（Swift）Guard语句中的调用函数

Question

我试图在一个保护语句中调用一个名为“nextPage”的函数，但它说“（）”不能转换为“Bool”。要调用此函数，我需要做什么

@IBAction func nextPressed(_ sender: Any) {
    let geoCoder = CLGeocoder()
    geoCoder.geocodeAddressString(address) { (placemarks, error) in
        guard
            let placemark = placemarks?.first,
            let latVar = placemark.location?.coordinate.latitude,
            let lonVar = placemark.location?.coordinate.longitude,
            nextPage() // Error - '()' is not convertible to 'Bool'
            else {
                print("no location found")
                return
        }
    }
}

Answer 1

您应该要么调用返回布尔值的函数，要么不要在Guard谓词语句中执行此类操作，因为它不是调用函数的合适位置。你应该这样做

guard variable != nil else {
    //handle nil case
}

// continue work with variable, it is guaranteed that it’s not nil.

Answer 2

guard 语句用于检查是否满足特定条件。您不能将不返回 true 或 false 的函数放在该语句中。

参考:https://developer . apple . com/library/content/documentation/Swift/Conceptual/Swift _ Programming _ Language/statements . html

我相信你想要完成的是

@IBAction func nextPressed(_ sender: Any) {
        let geoCoder = CLGeocoder()
        geoCoder.geocodeAddressString(address) { (placemarks, error) in
            guard
                let placemark = placemarks?.first,
                let latVar = placemark.location?.coordinate.latitude,
                let lonVar = placemark.location?.coordinate.longitude
                else {
                    print("no location found")
                    return
            }

            // will only get executed of all the above conditions are met
            nextPage() // moved outside the guard statement

        }
}