什么是未定义的引用/无法parsing的外部符号错误，我该如何解决？

什么是未定义的参考/未解决的外部符号错误？什么是常见原因以及如何修复/防止它们？

_{随意编辑/添加自己的。}

编译一个C ++程序需要几个步骤，如2.2 所指（Keith Thompson提供的参考文献）：

翻译的语法规则的优先顺序由以下几个阶段规定[见脚注] 。

如果需要，物理源文件字符以实现定义的方式映射到基本源字符集（为行尾指示符引入新行字符）。 [SNIP]

每个反斜线字符（\）紧接着一个换行符的实例都被删除，拼接物理源代码行以形成逻辑源代码行。 [SNIP]

源文件被分解成预处理令牌（2.5）和空白字符序列（包括注释）。 [SNIP]

执行预处理指令，扩展macros调用，并执行_Pragma一元运算符expression式。 [SNIP]

字符文字或string文字中的每个源字符集成员，以及字符文字或非原始string文字中的每个转义序列和通用字符名称被转换为执行字符集的对应成员; [SNIP]

相邻的string文字标记是连接的。

分隔令牌的空白字符不再重要。每个预处理令牌都被转换成令牌。（2.7）。由此产生的标记语法和语义分析翻译为翻译单元。 [SNIP]

翻译的翻译单元和实例化单元组合如下： [SNIP]

所有外部实体引用都被parsing。 链接库组件以满足对当前翻译中未定义的实体的外部引用。 所有这样的翻译器输出被收集到一个程序图像中，该程序图像包含执行环境中执行所需的信息。 （重点是我的）

[脚注]实现必须performance得好像这些分离的阶段一样，尽pipe在实践中不同的阶段可能会被折叠在一起。

指定的错误发生在编译的最后阶段，通常称为链接。它基本上意味着你编译了一堆实现文件到目标文件或库中，现在你想让它们一起工作。

假设你在a.cpp定义了符号a 。现在， b.cpp 声明这个符号并使用它。在链接之前，它只是假设那个符号是在某个地方定义的，但它并不关心在哪里。链接阶段负责find符号，并将其正确地链接到b.cpp （实际上是对象或使用它的库）。

如果您使用MSVS，您会看到项目生成.lib文件。这些包含一个导出符号表和一个导入符号表。导入的符号是针对链接的库进行parsing的，导出的符号是为使用.lib （如果有的话）的库提供的。

其他编译器/平台也有类似的机制。

常见错误消息是error LNK2001 error LNK1120 error LNK2019 ， MSVS error LNK2019和undefined reference to symbolName的gcc undefined reference to symbolName 。

代码：

 struct X { virtual void foo(); }; struct Y : X { void foo() {} }; struct A { virtual ~A() = 0; }; struct B: A { virtual ~B(){} }; extern int x; void foo(); int main() { x = 0; foo(); Y y; B b; }

会用gcc产生下列错误：

 /home/AbiSfw/ccvvuHoX.o: In function `main': prog.cpp:(.text+0x10): undefined reference to `x' prog.cpp:(.text+0x19): undefined reference to `foo()' prog.cpp:(.text+0x2d): undefined reference to `A::~A()' /home/AbiSfw/ccvvuHoX.o: In function `B::~B()': prog.cpp:(.text._ZN1BD1Ev[B::~B()]+0xb): undefined reference to `A::~A()' /home/AbiSfw/ccvvuHoX.o: In function `B::~B()': prog.cpp:(.text._ZN1BD0Ev[B::~B()]+0x12): undefined reference to `A::~A()' /home/AbiSfw/ccvvuHoX.o:(.rodata._ZTI1Y[typeinfo for Y]+0x8): undefined reference to `typeinfo for X' /home/AbiSfw/ccvvuHoX.o:(.rodata._ZTI1B[typeinfo for B]+0x8): undefined reference to `typeinfo for A' collect2: ld returned 1 exit status

和MSVS类似的错误：

 1>test2.obj : error LNK2001: unresolved external symbol "void __cdecl foo(void)" (?foo@@YAXXZ) 1>test2.obj : error LNK2001: unresolved external symbol "int x" (?x@@3HA) 1>test2.obj : error LNK2001: unresolved external symbol "public: virtual __thiscall A::~A(void)" (??1A@@UAE@XZ) 1>test2.obj : error LNK2001: unresolved external symbol "public: virtual void __thiscall X::foo(void)" (?foo@X@@UAEXXZ) 1>...\test2.exe : fatal error LNK1120: 4 unresolved externals

常见原因包括：

未能链接到适当的库/对象文件或编译实现文件
声明和未定义的variables或函数。
类types成员的常见问题
模板实现不可见。
符号在C程序中定义并用于C ++代码。
不正确地导入/导出模块/ DLL的方法/类。（特定于MSVS）
循环库依赖
未定义引用'WinMain @ 16'
相互依存的图书馆秩序
多个相同名称的源文件
在使用#pragma （MSVC）时不要使用.lib扩展名
模板朋友的问题
UNICODE定义不一致

class级成员：

一个纯粹的`virtual`析构函数需要一个实现。

声明一个析构函数仍然需要你定义它（不像常规函数）：

 struct X { virtual ~X() = 0; }; struct Y : X { ~Y() {} }; int main() { Y y; } //X::~X(){} //uncomment this line for successful definition

