“'CGFloat'不能转换为'Double'” SWIFT中的错误（iOS） [英] "'CGFloat' is not convertible to 'Double'" error in SWIFT (iOS)

问题描述

我试图在SWIFT中将图像剪切成9个部分。我收到此错误'CGFloat'不能转换为'Double'。当我在两个变量中加上i或j时出现此错误。

Im trying to cut an image into 9 pieces in SWIFT. Im getting this error "'CGFloat' is not convertible to 'Double'". I get this error when I put "i" or "j" in the two variables.

以下是用于剪切图像的代码的一部分。

Below is part of the code used for cutting the image.

for i in 1...3
    {
        for j in 1...3
        {
            var intWidth = ( i * (sizeOfImage.width/3.0))
            var fltHeight = ( j * (sizeOfImage.height/3.0))
        var portion = CGRectMake(intWidth,fltHeight, sizeOfImage.width/3.0, sizeOfImage.height/3.0);
            .
            .
      "Code goes on"

无法确定问题是什么。请帮忙。提前致谢。

Couldn't really figure out what the problem is. Please help. Thanks in advance.

-Riyazul Aboobucker

-Riyazul Aboobucker

推荐答案

在swift中没有隐式转换数字数据类型，您正在混合整数和浮点数。
您必须将索引显式转换为 CGFloat s：

In swift there is no implicit conversion of numeric data types, and you are mixing integers and floats. You have to explicitly convert the indexes to CGFloats:

var intWidth = ( CGFloat(i) * (sizeOfImage.width/3.0))
var fltHeight = ( CGFloat(j) * (sizeOfImage.height/3.0))

正如swift中经常发生的那样，错误信息具有误导性 - 它说：

As often happening in swift, the error message is misleading - it says:

'CGFloat'不能转换为'Double'

'CGFloat' is not convertible to 'Double'

而我希望：

'CGFloat'不能转换为'Int'

'CGFloat' is not convertible to 'Int'

这篇关于“'CGFloat'不能转换为'Double'” SWIFT中的错误（iOS）的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！