Take the 2-minute tour ×
Super User is a question and answer site for computer enthusiasts and power users. It's 100% free, no registration required.

Debian Lenny. For every user, including root:

# cat /proc/sys/fs/file-max

# sysctl fs.file-max
fs.file-max = 262144

# ulimit -Hn

# ulimit -Sn

File /etc/security/limits.conf has no uncommented lines.

Where does it get that 1024?

share|improve this question
add comment

1 Answer

up vote 5 down vote accepted

The fs.file-max sysctl shows how many file handles can be allocated system-wide, while ulimit resource limits are per-process (or per-UID). The former is described in Documentation/sysctl/fs.txt:90:

file-max & file-nr:

The value in file-max denotes the maximum number of file-
handles that the Linux kernel will allocate. When you get lots
of error messages about running out of file handles, you might
want to increase this limit.

The 1024 files rlimit is not explicitly set anywhere; it's hardcoded into the kernel as the default value for pid 1, at include/asm-generic/resource.h:81:

 * boot-time rlimit defaults for the init task:
#define INIT_RLIMITS                                                    \
{                                                                       \
        [RLIMIT_NOFILE]         = {   INR_OPEN_CUR,   INR_OPEN_MAX },   \

which references INR_OPEN_CUR and INR_OPEN_MAX from include/linux/fs.h:26:

#define INR_OPEN_CUR 1024       /* Initial setting for nfile rlimits */
#define INR_OPEN_MAX 4096       /* Hard limit for nfile rlimits */

Other processes simply inherit the limit from init (or whatever is pid 1).

Why does /proc/1/limits on Debian report 1024 as both soft and hard nfile limit? I don't know: neither the sysvinit sources nor Debian kernel patches change it. It could be the initramfs scripts, maybe. (I run Arch, which has the 1024/4096 default.)

share|improve this answer
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.