|USELIB(2)||Linux Programmer's Manual||USELIB(2)|
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 format.
On success, zero is returned. On error, -1 is returned, and errno is set to indicate the error.
- 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 reached.
- The file specified by library is not an executable of a known type; for example, it does not have the correct magic numbers.
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).
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 the binary.
Since Linux 3.15, this system call is available only when the kernel is configured with the CONFIG_USELIB option.
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