发生这种情况是因为在隐式地销毁对象时调用基类析构函数，所以需要定义。

`virtual`方法必须被实现或定义为纯粹的。

这与没有定义的非virtual方法类似，只是增加了一个推理，即纯声明会生成一个虚拟的vtable，并且可能会在不使用该函数的情况下获得链接器错误：

 struct X { virtual void foo(); }; struct Y : X { void foo() {} }; int main() { Y y; //linker error although there was no call to X::foo }

为此，将X::foo()声明为纯粹的：

 struct X { virtual void foo() = 0; };

非`virtual`类成员

即使不明确使用，也需要定义一些成员：

 struct A { ~A(); };

以下将产生错误：

 A a; //destructor undefined

实现可以在类定义本身内联：

 struct A { ~A() {} };

或外面：

 A::~A() {}

如果实现不在类定义之外，而是在一个头中，则必须将方法标记为inline以防止多重定义。

如果使用所有使用的成员方法，则需要定义。

一个常见的错误是忘记限定名称：

 struct A { void foo(); }; void foo() {} int main() { A a; a.foo(); }

定义应该是

 void A::foo() {}

`static`数据成员必须在单个翻译单元的类外定义：

 struct X { static int x; }; int main() { int x = X::x; } //int X::x; //uncomment this line to define X::x

可以为类定义中的整型或枚举types的static const数据成员提供初始化程序; 然而，这个成员的odr使用仍然需要一个名称空间作用域定义，如上所述。 C ++ 11允许在类内为所有static const数据成员进行初始化。

未能链接到适当的库/对象文件或编译实现文件

通常，每个翻译单元将生成一个包含该翻译单元中定义的符号定义的目标文件。要使用这些符号，您必须链接到这些对象文件。

在gcc下，你可以指定要在命令行中连接在一起的所有目标文件，或者一起编译实现文件。

 g++ -o test objectFile1.o objectFile2.o -lLibraryName

这里的libraryName只是库的名字，没有特定的平台。所以，例如，在Linux库文件通常被称为libfoo.so但你只能写-lfoo 。在Windows上，同一个文件可能被称为foo.lib ，但是你会使用相同的参数。您可能需要使用-L‹directory›添加可以find这些文件的-L‹directory› 。请确保在-l或-L之后不写入空格。

对于XCode ：添加用户头searchpath – >添加库searchpath – >将实际的库引用拖放到项目文件夹中。

在MSVS下 ，添加到项目的文件会自动将其目标文件链接在一起，并生成一个lib文件（常用）。要在单独的项目中使用这些符号，您需要在项目设置中包含lib文件。这是在项目属性的链接器部分的Input -> Additional Dependencies 。（ lib文件的path应该添加到Linker -> General -> Additional Library Directories ）当使用lib文件提供的第三方库时，如果不这样做通常会导致错误。

也可能发生忘记将文件添加到编译中的情况，在这种情况下将不会生成目标文件。在gcc中，你可以将这些文件添加到命令行中。在MSVS中，将文件添加到项目中会自动进行编译（尽pipe可以手动将文件从构build中单独排除）。

在Windows编程中，没有链接必要库的迹象表明，未parsing符号的名称以__imp_ 。在文档中查找函数的名称，并且应该说哪个库需要使用。例如，MSDN将信息放在每个函数底部的一个框中，称为“库”。

声明，但没有定义一个variables或函数。

一个典型的variables声明是

 extern int x;

由于这只是一个声明，所以需要一个单一的定义 。相应的定义是：

 int x;

例如，以下内容会产生一个错误：

 extern int x; int main() { x = 0; } //int x; // uncomment this line for successful definition

类似的说法适用于function。声明一个函数而不定义它会导致错误：

 void foo(); // declaration only int main() { foo(); } //void foo() {} //uncomment this line for successful definition

要小心你实现的function与你所声明的完全一致。例如，你可能有不匹配的cv-qualifiers：

 void foo(int& x); int main() { int x; foo(x); } void foo(const int& x) {} //different function, doesn't provide a definition //for void foo(int& x)

其他错配的例子包括

在另一个名字空间中定义的函数/variables。
作为类成员声明的函数/variables，定义为全局（反之亦然）。
函数返回types，参数号和types以及调用约定并不完全一致。

编译器的错误消息通常会给你声明但从未定义的variables或函数的完整声明。仔细比较你提供的定义。 确保每个细节都匹配。

相互依赖的链接库指定的顺序是错误的。

图书馆链接的顺序如果图书馆互相依赖的话，这一点很重要。通常，如果库A依赖于库B ，则libA 必须出现在链接器标志的libB之前。

例如：

 // Bh #ifndef B_H #define B_H struct B { B(int); int x; }; #endif // B.cpp #include "Bh" B::B(int xx) : x(xx) {} // Ah #include "Bh" struct A { A(int x); B b; }; // A.cpp #include "Ah" A::A(int x) : b(x) {} // main.cpp #include "Ah" int main() { A a(5); return 0; };

创build库：

 $ g++ -c A.cpp $ g++ -c B.cpp $ ar rvs libA.a Ao ar: creating libA.a a - Ao $ ar rvs libB.a Bo ar: creating libB.a a - Bo

编译：

 $ g++ main.cpp -L. -lB -lA ./libA.a(Ao): In function `A::A(int)': A.cpp:(.text+0x1c): undefined reference to `B::B(int)' collect2: error: ld returned 1 exit status $ g++ main.cpp -L. -lA -lB $ ./a.out

