如何解开C ++ lambdas的名称? [英] How to unmangle mangled names of C++ lambdas?
问题描述
使用 g ++ - 4.9.3 -std = c ++ 11
编译
#include <iostream>
#include <typeinfo>
using namespace std;
int main() { cout << typeid([]{}).name() << endl; }
输出 Z4mainEUlvE _
Linux x86_64上的给定lambda。但是, c ++ filt
工具无法取消它。它只是输出给它的输入, Z4mainEUlvE _
。
outputs Z4mainEUlvE_
as the mangled name of the given lambda on Linux x86_64. However, the c++filt
tool is unable to unmangle it. It just outputs the input given to it, Z4mainEUlvE_
.
如何解开它?
推荐答案
您可以使用GCC的 abi :: __ cxa_demangle
函数:
You can use GCC's special abi::__cxa_demangle
function:
#include <memory>
#include <cstdlib>
#include <cxxabi.h>
#include <iostream>
// delete malloc'd memory
struct malloc_deleter
{
void operator()(void* p) const { std::free(p); }
};
// custom smart pointer for c-style strings allocated with std::malloc
using cstring_uptr = std::unique_ptr<char, malloc_deleter>;
int main()
{
// special function to de-mangle names
int error;
cstring_uptr name(abi::__cxa_demangle(typeid([]{}).name(), 0, 0, &error));
if(!error)
std::cout << name.get() << '\n';
else if(error == -1)
std::cerr << "memory allocation failed" << '\n';
else if(error == -2)
std::cerr << "not a valid mangled name" << '\n';
else if(error == -3)
std::cerr << "bad argument" << '\n';
}
输出:
main::{lambda()#1}
b $ b
根据文档此函数返回使用零终止字符串> std :: malloc ,调用程序需要释放 std: :免费。此示例使用一个智能指针在范围结束时自动释放返回的字符串。
According to The Documentation this function returns a c-style zero-terminated string allocated using std::malloc which the caller needs to free using std::free. This example uses a smart pointer to free the returned string automatically at the end of the scope.
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