__cxa_demangle在开发工具集rhel6 (centos6)上失败-4 gcc-5.2

Question

我已经尝试过最小的测试用例。此病例以devtoolset-4 (gcc-5.2)的方式传递，而在流变-6下失败.Status -2表示"mangled_name不是C++ ABI损坏规则下的有效名称“https://gcc.gnu.org/onlinedocs/libstdc++/libstdc++-html-USERS-4.3/a01696.html。

如果我想在RHEL-6上使用gcc-5.2，我是做错了吗?还是说这是个bug？如果有，有什么建议吗？到目前为止，我最好的想法是在将"IJ“这个错误的名字重新定义为"II”之后，再将其交给__cxa_demangle()。如果这可能是相对可靠的，我可以接受它。

#include <iostream>
#include <string>
#include <tuple>
#include <typeinfo>
#include <cxxabi.h>

#define DUMP(x) std::cout << #x << " is " << x << std::endl

template<typename T>
void print_type(T t)
{
    int status;
    auto name = typeid(t).name();
    DUMP(name);
    auto thing2 = abi::__cxa_demangle(name, 0, 0, &status);
    if(status == 0)
    {
        DUMP(thing2);
    }
    else
    {
        std::cout << typeid(t).name() << " failed to demangle\n";
        DUMP(status);
    }
}

typedef std::tuple<foo_t, foo_t> b_t;

int main(int argc, char **argv)
{
    std::tuple<bool, bool> t;
    print_type(t);
}

Centos-6输出

name is St5tupleIJbbEE
St5tupleIJbbEE failed to demangle
status is -2

Centos-7输出

name is St5tupleIJbbEE
thing2 is std::tuple<bool, bool>

Centos-6输出与devtoolset-3 (gcc-4.9)

name is St5tupleIIbbEE
thing2 is std::tuple<bool, bool>

Answer 1

std::string typestring(typeid(T).name());
auto pos = typestring.find("IJ");
if(pos != std::string::npos)
{
    // gcc-5.2 uses IJ in typestring where
    // gcc-4.9 used II - this fugly hack makes
    // cxa_demangle work on centos/rhel-6 with gcc-5.2
    // eg std::tuple<bool, bool> 
    typestring[pos + 1] = 'I';
}
rv = abi::__cxa_demangle(typestring.c_str(), 0, 0, &status);