void do_something(void) 
{ 
    blah; 
} 

void main(void) 
{ 
    do_something(); 
} 

並沒有更多的錯誤。函數定義和對函數的調用都在同一個.c文件中。

有人可以解釋爲什麼一個工作，另一個不是？

（第二個相關的問題：我應該把inline函數，如果我想在一個以上的.c文件中使用它們？）

Answer 1

首先，編譯器並不總是內聯標記爲inline的函數;例如，如果關閉所有優化，它可能不會內聯它們。

當你定義一個內聯函數

inline void do_something(void) 
{ 
    blah 
} 

，並使用該功能，即使在同一個文件，調用該函數由鏈接不是編譯器解決，因爲它是隱含「外部」。但是這個定義本身並沒有提供該功能的外部定義。

如果包括在C文件，而無需inline

void do_something(void); 

的聲明，可以看到inline定義，編譯器將提供功能的外部定義，錯誤應該消失。

原因static inline作品是，它使得僅在該compilatioin單元可視功能，因此允許編譯器來解決調用函數（和優化），併發出代碼爲編譯單元內的功能。鏈接器不需要解析它，所以不需要外部定義。

最好的地方，把內聯函數是在頭文件，並聲明他們static inline。這消除了對外部定義的任何需求，因此它解決了鏈接器問題。但是，這會導致編譯器在使用它的每個編譯單元中爲該函數發出代碼，因此可能導致代碼膨脹。但由於該函數是內聯的，因此它可能很小，所以這通常不是問題。

的另一種選擇是定義其作爲頭extern inline，並且在一個C文件提供和extern聲明而不inline改性劑。因此

GCC手冊解釋它：

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

你必須把它們放在一個頭文件，如果你想從使用它們多個文件。

而對於鏈接器錯誤：函數的默認聲明意味着它是「extern」，但由於它被內聯，鏈接器可以找到編譯器生成的符號存根，因此出現錯誤。

Answer 3

對於inline功能與C99工作（他們纔來到那裏進的語言），你就必須給定義在頭文件

inline void do_something(void) 
{ 
    blah 
} 

和一個編譯單元（又名.C ）你放置某種「實例化」

void do_something(void); 

沒有inline。