objective-c 如何在Objective C中形成CGPoint数组

How to form CGPoint array in Objective C

提问by Mladen

I want to get this structure

我想得到这个结构

CGPoint addLines1[] =
{
    CGPointMake(30.0, 150.0),
    CGPointMake(41.67, 145.19),
    CGPointMake(53.33, 103.25),
    CGPointMake(65.0, 131.67),
    CGPointMake(76.67, 106.11),
    CGPointMake(88.33, 110.20),
    CGPointMake(100.0, 111.54),
    CGPointMake(111.67, 112.13),
    CGPointMake(123.33, 115.66),
    CGPointMake(135.0, 123.7),
    CGPointMake(146.67, 125.53),
    CGPointMake(158.33, 115.1),
    CGPointMake(170.0, 69.38),
    CGPointMake(181.67, 112.47),
    CGPointMake(193.33, 65.1),
    CGPointMake(205.0, 103.33),
    CGPointMake(216.67, 92.6),
    CGPointMake(228.33, 54.76),
    CGPointMake(240.0, 79.66),
    CGPointMake(251.67, 53.81),
    CGPointMake(263.33, 56.81),
    CGPointMake(275.0, 88.19),
    CGPointMake(286.67, 74.81),
    CGPointMake(298.33, 28.1),
    CGPointMake(310, 20.0),
};

In order to make some calculations and draw data.

为了进行一些计算和绘制数据。

I have CGPoint *lines = appDelegate.averageResponseTimePoints;

我有 CGPoint *lines = appDelegate.averageResponseTimePoints;

How to make array addLines[]from *lines?

如何使阵列addLines[]从*lines？

回答by Tibidabo

For iOS:

对于 iOS：

Create array:

创建数组：

NSArray *myCGPointArray = @[[NSValue valueWithCGPoint:CGPointMake(30.0, 150.0)],[NSValue valueWithCGPoint:CGPointMake(41.67, 145.19)]];

Get 1st CGPoint object:

获取第一个 CGPoint 对象：

CGPoint myPoint = [myCGPointArray[0] CGPointValue];

回答by Will Harris

C arrays are not really arrays at run time, where they are just pointers to a contiguous block of objects of the same type. When you see items[n], that's just syntactic sugar for *(items+n).

C 数组在运行时并不是真正的数组，它们只是指向相同类型的连续对象块的指针。当您看到时items[n]，这只是*(items+n).

In your example addLines[1]would be *(lines+1)and addLines[0]would be *(lines+0), which is *lines. So, addLinesis just lineswithout the pointer dereference. *linesis the first item in the array and linesis the whole array.

在您的示例addLines[1]中将是*(lines+1)并且addLines[0]将是*(lines+0)，即*lines. 所以，addLines只是lines没有指针取消引用。*lines是数组中的第一项并且lines是整个数组。

Arrays have some differences to pointers at compile time. For example, sizeof(addLines)would give you the size of the whole array.

数组在编译时与指针有一些不同。例如，sizeof(addLines)会给你整个数组的大小。

Array-ness is lost as soon as you pass the array somewhere where it's size might be variable, but you can still use the subscript operator. For example:

一旦您将数组传递到其大小可能可变的地方，数组性就会丢失，但您仍然可以使用下标运算符。例如：

#include <Foundation/Foundation.h>

#define showsize( expr ) ( printf(#expr " = %zd\n", ( expr ) ) )

CGPoint *
pass_back(CGPoint points[4])
{
    showsize(sizeof(points));
    return points;
}

int
main(void)
{
    CGPoint square[] = {CGPointMake(-1.0,  1.0),
                        CGPointMake( 1.0,  1.0),
                        CGPointMake( 1.0, -1.0),
                        CGPointMake(-1.0, -1.0)};
    CGPoint* returned;
    int i;

    showsize(sizeof(CGPoint));
    showsize(sizeof(CGPoint*));
    showsize(sizeof(square));
    returned = pass_back(square);
    showsize(sizeof(returned));

    for (i = 0; i < 4; ++i) {
        printf("returned[%d] = {%0.1f, %0.1f}\n", i, (float) returned[i].x,
                                                 (float) returned[i].y);
    }

    return 0;
}

This outputs the following on my Mac:

这会在我的 Mac 上输出以下内容：

sizeof(CGPoint) = 8
sizeof(CGPoint*) = 4
sizeof(square) = 32
sizeof(points) = 4
sizeof(returned) = 4
returned[0] = {-1.0, 1.0}
returned[1] = {1.0, 1.0}
returned[2] = {1.0, -1.0}
returned[3] = {-1.0, -1.0}

Here, squareis the size of four CGPoints, but once sent to the pass_backfunction, it's only the size of a pointer, because that's what it is. When the pointer comes back (and named returned) it can still be used like an array.

