Strange output of std::typeid::name()

Question

I used typeid to get the type names of the std::vector::size_type and a zero sized class A with the following code (cppreference):

#include<iostream>
#include <vector>
#include <typeinfo>

using namespace std;

class A {};

int main()
{
    vector<int> v(10); 

    vector<int>::size_type s = v.size(); 

    A a; 

    cout << typeid(s).name() << endl;
    cout << typeid(a).name() << endl;

};

And I got this as output:

m
1A

I guess that "1" before "A" is a result of the Empty Base Class Optimization, but what does "m" stand for and is this normal?

I am using the following gcc version: g++ (Ubuntu 4.4.3-4ubuntu5.1) 4.4.3

The names returned by `type_info::name` are implementation defined. Gcc chose `m` for `size_type`. Beats me :) — jrok, May 06 '13 at 10:03
You might want to check a reference on the [`std::type_info::name`](http://en.cppreference.com/w/cpp/types/type_info/name) function. — Some programmer dude, May 06 '13 at 10:07
@JoachimPileborg I did, but I guess I was expecting something better, than a mere "m" character. :) — tmaric, May 06 '13 at 10:10
The "1" before the "A" is because that's the length of the name `A`. — Mike Seymour, May 06 '13 at 10:39
Possible duplicate of [human-readable type\_info.name()](http://stackoverflow.com/questions/12877521/human-readable-type-info-name) — Ciro Santilli OurBigBook.com, Jul 22 '16 at 22:03

score 22 · Accepted Answer · answered May 06 '13 at 10:29

22

G++ uses implementation-defined naming for the types, but it also offers the utility c++filt to make them human-readable:

$ ./test | c++filt -t
unsigned long
A

answered May 06 '13 at 10:29

Cubbi

46,567
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So, the name "m" is then reserved and fixed for vector::size_type (for g++)? – tmaric May 06 '13 at 10:31
@tomislav-maric the name "m" stands for `unsigned long`, which happens to be typedef for `std::size_t`, which happens to be the typedef for `std::vector<>::size_type`. Typedefs do not create new types. – Cubbi May 06 '13 at 10:33
I know they don't create new types, I meant this: is "m" then for G++ a name for "unsigned_long", and it is defined somewhere (compiler implementation?), so c++filt looks it up there, and maps it back to "unsigned_long"? – tmaric May 06 '13 at 10:35
2

@tomislav-maric Yes, `/usr/bin/g++` and `/usr/bin/c++filt` are part of the same compiler implementation. – Cubbi May 06 '13 at 10:43

score 1 · Answer 2 · answered Feb 01 '22 at 20:57

The quick hack I use on gcc to check types is a templated function with a forwarding reference parameter, printing __PRETTY_FUNCTION__:

template<typename T>
void PrintType(T&&) {
    cout << __PRETTY_FUNCTION__ << endl;
}

This might print something like:

void PrintType(T&&) [with T = long unsigned int]

Strange output of std::typeid::name()

2 Answers2

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