“表达不可分配” - 问题分配float作为xCode中的另外两个浮点数的总和？ [英] "Expression is not assignable" -- Problem assigning float as sum of two other floats in xCode?

问题描述

在钢琴应用程序中，我正在分配黑键的坐标。
这里是导致错误的代码行。

blackKey和whiteKey都是customViews

  blackKey.center.x =（whiteKey.frame.origin.x + whiteKey.frame.size.width）; 
  

解决方案

其他答案并不完全解释发生了什么在这里，所以这是基本的问题：

当你写 blackKey.center.x 时， blackKey.center 和 center.x 都看起来像结构成员访问，但它们实际上是完全不同的东西。 blackKey.center 是一个属性访问，它解析成类似于 [blackKey center] 的东西，像 objc_msgSend（blackKey，@selector（center））。你不能修改一个函数的返回值，比如 objc_msgSend（blackKey，@selector（center））。x = 2 - 只是没有意义，因为返回值不是存储在任何有意义的地方。

因此如果你想修改结构，你必须存储属性的返回值在一个变量中，修改变量，然后将该属性设置为新的值。

In a piano app, I'm assigning the coordinates of the black keys. Here is the line of code causing the error.

'blackKey' and 'whiteKey' are both customViews

blackKey.center.x = (whiteKey.frame.origin.x + whiteKey.frame.size.width);

解决方案

The other answers don't exactly explain what's going on here, so this is the basic problem:

When you write blackKey.center.x, the blackKey.center and center.x both look like struct member accesses, but they're actually completely different things. blackKey.center is a property access, which desugars to something like [blackKey center], which in turn desugars to something like objc_msgSend(blackKey, @selector(center)). You can't modify the return value of a function, like objc_msgSend(blackKey, @selector(center)).x = 2 — it just isn't meaningful, because the return value isn't stored anywhere meaningful.

So if you want to modify the struct, you have to store the return value of the property in a variable, modify the variable, and then set the property to the new value.

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