Swift error: '&' used with non-inout argument of type 'UnsafeMutablePointer'

Question

我正在尝试从这个

转换以下Objective-C代码（source）

-(CGRect) dimensionsForAttributedString: (NSAttributedString *) asp {

    CGFloat ascent = 0, descent = 0, width = 0;
    CTLineRef line = CTLineCreateWithAttributedString( (CFAttributedStringRef) asp);

    width = CTLineGetTypographicBounds( line, &ascent, &descent, NULL );

    // ...
}

进入Swift:

func dimensionsForAttributedString(asp: NSAttributedString) -> CGRect {

    let ascent: CGFloat = 0
    let descent: CGFloat = 0
    var width: CGFloat = 0
    let line: CTLineRef = CTLineCreateWithAttributedString(asp)

    width = CTLineGetTypographicBounds(line, &ascent, &descent, nil)

    // ...
}

但是我在这一行中收到 &ascent 的错误：

width = CTLineGetTypographicBounds(line, &ascent, &descent, nil)

'&' used with non-inout argument of type 'UnsafeMutablePointer'

Xcode 建议我通过删除 & 来修复它。但是，当我这样做时，我得到了错误

Cannot convert value of type 'CGFloat' to expected argument type 'UnsafeMutablePointer'

Interacting with C APIs documentation 使用 & 语法，所以我看不出问题是什么。我该如何解决这个错误？

Answer 1

ascent和descent必须是变量才能被传递作为输入输出参数 &:

var ascent: CGFloat = 0
var descent: CGFloat = 0
let line: CTLineRef = CTLineCreateWithAttributedString(asp)
let width = CGFloat(CTLineGetTypographicBounds(line, &ascent, &descent, nil))

从 CTLineGetTypographicBounds() 开始的 return，这些变量将被设置为线的上升和下降。还要注意这个函数 returns Double，因此您需要将其转换为 CGFloat。

Swift error: '&' used with non-inout argument of type 'UnsafeMutablePointer'

Swift error: '&' used with non-inout argument of type 'UnsafeMutablePointer'

objective-c

unsafe-pointers

ios

swift