这里，square是四个CGPoints的大小，但是一旦发送到pass_back函数，它只是一个指针的大小，因为它就是这样。当指针返回（并命名returned）时，它仍然可以像数组一样使用。

Note the magic number 4in the loop. The pointer doesn't know the length of the array it's pointing to.

注意4循环中的幻数。指针不知道它指向的数组的长度。

Arrays cannot be reassigned with the =operator. If you really must populate addLineswith the points from lines, you can do that with something like the following:

不能使用=运算符重新分配数组。如果您确实必须addLines使用 from 中的点进行填充lines，您可以使用以下内容进行操作：

memcpy(addLines, lines, sizeof(CGPoint) * numberOfPoints);

You'll have to get numberOfPointsfrom somewhere, and addLineswill have to be large enough to handle those points. That's okay if the number of points is a constant, but it would be badif the number of points can vary at run time, especially if the points come from the outside world (think arbitrary code execution).

您必须numberOfPoints从某个地方出发，并且addLines必须足够大才能处理这些点。如果点数是常数，那没关系，但是如果点数在运行时会发生变化，特别是如果点来自外部世界（想想任意代码执行），那就不好了。

I'd change averageResponseTimePointsto return an NSArray rather than a C-style array. You'll need to encapsulate the CGPoints in objects - either your own object or NSValues.

我会更改averageResponseTimePoints为返回 NSArray 而不是 C 样式的数组。您需要将CGPoints封装在对象中 - 无论是您自己的对象还是NSValues。

Here's an example of how you could write averageResponseTimePoints:

这是一个如何编写的示例averageResponseTimePoints：

- (NSArray*) averageResponseTimePoints
{
    NSMutableArray* result = [[[NSMutableArray alloc] init] autorelease];

    for (int i = 0; i < numberOfPoints; ++i) {
        NSValue* point = [NSValue value:points+i
                           withObjCType:@encode(CGPoint)];
        [result addObject:point];
    }

    return result;
}

If your code runs with CocoaTouch, you can use this to create the point value instead:

如果您的代码使用 CocoaTouch 运行，您可以使用它来创建点值：

NSValue* point = [NSValue valueWithCGPoint:points[i]];

To get the CGPoints out of the array, you could write something like this:

要从CGPoint数组中取出 s，您可以这样写：

for (NSValue* value in result) {
    NSPoint pt;
    [value getValue:&pt];
    NSLog(@"%f %f", pt.x, pt.y);
}

Or with CocoaTouch:

或者使用 CocoaTouch：

CGPoint pt = [value CGPointValue];

回答by Baruch Kly

Try this

尝试这个

CGPoint A[10];
A[0] = CGPointMake(10.0, 10.0);
A[1] = CGPointMake(10.0, 12.0);
float x = A[1].y;
NSLog(@"%.0f", x);

It works. Good Luck.

有用。祝你好运。

回答by juggler

I tried Tibidabo's answer, and it didn't work for me. It said [myCGPointArray objectAtIndex:0] was of type id, so I couldn't call CGPointValue on it. hm.

我尝试了 Tibidabo 的答案，但对我不起作用。它说 [myCGPointArray objectAtIndex:0] 是 id 类型，所以我不能在它上面调用 CGPointValue。嗯。

this worked however:

然而，这有效：

NSArray *points = [NSArray arrayWithObjects:
                   [NSValue valueWithCGPoint:CGPointMake(5.5, 6.6)],
                   [NSValue valueWithCGPoint:CGPointMake(7.7, 8.8)],
                   nil];

NSValue *val = [points objectAtIndex:0];
CGPoint point = [val CGPointValue];
float X = point.x;
NSLog(@"cgpoint value is: %f", X);

or

或者

NSArray *points = [NSArray arrayWithObjects:
                        [NSValue valueWithCGPoint:CGPointMake(20, 6.6)],
                        [NSValue valueWithCGPoint:CGPointMake(7.7, 8.8)],
                        nil];
float xCoordinate = [[points objectAtIndex:0] CGPointValue].x;

I found it here: How can I add CGPoint objects to an NSArray the easy way?, posted by Jarret Hardie.

我在这里找到了：How can I add CGPoint objects to an NSArray the simple way? ，由 Jarret Hardie 发布。

回答by IgniteCoders

I solve my problem doing the next:

我解决了我的问题做下一个：

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NSMutableArray *myArray = [[NSMutableArray alloc] init];
[myArray addObject:[NSValue valueWithCGPoint:myPoint]];

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Thanks to michael-jensenfor the solution!

感谢michael-jensen的解决方案！