无法将类型 'Void' 的值转换为预期的参数类型“(String) -> Void”

Question

我正在编写应用程序时出现此错误： Cannot convert value of type 'Void' to expected argument type '(String) -> Void'

函数为：

func updateSymbolsPadding(symbolType: String, symbolSize:inout Double) -> Void {
        var size: Double
        if symbolType=="multiply" {
            size = 2.4
        } else if symbolType=="circle" {
            size = 2.0
        } else if symbolType=="triangle" {
            size = 2.0
        } else if symbolType=="square" {
            size = 2.0
        } else if symbolType=="pentagon" {
            size = 2.0
        } else if symbolType=="octagon" {
            size = 2.0
        }
        symbolSize = size
    }

然后我在选择器中调用了它：

Picker("symbol one", selection: $symbolOne, content: {
                        Image(systemName: "multiply")
                            .tag("multiply")
                        Image(systemName: "circle")
                            .tag("circle")
                        Image(systemName: "triangle")
                            .tag("triangle")
                        Image(systemName: "square")
                            .tag("square")
                        Image(systemName: "pentagon")
                            .tag("pentagon")
                        Image(systemName: "octagon")
                            .tag("octagon")
                    })
                    .onChange(of: symbolOne, perform: updateSymbolsPadding(symbolType: symbolOne, symbolSize: &symbolOneSize))

变量是：

@Binding var symbolOne: String
@Binding var symbolOneSize: Double

我该如何解决这个问题？

Answer 1

正如 Ptit Xav 所说：执行参数必须是闭包而不是函数调用的结果。

替换

            .onChange(of: symbolOne, perform: updateSymbolsPadding(symbolType: symbolOne, symbolSize: &symbolOneSize))

有

    .onChange(of: symbolOne) { symbolOne in
        updateSymbolsPadding(symbolType: symbolOne, symbolSize: &symbolOneSize)
    }

无法将类型 'Void' 的值转换为预期的参数类型“(String) -> Void”

Cannot convert value of type 'Void' to expected argument type '(String) -> Void'

ios

swift

swiftui