所以要重复一遍，顺序很重要！

如果一切都失败了，重新编译。

我最近刚刚通过重新编译有问题的文件，摆脱了Visual Studio 2012中无法parsing的外部错误。当我重build时，错误消失了。

这通常发生在两个（或更多）库具有循环依赖关系时。库A尝试使用B.lib中的符号，库B尝试使用A.lib中的符号。两者都不存在开始。当您尝试编译A时，链接步骤将失败，因为它无法findB.lib。 A.lib将生成，但没有DLL。然后编译B，它会成功并生成B.lib。重新编译A现在可以工作，因为现在可以findB.lib。

符号在C程序中定义并用于C ++代码。

函数（或variables） void foo()是在C程序中定义的，您可以尝试在C ++程序中使用它：

 void foo(); int main() { foo(); }

C ++链接器期望名称被修改，所以你必须声明这个函数为：

 extern "C" void foo(); int main() { foo(); }

等价地，不是在C程序中定义，函数（或variables） void foo()是在C ++中定义的，但是使用C链接：

 extern "C" void foo();

并尝试在C ++程序中使用它。

如果整个库被包含在头文件中（并被编译为C代码）; 包括将需要如下;

 extern "C" { #include "cheader.h" }

什么是“未定义的参考/未parsing的外部符号”

我将尝试解释什么是“未定义的引用/未parsing的外部符号”。

注意：我使用g ++和Linux，所有的例子都是为了它

比如我们有一些代码

 // src1.cpp void print(); static int local_var_name; // 'static' makes variable not visible for other modules int global_var_name = 123; int main() { print(); return 0; }

和

 // src2.cpp extern "C" int printf (const char*, ...); extern int global_var_name; //extern int local_var_name; void print () { // printf("%d%d\n", global_var_name, local_var_name); printf("%d\n", global_var_name); }

制作对象文件

 $ g++ -c src1.cpp -o src1.o $ g++ -c src2.cpp -o src2.o

在汇编阶段之后，我们有一个目标文件，其中包含任何要导出的符号。看符号

 $ readelf --symbols src1.o Num: Value Size Type Bind Vis Ndx Name 5: 0000000000000000 4 OBJECT LOCAL DEFAULT 4 _ZL14local_var_name # [1] 9: 0000000000000000 4 OBJECT GLOBAL DEFAULT 3 global_var_name # [2]

我拒绝了一些输出的内容，因为它们并不重要

所以，我们看到后面的符号出口。

 [1] - this is our static (local) variable (important - Bind has a type "LOCAL") [2] - this is our global variable

src2.cpp没有输出，我们也没有看到它的符号

链接我们的目标文件

 $ g++ src1.o src2.o -o prog

并运行它

 $ ./prog 123

链接器会看到导出的符号并将其链接。现在我们尝试取消注释src2.cpp中的行，就像这里一样

 // src2.cpp extern "C" int printf (const char*, ...); extern int global_var_name; extern int local_var_name; void print () { printf("%d%d\n", global_var_name, local_var_name); }

并重build一个对象文件

 $ g++ -c src2.cpp -o src2.o

OK（没有错误），因为我们只build立目标文件，连接还没有完成。尝试链接

 $ g++ src1.o src2.o -o prog src2.o: In function `print()': src2.cpp:(.text+0x6): undefined reference to `local_var_name' collect2: error: ld returned 1 exit status

发生这种情况是因为我们的local_var_name是静态的，即对其他模块不可见。现在更深入。获取翻译阶段输出

 $ g++ -S src1.cpp -o src1.s // src1.s look src1.s .file "src1.cpp" .local _ZL14local_var_name .comm _ZL14local_var_name,4,4 .globl global_var_name .data .align 4 .type global_var_name, @object .size global_var_name, 4 global_var_name: .long 123 .text .globl main .type main, @function main: ; assembler code, not interesting for us .LFE0: .size main, .-main .ident "GCC: (Ubuntu 4.8.2-19ubuntu1) 4.8.2" .section .note.GNU-stack,"",@progbits

所以，我们已经看到local_var_name没有标签，这就是为什么连接器没有find它。但是我们是黑客:)我们可以修复它。在文本编辑器中打开src1.s并更改

 .local _ZL14local_var_name .comm _ZL14local_var_name,4,4

至

  .globl local_var_name .data .align 4 .type local_var_name, @object .size local_var_name, 4 local_var_name: .long 456789

即你应该有如下

  .file "src1.cpp" .globl local_var_name .data .align 4 .type local_var_name, @object .size local_var_name, 4 local_var_name: .long 456789 .globl global_var_name .align 4 .type global_var_name, @object .size global_var_name, 4 global_var_name: .long 123 .text .globl main .type main, @function main: ; ...

