你可以像这样使用UIDevice类:

NSString *deviceType = [UIDevice currentDevice].model;

if([deviceType isEqualToString:@"iPhone"])
    // it's an iPhone

添加到Arash的代码，我不关心我的应用程序使用什么模型，我只想知道什么样的设备，所以，我可以测试如下:

    if (UI_USER_INTERFACE_IDIOM() == UIUserInterfaceIdiomPad)
        {
            NSLog(@"I'm definitely an iPad");
    } else {
    NSString *deviceType = [UIDevice currentDevice].model;
                if([deviceType rangeOfString:@"iPhone"].location!=NSNotFound)
                {
                    NSLog(@"I must be an iPhone");

                } else {
                    NSLog(@"I think I'm an iPod");

                }
}

NSString *deviceType = [UIDevice currentDevice].model;

为了简单的比较，我总是喜欢宏观:

#define IS_IPOD [[UIDevice currentDevice].model containsString:@"iPod"]
#define IS_IPAD (UI_USER_INTERFACE_IDIOM() == UIUserInterfaceIdiomPad)
#define IS_IPHONE (UI_USER_INTERFACE_IDIOM() == UIUserInterfaceIdiomPhone)

它简单易用。

if([UIDevice currentDevice].userInterfaceIdiom==UIUserInterfaceIdiomPad) {
    //Device is ipad 
}else{
    //Device is iphone
}

你可以像这样使用UIDevice类:

NSString *deviceType = [UIDevice currentDevice].model;

if([deviceType isEqualToString:@"iPhone"])
    // it's an iPhone