Question

a function named getTimerInterval returned a NSTimerInterval value

-(NSTimeInterval *)getInterval {
    NSTimeInterval interval = 1;
    return interval;
}

Xcode warn me

Returning ‘NSTimeInterval’ (aka ‘double’) from a function with
incompatible result type ‘NSTimeInterval *’ (aka ‘double *’); take the
address with &” on the line “return interval”

well, NSInteger type is fine:

-(NSIntefer)getInteger {
    NSIntefer result = 1;
    return result;
}

as well as return number directly:

 -(NSTimeInterval *)getInterval {
     return 1;
 }

OK, after that calling getInterval function got a warning:

[NSTimer scheduledTimerWithTimeInterval:[self getInterval] target:self selector:@selector(timerFired:) userInfo:nil repeats:YES];

Sending ‘NSTimeInterval *’ (aka ‘double *’) to parameter of
incompatible type ‘NSTimeInterval’ (aka ‘double’); dereference with *

I am confused with & and *, where should I have to use & or *?
what’s that meaning actually?
I just know it’s c++ syntax about pointer or something, thank in advanced.

Answer 1

In order to fix the compiler warning, you’d need to return the address of it, like this:

-(NSTimeInterval *)getInterval {
    NSTimeInterval interval = 1;
    return &interval;
}

However that is bad as you will return the address on the method’s stack frame which disappears after return. Why not just return it like this:

-(NSTimeInterval)getInterval {
    NSTimeInterval interval = 1;
    return interval;
}

Which I believe will also fix your other compiler error.