信号量可用于访问同一库的两个不同应用程序

Question

i have two different applications accessing the same code in the shared library in Linux.

Case1:- when first application access the code in the library, it acquires lock. But when the second application access the same code, how does it know that lock is already applied because another separate copy of code will be created for the second application.

Answer 1

Semaphore is a kernel variable.

A useful way to think of a semaphore is as a record of how many units of a particular resource are available, coupled with operations to safely (ie, without race conditions) adjust that record as units are required or become free, and, if necessary, wait until a unit of the resource becomes available link

so when you lock your semaphore its value will be 0(zero) which shows now its unavailable. So when other processes try to access that shared region due to lock the shared region will not be available. And when semaphore is unlocked the value will be 1(one) which means now its available.

since its a kernel variable we use it for synchronization

There are no methods for accessing the value of semaphore we rely on Semaphore's invariant to define its behavior.

On Unix-based systems, the code segment (.text) may be shared among multiple processes because it's immutable. The code for the shared library is mapped into memory by the operating system.

Basically, each shared library that contains static data (such as global variables) has a Global Offset Table GOT . On shared libraries, all references to static data (global) occur via. So even if the code segment is shared among multiple processes, each process has its exclusive mapping of other segments of the shared library, including the respective Global Offset Table, whose entries are relocated accordingly. The complexity is resolved by CPU's memory management unit.

You can go through this link Dynamic Linking and Loading memory management