Why I can't assign struct returned by a function to struct?

Question

I wrote "lightweight" time library and I have struct and typedef like this:

struct tmt {
    uint16_t year;
    uint8_t month;
    uint8_t day;
    uint8_t hour;
    uint8_t minute;
    uint8_t second;
    uint8_t weekday;
    uint8_t is_dst; 
};

typedef struct tmt tm_t;

I have a function that returns tm_t:

tm_t rtc_get_current_time(void){
    tm_t tm;
    xSemaphoreTake(rtc_mutex, RTC_MUTEX_MAX_WAIT);
    tm = rtc.current_time;
    xSemaphoreGive(rtc_mutex);
    return tm;
}

And I want to use it like this:

tm_t s;
s = rtc_get_current_time();    // error is here

I'm getting this error:

incompatible types when assigning to type 'tm_t' {aka 'struct tmt'} from type 'int'

---

I have also tried to change function and variable like this:

struct tmt rtc_get_current_time(void){
    tm_t tm;
    xSemaphoreTake(rtc_mutex, RTC_MUTEX_MAX_WAIT);
    tm = rtc.current_time;
    xSemaphoreGive(rtc_mutex);
    return tm;
}

struct tmt tm = rtc_get_current_time();

What I am doing wrong?

Answer 1

The compiler is complaining that you are trying to assign an int to a tm_t, but since you implemented rtc_get_current_time() as returning a tm_t, this implies that the compiler thinks rtc_get_current_time() is returning an int instead of a tm_t.

That can happen if rtc_get_current_time() has not been declared yet in the context where it is being called. The C language allows a function to be called without a prior declaration, but then you end up with the compiler assuming a default declaration for the function, and that default uses int as the return type. You don't want that in this case.

You need to declare rtc_get_current_time() before you call it.