我们已经改变了local_var_name的可见性，并将其值设置为456789.尝试从它build立一个对象文件

 $ g++ -c src1.s -o src2.o

好吧，看看readelf输出（符号）

 $ readelf --symbols src1.o 8: 0000000000000000 4 OBJECT GLOBAL DEFAULT 3 local_var_name

现在local_var_name具有绑定全局（是本地）

链接

 $ g++ src1.o src2.o -o prog

并运行它

 $ ./prog 123456789

好的，我们砍了:)

因此，当链接器无法在对象文件中find全局符号时，会发生“未定义的引用/未parsing的外部符号错误”。

错误地导入/导出方法/类跨模块/ DLL（编译器特定）。

MSVS要求您使用__declspec(dllexport)和__declspec(dllimport)指定要导出和导入哪些符号。

这个双重function通常是通过使用macros获得的：

 #ifdef THIS_MODULE #define DLLIMPEXP __declspec(dllexport) #else #define DLLIMPEXP __declspec(dllimport) #endif

macrosTHIS_MODULE只能在导出函数的模块中定义。这样，声明：

 DLLIMPEXP void foo();

扩展到

 __declspec(dllexport) void foo();

并告诉编译器导出函数，因为当前模块包含它的定义。将声明包含在不同的模块中时，会扩展到

 __declspec(dllimport) void foo();

并告诉编译器，这个定义是在你链接的库之一中的（参见1））。

你可以类似的导入/导出类：

 class DLLIMPEXP X { };

模板实现不可见。

未专用模板的定义必须对使用它们的所有翻译单元可见。这意味着您不能将模板的定义分离到实现文件。如果你必须分开实现，通常的解决方法是在你声明模板的头部末尾包含一个impl文件。常见的情况是：

 template<class T> struct X { void foo(); }; int main() { X<int> x; x.foo(); } //differentImplementationFile.cpp template<class T> void X<T>::foo() { }

为了解决这个问题，你必须将X::foo的定义移动到头文件或使用它的翻译单元可见的地方。

专门的模板可以在一个实现文件中实现，并且实现不必是可见的，但是专门化必须事先声明。

为了进一步解释和另一个可能的解决scheme（显式实例化）看到这个问题和答案。

这是每个VC ++程序员一次又一次看到的最混乱的错误信息之一。让我们先把事情弄清楚。

A.什么是符号？ 总之，一个符号是一个名字。它可以是variables名称，函数名称，类名称，typedef名称，或者除属于C ++语言的名称和符号以外的任何名称。它是用户定义的或由依赖库（另一个用户定义的）引入的。

B.什么是外部？ 在VC ++中，每个源文件（.cpp，.c等）都被认为是一个翻译单元，编译器一次编译一个单元，并为当前翻译单元生成一个目标文件（.obj）。（请注意，包含此源文件的每个头文件都将被预处理，并将被视为此翻译单元的一部分）翻译单元中的所有内容均被视为内部，其他所有内容均视为外部。在C ++中，可以使用extern ， __declspec (dllimport)等关键字引用外部符号。

C.什么是“决心”？ 解决是一个链接时间的术语。在链接时，链接器尝试为内部找不到定义的对象文件中的每个符号查找外部定义。这个search过程的范围包括：

编译时生成的所有对象文件
所有库（.lib）明确或隐式地指定为此build筑应用程序的附加依赖项。

这个search过程被称为parsing。

D.最后，为什么没有解决的外部符号？ 如果链接程序找不到内部没有定义的符号的外部定义，则会报告“未parsing的外部符号”错误。

E. LNK2019的可能原因：无法parsing的外部符号错误。我们已经知道这个错误是由于连接器没有find外部符号的定义，可能的原因可以分类为：

定义存在

例如，如果我们在a.cpp中定义了一个名为foo的函数：

 int foo() { return 0; }

在b.cpp中，我们要调用函数foo，所以我们添加

 void foo();

声明函数foo（），并在另一个函数体中调用它，比如bar() ：

 void bar() { foo(); }

现在，当你构build这个代码时，你会得到一个LNK2019错误，抱怨foo是一个未解决的符号。在这种情况下，我们知道foo（）在a.cpp中有其定义，但与我们所调用的不同（返回值不同）。这是定义存在的情况。

定义不存在

对WinMain@16或类似的“exception” main()入口点引用 （特别是对于visual-studio ）的未定义引用 。

您可能错过了用您的实际IDEselect正确的项目types。 IDE可能希望将Windows应用程序项目绑定到这样的入口函数（如上面缺less的引用所指定的那样），而不是常用的int main(int argc, char** argv); 签名。

如果您的IDE支持纯控制台项目 ，则可能需要select此项目types，而不是Windows应用程序项目。

另外，如果您使用第三方库，请确保您拥有正确的32/64位二进制文件

Microsoft提供了一个#pragma来在链接时引用正确的库;

 #pragma comment(lib, "libname.lib")

除了库path（包括库的目录）之外，这应该是库的全名。

Visual Studio NuGet包需要更新为新的工具集版本

我只是有这个问题试图链接libpng与Visual Studio 2013.问题是，包文件只有Visual Studio 2010和2012的库。

正确的解决scheme是希望开发人员发布一个更新的软件包然后升级，但是它通过在VS2013中额外的设置来攻击我，指向VS2012库文件。

我通过查找包名packagename\build\native\packagename.targets并在该文件内编辑包（在解决scheme目录内的packages文件夹中），复制所有v110部分。我在条件字段中将v110更改为v120 ，只是非常小心地将文件名path全部保留为v110 。这只是允许Visual Studio 2013链接到2012年的库，在这种情况下，它的工作。

假设你有一个用c ++编写的大项目，它有一千个.cpp文件和一千个.h文件。我们说这个项目也依赖于十个静态库。让我们说，我们在Windows上，我们在Visual Studio 20xx中build立我们的项目。当您按Ctrl + F7 Visual Studio开始编译整个解决scheme（假设我们在解决scheme中只有一个项目）

