Hi! I have some doubts on semahore use for lock a resource (EEPROM). I try to define the semaphore, let’s say in eeprom.c and initialize it: static xSemaphoreHandle xSemaphoreEEPROM; char eepromInit(void) {     vSemaphoreCreateBinary( xSemaphoreEEPROM );     if( xSemaphoreEEPROM != NULL ) return 0;     return 1; } Then, can I have two files, each one with one different task that can test the semaphore xSemaphoreEEPROM before addressing the EEPROM, like:     …     if( xSemaphoreTake( xSemaphoreEEPROM, ( portTickType ) timeout ) )     {         /* Call function to write data on eeprom… */         xSemaphoreGive( xSemaphoreEEPROM );     }     … This can lock the resource, or the last code must be included in eeprom.c like a driver: char writeEEPROM (int addr, int n, char *data, int timeout) {     if( xSemaphoreTake( xSemaphoreEEPROM, ( portTickType ) timeout ) )     {         /* Write data… */         xSemaphoreGive( xSemaphoreEEPROM );     } } Please give me some hint about this.

What you say seems to be correct.   You are using the semaphore to ensure mutual exclusion of the eeprom resource.

I don’t have experience in these kinf of OS’s but there will eventually two task accessing the same function: char writeEEPROM (int addr, int n, char *data, int timeout) { if( xSemaphoreTake( xSemaphoreEEPROM, ( portTickType ) timeout ) ) { /* Write data… */ xSemaphoreGive( xSemaphoreEEPROM ); } } There is any problem when two tasks call the same function with a lock semaphore like the previous one? Or there is a better way to do this?

Nope, this is exactly how semaphores should be used.