- (BOOL)deviceiPhoneOriPod
  {
    NSString *deviceType = [UIDevice currentDevice].model;
    if([deviceType rangeOfString:@"iPhone"].location!=NSNotFound)
      return YES;
    else
      return NO;
  }

更有用

#include <sys/types.h>
#include <sys/sysctl.h>

@interface UIDevice(Hardware)

- (NSString *) platform;

- (BOOL)hasRetinaDisplay;

- (BOOL)hasMultitasking;

- (BOOL)hasCamera;

@end

@implementation UIDevice(Hardware)

- (NSString *) platform{
    int mib[2];
size_t len;
char *machine;

mib[0] = CTL_HW;
mib[1] = HW_MACHINE;
sysctl(mib, 2, NULL, &len, NULL, 0);
machine = malloc(len);
sysctl(mib, 2, machine, &len, NULL, 0);

    NSString *platform = [NSString stringWithCString:machine encoding:NSASCIIStringEncoding];
    free(machine);
return platform;
}

- (BOOL)hasRetinaDisplay {
    NSString *platform = [self platform];
    BOOL ret = YES;
    if ([platform isEqualToString:@"iPhone1,1"]) {
        ret = NO;
    }
    else
        if ([platform isEqualToString:@"iPhone1,2"])    ret = NO;
    else 
        if ([platform isEqualToString:@"iPhone2,1"])    ret = NO;
    else 
        if ([platform isEqualToString:@"iPod1,1"])      ret = NO;
    else
        if ([platform isEqualToString:@"iPod2,1"])      ret = NO;
    else
        if ([platform isEqualToString:@"iPod3,1"])      ret = NO;
    return ret;
}

- (BOOL)hasMultitasking {
    if ([self respondsToSelector:@selector(isMultitaskingSupported)]) {
        return [self isMultitaskingSupported];
    }
    return NO;
}

- (BOOL)hasCamera {
   BOOL ret = NO;
   // check camera availability
   return ret;
}

@end

您可以读取属性

NSLog(@"platform %@, retita %@, multitasking %@", [[UIDevice currentDevice] platform], [[UIDevice currentDevice] hasRetinaDisplay] ? @"YES" : @"NO" , [[UIDevice currentDevice] hasMultitasking] ? @"YES" : @"NO");

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");

                }
}

以下是新模型的小更新:

- (NSString *) platformString{
    NSString *platform = [self platform];
    if ([platform isEqualToString:@"iPhone1,1"]) return @"iPhone 1G";
    if ([platform isEqualToString:@"iPhone1,2"]) return @"iPhone 3G";
    if ([platform isEqualToString:@"iPhone2,1"]) return @"iPhone 3GS";
    if ([platform isEqualToString:@"iPhone3,1"]) return @"iPhone 4";
    if ([platform isEqualToString:@"iPod1,1"])   return @"iPod Touch 1G";
    if ([platform isEqualToString:@"iPod2,1"])   return @"iPod Touch 2G";
    if ([platform isEqualToString:@"iPod3,1"])   return @"iPod Touch 3G";
    if ([platform isEqualToString:@"i386"])   return @"iPhone Simulator";
    return platform;
}

你可以像这样使用UIDevice类:

NSString *deviceType = [UIDevice currentDevice].model;

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

下面提到的代码片段应该有帮助:

 if ([[UIDevice currentDevice] userInterfaceIdiom] == UIUserInterfaceIdiomPhone) {
   // iPhone device
 }
 else if ([[UIDevice currentDevice] userInterfaceIdiom] == UIUserInterfaceIdiomPad) {
   // iPad device
 }
 else {
  // Other device i.e. iPod
 }