汇编的意义是什么？

Visual Studiosearch文件.vcxproj并开始编译每个扩展名为.cpp的文件。编译顺序是不确定的。所以你不能假定main.cpp文件是先编译的
如果.cpp文件依赖于附加的.h文件，以便查找文件.cpp中可能定义或可能不定义的符号
如果存在一个.cpp文件，其中编译器找不到一个符号，则编译器时间错误会引发该消息符号x找不到
对于扩展名为.cpp的每个文件都生成一个对象文件.o，并且Visual Studio将输出写入一个名为ProjectName.Cpp.Clean.txt的文件中，该文件包含必须由链接器处理的所有对象文件。

编译的第二步由Linker完成.Linker应该合并所有的目标文件并最终生成输出（可能是一个可执行文件或库）

链接项目的步骤

Parse all the object files and find the definition which was only declared in headers ( eg: The code of one method of a class as is mentioned in previous answers, or event the initialization of a static variable which is member inside a class)
If one symbol could not be found in object files he also is searched in Additional Libraries.For adding a new library to a project Configuration properties -> VC++ Directories -> Library Directories and here you specified additional folder for searching libraries and Configuration properties -> Linker -> Input for specifying the name of the library. -If the Linker could not find the symbol which you write in one .cpp he raises a linker time error which may sound like error LNK2001: unresolved external symbol "void __cdecl foo(void)" (?foo@@YAXXZ)

意见

Once the Linker find one symbol he doesn't search in other libraries for it
The order of linking libraries does matter .
If Linker finds an external symbol in one static library he includes the symbol in the output of the project.However, if the library is shared( dynamic ) he doesn't include the code ( symbols ) in output, but Run-Time crashes may occur

How To Solve this kind of error

Compiler Time Error :

Make sure you write your c++ project syntactical correct.

Linker Time Error

Define all your symbol which you declare in your header files
Use #pragma once for allowing compiler not to include one header if it was already included in the current .cpp which are compiled
Make sure that your external library doesn't contain symbols that may enter into conflict with other symbols you defined in your header files
When you use the template to make sure you include the definition of each template function in the header file for allowing the compiler to generate appropriate code for any instantiations.

A bug in the compiler/IDE

I recently had this problem, and it turned out it was a bug in Visual Studio Express 2013 . I had to remove a source file from the project and re-add it to overcome the bug.

Steps to try if you believe it could be a bug in compiler/IDE:

Clean the project (some IDEs have an option to do this, you can also manually do it by deleting the object files)
Try start a new project, copying all source code from the original one.

Linked .lib file is associated to a .dll

我遇到过同样的问题。 Say i have projects MyProject and TestProject. I had effectively linked the lib file for MyProject to the TestProject. However, this lib file was produced as the DLL for the MyProject was built. Also, I did not contain source code for all methods in the MyProject, but only access to the DLL's entry points.

To solve the issue, i built the MyProject as a LIB, and linked TestProject to this .lib file (i copy paste the generated .lib file into the TestProject folder). I can then build again MyProject as a DLL. It is compiling since the lib to which TestProject is linked does contain code for all methods in classes in MyProject.

Use the linker to help diagnose the error

Most modern linkers include a verbose option that prints out to varying degrees;

Link invocation (command line),
Data on what libraries are included in the link stage,
The location of the libraries,
Search paths used.

For gcc and clang; you would typically add -v -Wl,--verbose or -v -Wl,-v to the command line. More details can be found here;

Linux ld man page .
LLVM linker page .
"An introduction to GCC" chapter 9 .

For MSVC, /VERBOSE (in particular /VERBOSE:LIB ) is added to the link command line.

The MSDN page on the /VERBOSE linker option .

Since people seem to be directed to this question when it comes to linker errors I am going to add this here.

One possible reason for linker errors with GCC 5.2.0 is that a new libstdc++ library ABI is now chosen by default.

If you get linker errors about undefined references to symbols that involve types in the std::__cxx11 namespace or the tag [abi:cxx11] then it probably indicates that you are trying to link together object files that were compiled with different values for the _GLIBCXX_USE_CXX11_ABI macro. This commonly happens when linking to a third-party library that was compiled with an older version of GCC. If the third-party library cannot be rebuilt with the new ABI then you will need to recompile your code with the old ABI.

So if you suddenly get linker errors when switching to a GCC after 5.1.0 this would be a thing to check out.

A wrapper around GNU ld that doesn't support linker scripts

Some .so files are actually GNU ld linker scripts , eg libtbb.so file is an ASCII text file with this contents:

 INPUT (libtbb.so.2)

Some more complex builds may not support this. For example, if you include -v in the compiler options, you can see that the mainwin gcc wrapper mwdip discards linker script command files in the verbose output list of libraries to link in. A simple work around is to replace the linker script input command file with a copy of the file instead (or a symlink), eg

 cp libtbb.so.2 libtbb.so

Or you could replace the -l argument with the full path of the .so, eg instead of -ltbb do /home/foo/tbb-4.3/linux/lib/intel64/gcc4.4/libtbb.so.2

Befriending templates…

Given the code snippet of a template type with a friend operator (or function);

 template <typename T> class Foo { friend std::ostream& operator<< (std::ostream& os, const Foo<T>& a); };

The operator<< is being declared as a non-template function. For every type T used with Foo , there needs to be a non-templated operator<< . For example, if there is a type Foo<int> declared, then there must be an operator implementation as follows;

 std::ostream& operator<< (std::ostream& os, const Foo<int>& a) {/*...*/}

Since it is not implemented, the linker fails to find it and results in the error.

