uselib - load shared library
int uselib(const char *library);
Note: No declaration of this system call is provided in
glibc headers; see NOTES.
The system call uselib() serves to load a shared library to be used by
the calling process. It is given a pathname. The address where to load is
found in the library itself. The library can have any recognized binary
On success, zero is returned. On error, -1 is returned, and errno is set
to indicate the error.
In addition to all of the error codes returned by open(2) and
mmap(2), the following may also be returned:
uselib() is Linux-specific, and should not be used in programs intended
to be portable.
This obsolete system call is not supported by glibc. No declaration is provided
in glibc headers, but, through a quirk of history, glibc versions before 2.23
did export an ABI for this system call. Therefore, in order to employ this
system call, it was sufficient to manually declare the interface in your code;
alternatively, you could invoke the system call using syscall(2).
- The library specified by library does not have read or execute
permission, or the caller does not have search permission for one of the
directories in the path prefix. (See also path_resolution(7).)
- The system-wide limit on the total number of open files has been
- The file specified by library is not an executable of a known type;
for example, it does not have the correct magic numbers.
In ancient libc versions (before glibc 2.0), uselib() was
used to load the shared libraries with names found in an array of names in
Since Linux 3.15, this system call is available only when the
kernel is configured with the CONFIG_USELIB option.
This page is part of release 5.13 of the Linux man-pages project. A
description of the project, information about reporting bugs, and the latest
version of this page, can be found at https://www.kernel.org/doc/man-pages/.