Calculate the distance between two CGPoints

问题

i need to calculate the distance between two CGPoints. I refered this and this, but I don't get it.

回答1:

Well, with stuff your refering too where is the full code:

CGPoint p2; //[1]
CGPoint p1;
//Assign the coord of p2 and p1...
//End Assign...
CGFloat xDist = (p2.x - p1.x); //[2]
CGFloat yDist = (p2.y - p1.y); //[3]
CGFloat distance = sqrt((xDist * xDist) + (yDist * yDist)); //[4]

The distance is the variable distance.

What is going on here:

So first off we make two points...
Then we find the distance between x coordinates of the points.
Now we find the distance between the y coordinates.
These lengths are two sides of the triangle, infact they are the legs, time to find the hypotenuse which means after doing some math to rearragne c^2 = a^2 + b^2 we get the hypotenuse to equal sqrt((xDist^2) + (yDist^2)). xDist^2 = (xDist * xDist). And likewise: yDist^2 = (yDist * yDist)

You can't really make a CGPoint be the distance, distance doesn't have an x and y component. It is just 1 number.

If you think CGPoint is a unit of measurement (for example feet is a unit of measurement) it is not.

回答2:

I've had to do this by hand 10,000 times so I wrote a function for it and stuck it in my personal library that I always dump in at the beginning of a new program so I forget it's not cannon.

- (float)distanceBetween:(CGPoint)p1 and:(CGPoint)p2
{
    return sqrt(pow(p2.x-p1.x,2)+pow(p2.y-p1.y,2));
}

so you call it like this (say you want to know how far you moved your finger):

float moveDistance = [self distanceBetween:touchStart and:touchEnd];

This is useful in movement functions as well for spot checking in a scrolling menu:

if([self distanceBetween:touchStart and:touchAt] > 20*scalePoints)
    isItATap = FALSE;

Set "isItATap" true in touchesBegan, put the above in touchesMoved, then you know the player moved their finger too far for the touch to be a "tap", so you can have it NOT select the object the player touched and instead scroll the object around.

As for scale, that should be based on whether or not you have retina display and what size of a device you're on (divide by 2 for retina display since a physical distance of 5 "points" on a regular screen as the user's finger feels it will come up as 10 "pixels" on a retina display screen, since each point is 4 pixels, so you'll wind up with a situation where the player has a very hard time tapping on retina display (which is a common oversight)

回答3:

Sounds like you probably want the vector from p1 to p2 (or difference) rather than the distance.

const CGPoint p1 = {10, 10};
const CGPoint p2 = {510, 310};

const CGPoint diff = {p2.x - p1.x, p2.y - p1.y} // == (CGPoint){500, 300}

回答4:

Short Answer

CGPoint p1, p2; // Having two points
CGFloat distance = hypotf((p1.x-p2.x), (p1.y-p2.y));

Longer Explination

If you have two points p1 and p2 it is obviously easy to find the difference between their height and width (e.g. ABS(p1.x - p2.x)) but to find a true representation of their distance you really want the hypothenuse (H below).

 p1
  |\  
  | \  
  |  \ H
  |   \
  |    \
  |_ _ _\
         p2

Thankfully there is a built in macro for this: hypotf (or hypot for doubles):

// Returns the hypothenuse (the distance between p1 & p2)
CGFloat dist = hypotf((p1.x-p2.x), (p1.y-p2.y));

(original reference)

回答5:

In Swift, you can add an extension to CGPoint:

extension CGPoint {
    func distance(to point: CGPoint) -> CGFloat {
        return sqrt(pow((point.x - x), 2) + pow((point.y - y), 2))
    }
}

and use it like this:

let distance = p1.distance(to: p2)

回答6:

only this...

    float distance = ccpLength(ccpSub(p1,p2));

where p1 and p2 are objects of CGPoint

回答7:

Swift 4, Swift 3 solution

 extension CGPoint {
        static func distanceBetween(point p1: CGPoint,
                                    andPoint p2: CGPoint) -> CGFloat {
            return sqrt(pow((p2.x - p1.x), 2) + pow((p2.y - p1.y), 2))
        }
    }

来源：https://stackoverflow.com/questions/6416101/calculate-the-distance-between-two-cgpoints

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