Why does typeid.name() return weird characters using GCC and how to make it print unmangled names?

Question

How come when I run this main.cpp:

#include <iostream>
#include <typeinfo>

using namespace std;

struct Blah {};

int main() {
  cout << typeid(Blah).name() << endl;
  return 0;
}

By compiling it with GCC version 4.4.4:

g++ main.cpp

I get this:

4Blah

On Visual C++ 2008, I would get:

struct Blah

Is there a way to make it just print Blah or struct Blah?

Benedictus

possible duplicate of Print variable type in C++ – Gae

Possible duplicate of Unmangling the result of std::type_info::name – Britannic

Answer 1

The return of name is implementation defined : an implementation is not even required to return different strings for different types.

What you get from g++ is a decorated name, that you can "demangle" using the c++filt command or __cxa_demangle.

Histone

specifically c++filt -t – Sublet

Ugh. Goddamn it GCC. – Zoroastrian

Answer 2

The string returned is implementation defined.

What gcc is doing is returning the mangled name.
You can convert the mangled name into plain text with c++filt

> a.out | c++filt

Casey

And here's a example: assume the content of stdout is PFPFiiEiiE, we can use the shell command echo "PFPFiiEiiE" | c++filt, and the result would be int (* (*)(int, int))(int). – Huey

You might (e.g on Ubuntu) also need to add --types to demangle type endcodings: echo "PFPFiiEiiE" | c++filt --types – Edva

Answer 3

Is there a way to make it just print

Blah or struct Blah?

No. The result of std::typeinfo::name() is unspecified. It might even return the same string for all types (or, indeed, empty strings for all types) and the implementation would still be standard-conforming. You must not rely on its result. Really, the only thing I found it useful for was debugging.

Tell us what what you need it for. Often traits is what you use instead.

Scrupulous

in my program, I just want to log my struct name to a logfile. Following the above example, I could have typed something like: LOG("Blah");. But I want to account for possibly changing the struct name in the future, so I'm trying to do: LOG(typeinfo(Blah).name(); – Benedictus

@ShaChris23: Well, I guess then you'll just have to live with not all compilers doing it in the same way. Have a look, however, what the runtime lib emits if, instead of blah, you have a map of strings to vectors of lists of strings. That can very easily exceed any sensible limits for log file size. – Scrupulous

Answer 4

As others have said, the result here is implementation-defined, meaning that the implementation (i.e., the compiler toolchain) is free to define it how it wants, so long as it documents that somewhere.

From the C++ standard, section 18.5.1/1 [lib.type.info]:

The class type_info describes type information generated by the implementation. Objects of this class effectively store a pointer to a name for the type, and an encoded value suitable for comparing two types for equality or collating order. The names, encoding rule, and collating sequence for types are all unspecified and may differ between programs.

Furness

Answer 5

in 4Blah, 4 is the number of letters in your class name. For example if your class name is myEmptyClass then it would print 12myEmptyClass.

Persist

you are right bro. – Noma

Answer 6

typeid().name() is implementation dependent. It may even return empty string for every type. That would not be very useful implementation, but it would be valid.

Stringfellow

No. It may not return empty string for every type. The way the string is represented is application defined but according to C++ standard it must differ a type from others. With empty strings you can't do it. – Devault

@RodrigoGurgel No guarantees are given; in particular, the returned string can be identical for several types and change between invocations of the same program. – Brunn

Answer 7

Use ./a.out | c++filt --types for GCC.

For more information follow the links below.

https://sourceware.org/binutils/docs/binutils/c_002b_002bfilt.html#:~:text=The%20c%2B%2Bfilt%201,that%20they%20can%20be%20read.&text=If%20the%20name%20decodes%20into,the%20original%20word%20is%20output.

Genseric