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iphone_os_version_max_allowed> = 90000在Xcode 6.3.1中返回false

[英]iphone_os_version_max_allowed >= 90000 returns false in Xcode 6.3.1

I am developing an application using Google drive sdk, in Xcode 5 its working fine, but now I open it in xcode6 and write this code for Safariviewcontroller 我正在使用Google驱动器sdk开发应用程序,在Xcode 5中可以正常工作,但是现在我在xcode6中打开它,并为Safariviewcontroller编写此代码

NSURL *URL = [NSURL URLWithString:@"https://accounts.google.com/SignUp"];
#if __IPHONE_OS_VERSION_MAX_ALLOWED >= 90000
if (URL) {

    if ([SFSafariViewController class] != nil) {
        SFSafariViewController *sfvc = [[SFSafariViewController alloc] initWithURL:URL];
        sfvc.delegate = self;
        [self presentViewController:sfvc animated:YES completion:nil];
    } else {
        if (![[UIApplication sharedApplication] openURL:URL]) {
            NSLog(@"%@%@",@"Failed to open url:",[URL description]);
        }
    }
}
#else
 [[UIApplication sharedApplication] openURL:URL];
#endif

but problem is in Xcode6.3.1 __IPHONE_OS_VERSION_MAX_ALLOWED is __IPHONE_8_3 and it always make it false. 但问题出在Xcode6.3.1 __IPHONE_OS_VERSION_MAX_ALLOWED__IPHONE_8_3 ,并且始终使它为假。 I tested in my iphone5c with iOS9.2 but it always return False and got to the else part. 我在iOS9.2的iOS9.2进行了测试,但始终返回False并转到else部分。 please help me... 请帮我...

Thanks in advance 提前致谢

See here for a discussion on __IPHONE_OS_VERSION_MAX_ALLOWED 请参阅此处以获取有关__IPHONE_OS_VERSION_MAX_ALLOWED的讨论

Basically, as you mentioned, this is a question for Xcode, which is answered by it at compile time, not for your device. 基本上,正如您提到的,这是Xcode的问题,它是在编译时回答的,而不是针对您的设备的。 You should be instead doing a check one of the ways implemented here . 相反,您应该检查此处实现的方法之一。

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