Linux User NameSpaces - linux

Question

I am experimenting with user namespaces using Go on Linux. The thing that I cannot figure out is that although am setting the uid and gid mappings when creating the namespace it still identifies as the nobody user when I launch the binary using sudo but when I launch it using the normal user everything works fine. For reference please see my code below
...
cmd := exec.Command("/bin/sh")
cmd.Stdout = os.Stdout
cmd.Stdin = os.Stdin
cmd.Stderr = os.Stderr
cmd.SysProcAttr = &syscall.SysProcAttr{
Cloneflags: syscall.CLONE_NEWUSER,
UidMappings: []syscall.SysProcIDMap{
{
ContainerID: 0,
HostID: 1000,
Size: 1,
},
},
GidMappings: []syscall.SysProcIDMap{
{
ContainerID: 0,
HostID: 1000,
Size: 1,
},
},
}
cmd.Run()
....
...
From the host I can confirm that indeed the user and group mappings were successful. The current pid is 87751
sudo cat /proc/87751/uid_map
0 1000 1
sudo cat /proc/87751/gid_map
0 1000 1
But when I run the binary after building
go build -o user_n
sudo ./user_n
sh-5.0$ whoami
nobody
sh-5.0$ id
uid=65534(nobody) gid=65534(nobody) groups=65534(nobody) context=unconfined_u:unconfined_r:unconfined_t:s0-s0:c0.c1023
But when I run the binary using the normal user it works as expected
./user_n
sh-5.0# whoami
root
sh-5.0# id
uid=0(root) gid=0(root) groups=0(root),65534(nobody) context=unconfined_u:unconfined_r:unconfined_t:s0-s0:c0.c1023
While running the binary using the normal user is an option I would like to know why running using sudo does not give the expected results. Any pointers will be greatly appreciated.
More info
Fedora 31
Kernel 5.3.11-100.fc29.x86_64
go version go1.14.3 linux/amd64

Answer 1

In the first case, you are running as root user (through sudo) for which there is no mapping specified in the child user namespace. Hence, the resulting "nobody" id.
In the second case, you run the program as user id 1000 for which the mapping says : 1000 becomes root in the child user namespace. Hence, the resulting "root" id.