C++中的const限定符导致的链接问题
问题
如下所示有两个cxx源文件, 分别定义与使用一个const变量. 将其一起编译时报错: undefined reference of 'meow::miao'.
[01:06:44] hansy@hansy:~/testcase$ cat 1.cc
namespace meow {
const int miao = 1;
}
[01:06:47] hansy@hansy:~/testcase$ cat 2.cc
namespace meow {
extern const int miao;
}
using namespace meow;
int main() {
return miao;
}
[01:06:49] hansy@hansy:~/testcase$
[01:06:52] hansy@hansy:~/testcase$ g++ 1.cc -c -o 1.o && g++ 2.cc -c -o 2.o && gcc 1.o 2.o
2.o：在函数‘main’中：
2.cc:(.text+0x6)：对‘meow::miao’未定义的引用
collect2: error: ld returned 1 exit status
原因
nm看下链接的符号, 发现两个文件中的符号并不一致, 怪不得linker找不到符号, 问题是c++filt打印的名字却是同一个.
[01:07:05] hansy@hansy:~/testcase$ nm 1.o | grep miao
0000000000000000 r _ZN4meowL4miaoE
[01:07:11] hansy@hansy:~/testcase$ nm 2.o | grep miao
U _ZN4meow4miaoE
[01:07:15] hansy@hansy:~/testcase$ c++filt _ZN4meowL4miaoE
meow::miao
[01:07:20] hansy@hansy:~/testcase$ c++filt _ZN4meow4miaoE
meow::miao
尝试去掉const重新编译发现结果发生变化:
[01:07:40] hansy@hansy:~/testcase$ g++ 1.cc -c -o 1.o && g++ 2.cc -c -o 2.o && gcc 1.o 2.o
[01:07:42] hansy@hansy:~/testcase$ nm 1.o | grep miao
0000000000000000 D _ZN4meow4miaoE
所以问题出在const上, 查了下c++ standard, 在3.5节找到了答案.
A name is said to have linkage when it might denote the same object, reference, function, type, template,
namespace or value as a name introduced by a declaration in another scope:
— When a name has external linkage , the entity it denotes can be referred to by names from scopes of
other translation units or from other scopes of the same translation unit.
— When a name has internal linkage , the entity it denotes can be referred to by names from other scopes
in the same translation unit.
— When a name has no linkage , the entity it denotes cannot be referred to by names from other scopes.
A name having namespace scope (3.3.6) has internal linkage if it is the name of
— a variable, function or function template that is explicitly declared static; or,
— a variable that is explicitly declared const or constexpr and neither explicitly declared extern nor
previously declared to have external linkage; or
— a data member of an anonymous union.
即在C++中const限定符会导致被声明const的变量在链接时仅作用于本编译单元内, 如果要使其生效则需要显示声明为extern或之前声明时已声明为外部链接(这和C标准有点不一样).
那么compiler是如何实现的呢? 从上面实验可以看到是通过name mangling实现的: 定义const变量的符号名比定义非const的变量多一个字符L, 而未定义的变量的符号都是不带L的.
关于name mangling的实现之前没有具体研究过, 参考wiki上的介绍说下自己的理解:
首先mangling的方式不止一种, 不同的mangling之间不能兼容(很明显不同的名字导致链接时找不到对应的符号). 其中最常见的两种mangling分别是用于Windows的Visual C++规范与用于linux的Itanium规范.
因为手边没有gcc源码, 就以llvm代码为例看下compiler是如何处理mangling来解决linkage visibility问题的.
在clang/lib/AST目录下有两个文件MicrosoftMangle.cpp与ItaniumMangle.cpp(分别对应两类mangling). 我们可以看下ItaniumMangle.cpp中CXXNameMangler::mangleUnqualifiedName()的实现:
switch (Name.getNameKind()) {
case DeclarationName::Identifier: {
const IdentifierInfo *II = Name.getAsIdentifierInfo();
if (II) {
// Match GCC's naming convention for internal linkage symbols, for
// symbols that are not actually visible outside of this TU. GCC
// distinguishes between internal and external linkage symbols in
// its mangling, to support cases like this that were valid C++ prior
// to DR426:
//
// void test() { extern void foo(); }
// static void foo();
//
// Don't bother with the L marker for names in anonymous namespaces; the
// 12_GLOBAL__N_1 mangling is quite sufficient there, and this better
// matches GCC anyway, because GCC does not treat anonymous namespaces as
// implying internal linkage.
if (ND && ND->getFormalLinkage() == InternalLinkage &&
!ND->isExternallyVisible() &&
getEffectiveDeclContext(ND)->isFileContext() &&
!ND->isInAnonymousNamespace())
Out << 'L';
auto *FD = dyn_cast<FunctionDecl>(ND);
bool IsRegCall = FD &&
FD->getType()->castAs<FunctionType>()->getCallConv() ==
clang::CC_X86RegCall;
if (IsRegCall)
mangleRegCallName(II);
else
mangleSourceName(II);
writeAbiTags(ND, AdditionalAbiTags);
break;
}
......
}
有趣的是注释已经解释了L的作用: 传统的Gcc通过mangling来判断外部链接的符号与内部链接的符号, 对于内部链接符号默认给名字加L处理. 而如果只声明不定义的符号默认是外部链接的(本文件内没有定义), 所以导致符号名不一致.
总结
好像没什么要总结的, 这么一个破问题研究了半天, 人wo生shi寂zai寞shi如tai雪cai啊le.
