¿De dónde toma Debian la umask predeterminada?

¿De dónde toma Debian la umask predeterminada?

Con un nuevo inicio de sesión ssh:

$ umask
0007

Pero:

$ find . -maxdepth 1 -name '.*' -type f | xargs grep 007 | less
$ 

# grep 007 /etc/profile
#
# grep -i umask /etc/login.defs
#       UMASK           Default "umask" value.
# UMASK is the default umask value for pam_umask and is used by
# 022 is the "historical" value in Debian for UMASK
UMASK           027
# Other former uses of this variable such as setting the umask when
# grep -i umask /etc/pam.d/common-session
session optional pam_umask.so usergroups

¿Qué carajo? Debianinventa¿Una umask arbitraria? ¡Y no hay otra fuente en /etc que coincida con 007!

El sistema utiliza LDAP (nscld, pam_ldap) para la autenticación

Más información: Ocurre solo para un usuario. No para root o un usuario diferente. Tan pronto como hago 'su' con un usuario, obtengo nuevamente 007.

Como raíz:

# strace -f -o basz.log su baduser

Algo está configurando esta umask pero no sé qué:

[...]
2622  open("/etc/group", O_RDONLY|O_CLOEXEC) = 4
2622  _llseek(4, 0, [0], SEEK_CUR)      = 0
2622  fstat64(4, {st_mode=S_IFREG|0644, st_size=1719, ...}) = 0
2622  mmap2(NULL, 1719, PROT_READ, MAP_SHARED, 4, 0) = 0xb7786000
2622  _llseek(4, 1719, [1719], SEEK_SET) = 0
2622  fstat64(4, {st_mode=S_IFREG|0644, st_size=1719, ...}) = 0
2622  munmap(0xb7786000, 1719)          = 0
2622  close(4)                          = 0
2622  socket(PF_FILE, SOCK_STREAM, 0)   = 4
2622  connect(4, {sa_family=AF_FILE, path="/var/run/nslcd/socket"}, 23) = 0
2622  gettimeofday({1404702848, 321946}, NULL) = 0
2622  gettimeofday({1404702848, 322029}, NULL) = 0
2622  poll([{fd=4, events=POLLOUT}], 1, 10000) = 1 ([{fd=4, revents=POLLOUT}])
2622  send(4, "\1\0\0\0\212\23\0\0\361\3\0\0", 12, MSG_NOSIGNAL) = 12
2622  gettimeofday({1404702848, 322363}, NULL) = 0
2622  gettimeofday({1404702848, 322464}, NULL) = 0
2622  poll([{fd=4, events=POLLIN}], 1, 60000) = 1 ([{fd=4, revents=POLLIN|POLLHUP}])
2622  read(4, "\1\0\0\0\212\23\0\0\0\0\0\0\4\0\0\0baduser\1\0\0\0*\361\3\0\0\2\0\0"..., 1024) = 57
2622  gettimeofday({1404702848, 323811}, NULL) = 0
2622  gettimeofday({1404702848, 323898}, NULL) = 0
2622  gettimeofday({1404702848, 323983}, NULL) = 0
2622  gettimeofday({1404702848, 324067}, NULL) = 0
2622  gettimeofday({1404702848, 324170}, NULL) = 0
2622  gettimeofday({1404702848, 324256}, NULL) = 0
2622  gettimeofday({1404702848, 324340}, NULL) = 0
2622  gettimeofday({1404702848, 324434}, NULL) = 0
2622  gettimeofday({1404702848, 324518}, NULL) = 0
2622  gettimeofday({1404702848, 324602}, NULL) = 0
2622  gettimeofday({1404702848, 324686}, NULL) = 0
2622  gettimeofday({1404702848, 324772}, NULL) = 0
2622  poll([{fd=4, events=POLLIN}], 1, 0) = 1 ([{fd=4, revents=POLLIN|POLLHUP}])
2622  read(4, "", 1024)                 = 0
2622  gettimeofday({1404702848, 325036}, NULL) = 0
2622  close(4)                          = 0
2622  umask(0777)                       = 027
2622  umask(07)                         = 0777
[...]

Más información:

  • Si escribo "umask 0027" en /etc/profile, ¡también funciona!
  • Si elimino el directorio de inicio y lo vuelvo a crear, el problema persiste

Respuesta1

La respuesta es:

man pam_umask
usergroups
    If the user is not root, and the user ID is equal to the group ID, and the
    username is the same as primary group name, the umask group bits are set to
    be the same as owner bits (examples: 022 -> 002, 077 -> 007). 

Hombre ...

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