Xcode 6.3中的“打开设置”警告问题：“UIApplicationOpenSettingsURLString”的地址与“空指针”的地址比较始终为true [英] Open Settings warning issue in Xcode 6.3: Comparison of address of 'UIApplicationOpenSettingsURLString' not equal to a null pointer is always true

问题描述

我不是在发明轮子。在iOS8中，要从应用程序内部打开设置我正在使用此代码：

I'm not inventing the wheel. In iOS8, to open Settings from inside the app I'm using this code:

BOOL canOpenSettings = (&UIApplicationOpenSettingsURLString != NULL);

if (canOpenSettings)
{
    NSURL *url = [NSURL URLWithString:UIApplicationOpenSettingsURLString];
    [[UIApplication sharedApplication] openURL:url];
}

代码在stackoverflow中有很多答案和问题。

The code is in a lot of answers and questions in stackoverflow.

问题出现在Xcode 6.3上，我有一个警告说：

The problem came out with Xcode 6.3, I've got a warning saying:

比较'UIApplicationOpenSettingsURLString'的地址不等于空指针总是为真

有趣的是Apple在他们的示例代码中使用它：

https://developer.apple.com/library/ ios / samplecode / AppPrefs / Listings / RootViewController_m.html

What is interesting is that Apple is using it in their example code:
https://developer.apple.com/library/ios/samplecode/AppPrefs/Listings/RootViewController_m.html

关于如何避免警告并仍然检查我是否可以打开设置的一些想法？

Some idea about how to avoid the warning and still checking if I can open Settings?

推荐答案

已解决：

问题是相关的使用应用程序中的部署目标。

The problem is related with the Deployment Target in the App.

如果是Targ et是8.0或更高版本，比较将始终为真，因为您总是超过8.0。所以我们不需要if验证：

If the Target is 8.0 or above, the comparison will be always true because you are always over 8.0. So we do not need the if verification:

NSURL *url = [NSURL URLWithString:UIApplicationOpenSettingsURLString];
[[UIApplication sharedApplication] openURL:url];

另一种选择可以是：

NSURL *settings = [NSURL URLWithString:UIApplicationOpenSettingsURLString];
if ([[UIApplication sharedApplication] canOpenURL:settings])
{
    [[UIApplication sharedApplication] openURL:settings];
}

这篇关于Xcode 6.3中的“打开设置”警告问题：“UIApplicationOpenSettingsURLString”的地址与“空指针”的地址比较始终为true的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！