快速链接错误:为Module.Class输入元数据访问器

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问题描述

在 Swift 项目中工作时，我有两个模块，比如

While working in Swift project, I have two modules, let's say

基本
功能

Base 有一个 SHService 类，从中我可以从其中调用 Feature 模块(类是 SHCommon ).功能模块的构建没有错误，但是 Base 在链接阶段提出了错误.

Base has a SHService class from where I am calling a function of Feature module (of class SHCommon). Feature module is building without error but Base raises error in linking phase.

import Foundation
import Feature

class SHService {

    public func printMe(printString: String){
        SHCommon().printMe(printString: printString)
    }
}

Feature.SHCommon

import Foundation

public class SHCommon {

    public init(){}
    public func printMe(printString: String) {
        print(printString)
    }
}

链接错误:

知道为什么会这样吗?

推荐答案

查看您的屏幕快照，我试图复制此设置并遇到相同的问题(我不得不说错误消息有点含糊).因此，问题在于您在工作空间中有两个iOS应用程序项目.而且，尽管iOS应用是一个快速模块，但无法将一个iOS应用导入另一个应用中.但是，您可以做的是将您的 Features 功能转换为框架，然后将该框架导入到 Base 应用程序中.或将 SHCommon 类提取到同时导入 Feature 和 Base 的框架中.

Looking at your screenshot I tried to replicate this setup and got the same problem (I have to say that error message is a bit cryptic). So the problem is that you have two iOS app projects inside workspace. And while iOS app is a swift module, it is impossible to import one iOS app inside another one. What you can do, though, is to convert your Feature into framework, and then import that framework into Base app. Or extract SHCommon class into framework that both Feature and Base will import.

More about frameworks: https://developer.apple.com/library/archive/documentation/MacOSX/Conceptual/BPFrameworks/Tasks/CreatingFrameworks.html

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