To correct this, you can declare a template operator before the Foo type and then declare as a friend, the appropriate instantiation. The syntax is a little awkward, but is looks as follows;

 // forward declare the Foo template <typename> class Foo; // forward declare the operator << template <typename T> std::ostream& operator<<(std::ostream&, const Foo<T>&); template <typename T> class Foo { friend std::ostream& operator<< <>(std::ostream& os, const Foo<T>& a); // note the required <> ^^^^ // ... }; template <typename T> std::ostream& operator<<(std::ostream&, const Foo<T>&) { // ... implement the operator }

The above code limits the friendship of the operator to the corresponding instantiation of Foo , ie the operator<< <int> instantiation is limited to access the private members of the instantiation of Foo<int> .

Alternatives include;

Allowing the friendship to extend to all instantiations of the templates, as follows;

 template <typename T> class Foo { template <typename T1> friend std::ostream& operator<<(std::ostream& os, const Foo<T1>& a); // ... };

Or, the implementation for the operator<< can be done inline inside the class definition;

 template <typename T> class Foo { friend std::ostream& operator<<(std::ostream& os, const Foo& a) { /*...*/ } // ... };

Note , when the declaration of the operator (or function) only appears in the class, the name is not available for "normal" lookup, only for argument dependent lookup, from cppreference ;

A name first declared in a friend declaration within class or class template X becomes a member of the innermost enclosing namespace of X, but is not accessible for lookup (except argument-dependent lookup that considers X) unless a matching declaration at the namespace scope is provided…

There is further reading on template friends at cppreference and the C++ FAQ .

Code listing showing the techniques above .

As a side note to the failing code sample; g++ warns about this as follows

warning: friend declaration 'std::ostream& operator<<(...)' declares a non-template function [-Wnon-template-friend]

note: (if this is not what you intended, make sure the function template has already been declared and add <> after the function name here)

Inconsistent `UNICODE` definitions

A Windows UNICODE build is built with TCHAR etc. being defined as wchar_t etc. When not building with UNICODE defined as build with TCHAR defined as char etc. These UNICODE and _UNICODE defines affect all the " T " string types ; LPTSTR , LPCTSTR and their elk.

Building one library with UNICODE defined and attempting to link it in a project where UNICODE is not defined will result in linker errors since there will be a mismatch in the definition of TCHAR ; char vs. wchar_t .

The error usually includes a function a value with a char or wchar_t derived type, these could include std::basic_string<> etc. as well. When browsing through the affected function in the code, there will often be a reference to TCHAR or std::basic_string<TCHAR> etc. This is a tell-tale sign that the code was originally intended for both a UNICODE and a Multi-Byte Character (or "narrow") build.

To correct this, build all the required libraries and projects with a consistent definition of UNICODE (and _UNICODE ).

This can be done with either;
```
 #define UNICODE #define _UNICODE 
```
Or in the project settings;

Project Properties > General > Project Defaults > Character Set
Or on the command line;
```
 /DUNICODE /D_UNICODE 
```

The alternative is applicable as well, if UNICODE is not intended to be used, make sure the defines are not set, and/or the multi-character setting is used in the projects and consistently applied.

Do not forget to be consistent between the "Release" and "Debug" builds as well.

Clean and rebuild

A "clean" of the build can remove the "dead wood" that may be left lying around from previous builds, failed builds, incomplete builds and other build system related build issues.

In general the IDE or build will include some form of "clean" function, but this may not be correctly configured (eg in a manual makefile) or may fail (eg the intermediate or resultant binaries are read-only).

Once the "clean" has completed, verify that the "clean" has succeeded and all the generated intermediate file (eg an automated makefile) have been successfully removed.

This process can be seen as a final resort, but is often a good first step ; especially if the code related to the error has recently been added (either locally or from the source repository).

Your linkage consumes libraries before the object files that refer to them

You are trying to compile and link your program with the GCC toolchain.
Your linkage specifies all of the necessary libraries and library search paths
If libfoo depends on libbar , then your linkage correctly puts libfoo before libbar .
Your linkage fails with undefined reference to something errors.
But all the undefined something s are declared in the header files you have #include d and are in fact defined in the libraries that you are linking.

Examples are in C. They could equally well be C++

A minimal example involving a static library you built yourself

my_lib.c

 #include "my_lib.h" #include <stdio.h> void hw(void) { puts("Hello World"); }

my_lib.h

 #ifndef MY_LIB_H #define MT_LIB_H extern void hw(void); #endif

eg1.c

 #include <my_lib.h> int main() { hw(); return 0; }

You build your static library:

 $ gcc -c -o my_lib.o my_lib.c $ ar rcs libmy_lib.a my_lib.o

You compile your program:

 $ gcc -I. -c -o eg1.o eg1.c

You try to link it with libmy_lib.a and fail:

 $ gcc -o eg1 -L. -lmy_lib eg1.o eg1.o: In function `main': eg1.c:(.text+0x5): undefined reference to `hw' collect2: error: ld returned 1 exit status

The same result if you compile and link in one step, like:

 $ gcc -o eg1 -I. -L. -lmy_lib eg1.c /tmp/ccQk1tvs.o: In function `main': eg1.c:(.text+0x5): undefined reference to `hw' collect2: error: ld returned 1 exit status

