我试图从 Swift 调用这个 C 函数并出现编译错误:// headerCURLcode curl_easy_setopt_callback(CURL *curl, CURLoption option, size_t (*param)(char*, size_t, size_t, vo...

我尝试从 Swift 调用此 C 函数并收到编译错误：

// header
CURLcode curl_easy_setopt_callback(CURL *curl, CURLoption option, size_t (*param)(char*, size_t, size_t, void*));

// c file
CURLcode curl_easy_setopt_callback(CURL *curl, CURLoption option, size_t (*param)(char*, size_t, size_t, void*)) {
    return curl_easy_setopt(curl, option, param);
}

错误信息：

Type of expression is ambiguous without a type annotation

使用此代码：

typealias Callback = @convention(c) (
    UnsafeMutableRawPointer?,
    Int32,
    Int32,
    UnsafeMutableRawPointer?
) -> Int32

func testConnection() {
    let curlHandle = curl_easy_init()
    
    // https://support.google.com/accounts/answer/185833?hl=en
    curl_easy_setopt_string(curlHandle, CURLOPT_USERNAME, "***@gmail.com")
    curl_easy_setopt_string(curlHandle, CURLOPT_PASSWORD, "***")
    
    curl_easy_setopt_string(curlHandle, CURLOPT_URL,
                         "imaps://imap.gmail.com:993//%5BGmail%5DSent%20Sent%20Mail");
    
    let callback: Callback = { ptr, size, nmemb, userdata in
        return 0
    }
    
    // Error on the next line: Type of expression is ambiguous without a type annotation
    curl_easy_setopt_callback(curlHandle, CURLOPT_WRITEFUNCTION, callback); 
    
    print("curl_easy_perform")
    let res = curl_easy_perform(curlHandle);
 
    if(res != CURLE_OK) {
        print("curl_easy_perform failed\(res)")
    } else {
        print("success!")
    }
    
    curl_easy_cleanup(curlHandle);
}

不确定我做错了什么。

我包装了 libcurl 的 curl_easy_setopt，因为 Swift 不喜欢可变参数函数。如果有办法摆脱这个包装器 - 我很乐意这么做。

我使用的是 Xcode 15.4，目标是 iOS 17.5。静态 libcurl 的版本是 8.10.1