与 C++一起使用的常见“解决方案”GetProcAddress是“extern“C”，但这会破坏重载。名称修改允许多个函数共存，只要它们的签名不同。但是有没有办法找到这些修改后的名称GetProcAddress？
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4 回答
7
VC++ 编译器知道它自己的名称修饰方案，那么为什么不使用它呢？在内部template<typename T> T GetProcAddress(HMODULE h, const char* name)，宏__FUNCDNAME__包含GetProcAddress. 这包括T部分。因此，在 内部GetProcAddress<void(*)(int)，我们有一个名称为 的子字符串void(*)(int)。从中，我们可以简单地推导出void foo(int);
此代码依赖于 VC++ 宏__FUNCDNAME__。对于MinGW ，您需要基于此。__PRETTY_FUNCTION__
FARPROC GetProcAddress_CppImpl(HMODULE h, const char* name, std::string const& Signature)
{
// The signature of T appears twice in the signature of T GetProcAddress<T>(HMODULE, const char*)
size_t len = Signature.find("@@YA");
std::string templateParam = Signature.substr(0, len);
std::string returnType = Signature.substr(len+4);
returnType.resize(templateParam.size()); // Strip off our own arguments (HMODULE and const char*)
assert(templateParam == returnType);
// templateParam and returnType are _pointers_ to functions (P6), so adjust to function type (Y)
std::string funName = "?" + std::string(name) + "@@Y" + templateParam.substr(2);
return ::GetProcAddress(h, funName.c_str());
}
template <typename T>
T GetProcAddress(HMODULE h, const char* name)
{
// Get our own signature. We use `const char* name` to keep it simple.
std::string Signature = __FUNCDNAME__ + 18; // Ignore prefix "??$GetProcAddress@"
return reinterpret_cast<T>(GetProcAddress_CppImpl(h, name, Signature));
}
// Showing the result
struct Dummy { };
__declspec(dllexport) void foo( const char* s)
{
std::cout << s;
}
__declspec(dllexport) void foo( int i, Dummy )
{
std::cout << "Overloaded foo(), got " << i << std::endl;
}
__declspec(dllexport) void foo( std::string const& s )
{
std::cout << "Overloaded foo(), got " << s << std::endl;
}
__declspec(dllexport) int foo( std::map<std::string, double> volatile& )
{
std::cout << "Overloaded foo(), complex type\n";
return 42;
}
int main()
{
HMODULE h = GetModuleHandleW(0);
foo("Hello, ");
auto pFoo1 = GetProcAddress<void (*)( const char*)>(h, "foo");
// This templated version of GetProcAddress is typesafe: You can't pass
// a float to pFoo1. That is a compile-time error.
pFoo1(" world\n");
auto pFoo2 = GetProcAddress<void (*)( int, Dummy )>(h, "foo");
pFoo2(42, Dummy()); // Again, typesafe.
auto pFoo3 = GetProcAddress<void (*)( std::string const& )>(h, "foo");
pFoo3("std::string overload\n");
auto pFoo4 = GetProcAddress<int (*)( std::map<std::string, double> volatile& )>(h, "foo");
// pFoo4 != NULL, this overload exists.
auto pFoo5 = GetProcAddress<void (*)( float )>(h, "foo");
// pFoo5==NULL - no such overload.
}
于 2013-04-15T13:47:10.953 回答
4
用于dumpbin /exports 'file.dll'获取所有符号的装饰/未装饰名称。
于 2013-04-15T13:46:48.903 回答
2
仅仅通过使用是不可能做到的GetProcAddress。但是，一种方法是枚举该特定模块的所有导出函数，并进行模式匹配以查找所有损坏的名称。
更具体地说，请在此处参考此答案。您需要做的唯一更改是将TRUEfor参数和forMappedAsImage参数的返回值传递给函数调用。GetModuleHandleBaseImageDirectoryEntryToData
void EnumerateExportedFunctions(HMODULE hModule, vector<string>& slListOfDllFunctions)
{
DWORD *dNameRVAs(0);
_IMAGE_EXPORT_DIRECTORY *ImageExportDirectory;
unsigned long cDirSize;
_LOADED_IMAGE LoadedImage;
string sName;
slListOfDllFunctions.clear();
ImageExportDirectory = (_IMAGE_EXPORT_DIRECTORY*)
ImageDirectoryEntryToData(hModule,
TRUE, IMAGE_DIRECTORY_ENTRY_EXPORT, &cDirSize);
if (ImageExportDirectory != NULL)
{
dNameRVAs = (DWORD *)ImageRvaToVa(LoadedImage.FileHeader,
LoadedImage.MappedAddress,
ImageExportDirectory->AddressOfNames, NULL);
for(size_t i = 0; i < ImageExportDirectory->NumberOfNames; i++)
{
sName = (char *)ImageRvaToVa(LoadedImage.FileHeader,
LoadedImage.MappedAddress,
dNameRVAs[i], NULL);
slListOfDllFunctions.push_back(sName);
}
}
}
于 2013-04-15T13:51:06.470 回答
0
我不明白为什么你会想要/需要MSalters 解决方案的 constexpr 版本，但它就是这样，完成了命名空间修改。用于
using F = int(double*);
constexpr auto f = mangled::name<F>([]{ return "foo::bar::frobnicate"; });
constexpr const char* cstr = f.data();
其中F是函数签名，foo::bar::frobnicate是函数的（可能是限定的）名称。
#include<string_view>
#include<array>
namespace mangled
{
namespace detail
{
template<typename F>
inline constexpr std::string_view suffix()
{
auto str = std::string_view(__FUNCDNAME__);
return str.substr(14, str.size() - 87);
}
template<typename L>
struct constexpr_string
{
constexpr constexpr_string(L) {}
static constexpr std::string_view data = L{}();
};
template<typename Name>
inline constexpr int qualifiers()
{
int i = -2, count = -1;
while(i != std::string_view::npos)
{
i = Name::data.find("::", i + 2);
count++;
}
return count;
}
template<typename Name>
inline constexpr auto split()
{
std::array<std::string_view, qualifiers<Name>() + 1> arr = {};
int prev = -2;
for(int i = arr.size() - 1; i > 0; i--)
{
int cur = Name::data.find("::", prev + 2);
arr[i] = Name::data.substr(prev + 2, cur - prev - 2);
prev = cur;
}
arr[0] = Name::data.substr(prev + 2);
return arr;
}
template<typename F, typename Name>
struct name_builder
{
static constexpr auto suf = detail::suffix<F>();
static constexpr auto toks = split<Name>();
static constexpr auto len = Name::data.size() + suf.size() - toks.size() + 6;
static constexpr auto str = [] {
std::array<char, len> arr = {};
arr[0] = '?';
int i = 1;
for(int t = 0; t < toks.size(); t++)
{
if(t > 0)
arr[i++] = '@';
for(auto c : toks[t])
arr[i++] = c;
}
arr[i++] = '@';
arr[i++] = '@';
arr[i++] = 'Y';
for(auto c : suf)
arr[i++] = c;
return arr;
}();
};
}
template<typename F, typename LambdaString>
inline constexpr std::string_view name(LambdaString)
{
using Cs = detail::constexpr_string<LambdaString>;
using N = detail::name_builder<F, Cs>;
return {N::str.data(), N::len};
}
}
于 2020-09-26T11:26:06.373 回答