increment variable value in function parameter - ios

Question

Is it possible to increment a variable's value by passing this variable to some function parameter, so that variable's value changes e.g.
int y=0;
void someFunc(int a);
somefunc(y+50);
NSLog(#"y=#f",y);
prints "y = 50"
You can do it this way
someFunc(y=y+50);
someFunc(y=+50);
but is there more elegant way accomplishing this, using pointers maybe? e.g
someFunc(*y+50);
This is the snippet of my code where I want to accomplish this
-(void)drawRect:(CGRect)rect
{
CGFloat currentX = 0;
CGFloat currentY = height / 2;
CGContextRef ctx = UIGraphicsGetCurrentContext();
CGContextSetLineWidth(ctx, 5);
CGContextSetStrokeColorWithColor(ctx, [UIColor blackColor].CGColor);
CGContextSetFillColorWithColor(ctx, [UIColor redColor].CGColor);
CGContextMoveToPoint(ctx,currentX, currentY);
CGContextAddLineToPoint(ctx, currentX=currentX+50, currentY - 200);
CGContextAddLineToPoint(ctx, currentX=+50, currentY - 200);
CGContextAddLineToPoint(ctx, currentX=+50, currentY - 200);
CGContextAddLineToPoint(ctx, currentX=+50, currentY - 200);
CGContextDrawPath(ctx, kCGPathStroke);
}
I want to keep that currentX's value being updated everytime I increment it in the function parameter. Get it?

Answer 1

You can do this by passing a pointer to the variable. This is called an "out parameter" or "call by reference":
void someFunc( int * a, int o )
{
…
*a = *a + o;
}
int y = 0;
someFunc( &y, 50 );
// y is y+o

Answer 2

Either you need to change the value before calling the function, in which case you do:
x = x + something;
func(x);
or if you need to change the value after calling the function, you do like this:
func(x);
x = x + something;
Common sense.
NOTE: never write code like func(x=something) with the assignment inside the parameter list. Because in case you get a habit of doing so, you will soon be writing order-of-evaluation bugs such as func(x, x=something) // BAD, unspecified behavior.
Alternatively, if you need to change the value inside the function, you could pass its address:
void func (int* param)
{
*param = something;
}
...
func(&x);
But you can obviosuly only do this if the function is written by you. This doesn't seem to be the case here, since those functions are apparently part of some Apple library.