A minimal example involving a shared system library, the compression library `libz`

eg2.c

 #include <zlib.h> #include <stdio.h> int main() { printf("%s\n",zlibVersion()); return 0; }

Compile your program:

 $ gcc -c -o eg2.o eg2.c

Try to link your program with libz and fail:

 $ gcc -o eg2 -lz eg2.o eg2.o: In function `main': eg2.c:(.text+0x5): undefined reference to `zlibVersion' collect2: error: ld returned 1 exit status

Same if you compile and link in one go:

 $ gcc -o eg2 -I. -lz eg2.c /tmp/ccxCiGn7.o: In function `main': eg2.c:(.text+0x5): undefined reference to `zlibVersion' collect2: error: ld returned 1 exit status

And a variation on example 2 involving pkg-config :

 $ gcc -o eg2 $(pkg-config --libs zlib) eg2.o eg2.o: In function `main': eg2.c:(.text+0x5): undefined reference to `zlibVersion'

What are you doing wrong?

In the sequence of object files and libraries you want to link to make your program, you are placing the libraries before the object files that refer to them. You need to place the libraries after the object files that refer to them.

Link example 1 correctly:

 $ gcc -o eg1 eg1.o -L. -lmy_lib

Success:

 $ ./eg1 Hello World

Link example 2 correctly:

 $ gcc -o eg2 eg2.o -lz

Success:

 $ ./eg2 1.2.8

Link the example 2 pkg-config variation correctly:

 $ gcc -o eg2 eg2.o $(pkg-config --libs zlib) $ ./eg2 1.2.8

The explanation

Reading is optional from here on .

By default, a linkage command generated by GCC, on your distro, consumes the files in the linkage from left to right in commandline sequence. When it finds that a file refers to something and does not contain a definition for it, to will search for a definition in files further to the right. If it eventually finds a definition, the reference is resolved. If any references remain unresolved at the end, the linkage fails: the linker does not search backwards.

First, example 1 , with static library my_lib.a

A static library is an indexed archive of object files. When the linker finds -lmy_lib in the linkage sequence and figures out that this refers to the static library ./libmy_lib.a , it wants to know whether your program needs any of the object files in libmy_lib.a .

There is only object file in libmy_lib.a , namely my_lib.o , and there's only one thing defined in my_lib.o , namely the function hw .

The linker will decide that your program needs my_lib.o if and only if it already knows that your program refers to hw , in one or more of the object files it has already added to the program, and that none of the object files it has already added contains a definition for hw .

If that is true, then the linker will extract a copy of my_lib.o from the library and add it to your program. Then, your program contains a definition for hw , so its references to hw are resolved .

When you try to link the program like:

 $ gcc -o eg1 -L. -lmy_lib eg1.o

the linker has not added eg1.o to the program when it sees -lmy_lib . Because at that point, it has not seen eg1.o . Your program does not yet make any references to hw : it does not yet make any references at all , because all the references it makes are in eg1.o .

So the linker does not add my_lib.o to the program and has no further use for libmy_lib.a .

Next, it finds eg1.o , and adds it to be program. An object file in the linkage sequence is always added to the program. Now, the program makes a reference to hw , and does not contain a definition of hw ; but there is nothing left in the linkage sequence that could provide the missing definition. The reference to hw ends up unresolved , and the linkage fails.

Second, example 2 , with shared library libz

A shared library isn't an archive of object files or anything like it. It's much more like a program that doesn't have a main function and instead exposes multiple other symbols that it defines, so that other programs can use them at runtime.

Many Linux distros today configure their GCC toolchain so that its language drivers ( gcc , g++ , gfortran etc) instruct the system linker ( ld ) to link shared libraries on an as-needed basis. You have got one of those distros.

This means that when the linker finds -lz in the linkage sequence, and figures out that this refers to the shared library (say) /usr/lib/x86_64-linux-gnu/libz.so , it wants to know whether any references that it has added to your program that aren't yet defined have definitions that are exported by libz

If that is true, then the linker will not copy any chunks out of libz and add them to your program; instead, it will just doctor the code of your program so that:-

At runtime, the system program loader will load a copy of libz into the same process as your program whenever it loads a copy of your program, to run it.
At runtime, whenever your program refers to something that is defined in libz , that reference uses the definition exported by the copy of libz in the same process.

Your program wants to refer to just one thing that has a definition exported by libz , namely the function zlibVersion , which is referred to just once, in eg2.c . If the linker adds that reference to your program, and then finds the definition exported by libz , the reference is resolved

But when you try to link the program like:

 gcc -o eg2 -lz eg2.o

the order of events is wrong in just the same way as with example 1. At the point when the linker finds -lz , there are no references to anything in the program: they are all in eg2.o , which has not yet been seen. So the linker decides it has no use for libz . When it reaches eg2.o , adds it to the program, and then has undefined reference to zlibVersion , the linkage sequence is finished; that reference is unresolved, and the linkage fails.

Lastly, the pkg-config variation of example 2 has a now obvious explanation. After shell-expansion:

 gcc -o eg2 $(pkg-config --libs zlib) eg2.o

变为：

 gcc -o eg2 -lz eg2.o

which is just example 2 again.

I can reproduce the problem in example 1, but not in example 2

The linkage:

 gcc -o eg2 -lz eg2.o

works just fine for you!

(Or: That linkage worked fine for you on, say, Fedora 23, but fails on Ubuntu 16.04)

That's because the distro on which the linkage works is one of the ones that does not configure its GCC toolchain to link shared libraries as-needed .

