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
 }

只是将iPhone 4S设备代码添加到这个线程…

iPhone 4S将返回字符串@"iPhone4,1"。

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");

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
 }

请随意使用这个类(gist @ github)

代码已删除并重新定位到 https://gist.github.com/1323251

更新(01/14/11)

显然，这段代码现在有点过时了，但它肯定可以使用Brian Robbins提供的这个线程上的代码进行更新，其中包括更新模型的类似代码。谢谢你在这个帖子上的支持。

要识别iPhone 4S，只需检查以下内容:

var isIphone4S: Bool {

    let width = UIScreen.main.bounds.size.width
    let height = UIScreen.main.bounds.size.height
    let proportions = width > height ? width / height : height / width

    return proportions == 1.5 && UIDevice.current.model == "iPhone"
}