Back in the day, it was normal for unix-like systems to link static and shared libraries by different rules. Static libraries in a linkage sequence were linked on the as-needed basis explained in example 1, but shared libraries were linked unconditionally.

This behaviour is economical at linktime because the linker doesn't have to ponder whether a shared library is needed by the program: if it's a shared library, link it. And most libraries in most linkages are shared libraries. But there are disadvantages too:-

It is uneconomical at runtime , because it can cause shared libraries to be loaded along with a program even if doesn't need them.
The different linkage rules for static and shared libraries can be confusing to inexpert programmers, who may not know whether -lfoo in their linkage is going to resolve to /some/where/libfoo.a or to /some/where/libfoo.so , and might not understand the difference between shared and static libraries anyway.

This trade-off has led to the schismatic situation today. Some distros have changed their GCC linkage rules for shared libraries so that the as-needed principle applies for all libraries. Some distros have stuck with the old way.

Why do I still get this problem even if I compile-and-link at the same time?

If I just do:

 $ gcc -o eg1 -I. -L. -lmy_lib eg1.c

surely gcc has to compile eg1.c first, and then link the resulting object file with libmy_lib.a . So how can it not know that object file is needed when it's doing the linking?

Because compiling and linking with a single command does not change the order of the linkage sequence.

When you run the command above, gcc figures out that you want compilation + linkage. So behind the scenes, it generates a compilation command, and runs it, then generates a linkage command, and runs it, as if you had run the two commands:

 $ gcc -I. -c -o eg1.o eg1.c $ gcc -o eg1 -L. -lmy_lib eg1.o

So the linkage fails just as it does if you do run those two commands. The only difference you notice in the failure is that gcc has generated a temporary object file in the compile + link case, because you're not telling it to use eg1.o . We see:

 /tmp/ccQk1tvs.o: In function `main'

代替：

 eg1.o: In function `main':

也可以看看

The order in which interdependent linked libraries are specified is wrong

Putting interdependent libraries in the wrong order is just one way in which you can get files that need definitions of things coming later in the linkage than the files that provide the definitions. Putting libraries before the object files that refer to them is another way of making the same mistake.

When your include paths are different

Linker errors can happen when a header file and its associated shared library (.lib file) go out of sync. 让我解释。

How do linkers work? The linker matches a function declaration (declared in the header) with its definition (in the shared library) by comparing their signatures. You can get a linker error if the linker doesn't find a function definition that matches perfectly.

Is it possible to still get a linker error even though the declaration and the definition seem to match? 是! They might look the same in source code, but it really depends on what the compiler sees. Essentially you could end up with a situation like this:

 // header1.h typedef int Number; void foo(Number); // header2.h typedef float Number; void foo(Number); // this only looks the same lexically

Note how even though both the function declarations look identical in source code, but they are really different according to the compiler.

You might ask how one ends up in a situation like that? Include paths of course! If when compiling the shared library, the include path leads to header1.h and you end up using header2.h in your own program, you'll be left scratching your header wondering what happened (pun intended).

An example of how this can happen in the real world is explained below.

Further elaboration with an example

I have two projects: graphics.lib and main.exe . Both projects depend on common_math.h . Suppose the library exports the following function:

 // graphics.lib #include "common_math.h" void draw(vec3 p) { ... } // vec3 comes from common_math.h

And then you go ahead and include the library in your own project.

 // main.exe #include "other/common_math.h" #include "graphics.h" int main() { draw(...); }

繁荣！ You get a linker error and you have no idea why it's failing. The reason is that the common library uses different versions of the same include common_math.h (I have made it obvious here in the example by including a different path, but it might not always be so obvious. Maybe the include path is different in the compiler settings).

Note in this example, the linker would tell you it couldn't find draw() , when in reality you know it obviously is being exported by the library. You could spend hours scratching your head wondering what went wrong. The thing is, the linker sees a different signature because the parameter types are slightly different. In the example, vec3 is a different type in both projects as far as the compiler is concerned. This could happen because they come from two slightly different include files (maybe the include files come from two different versions of the library).

Debugging the linker

DUMPBIN is your friend, if you are using Visual Studio. I'm sure other compilers have other similar tools.

过程如下所示：

Note the weird mangled name given in the linker error. (eg. draw@graphics@XYZ).
Dump the exported symbols from the library into a text file.
Search for the exported symbol of interest, and notice that the mangled name is different.
Pay attention to why the mangled names ended up different. You would be able to see that the parameter types are different, even though they look the same in the source code.
Reason why they are different. In the example given above, they are different because of different include files.

[1] By project I mean a set of source files that are linked together to produce either a library or an executable.

EDIT 1: Rewrote first section to be easier to understand. Please comment below to let me know if something else needs to be fixed. 谢谢！

Missing "extern" in `const` variable declarations/definitions (C++ only)

For people coming from C it might be a surprise that in C++ global const variables have internal (or static) linkage. In C this was not the case, as all global variables are implicitly extern (ie when the static keyword is missing).

例：

 // file1.cpp const int test = 5; // in C++ same as "static const int test = 5" int test2 = 5; // file2.cpp extern const int test; extern int test2; void foo() { int x = test; // linker error in C++ , no error in C int y = test2; // no problem }

correct would be to use a header file and include it in file2.cpp and file1.cpp

 extern const int test; extern int test2;

Alternatively one could declare the const variable in file1.cpp with explicit extern