判断当前iOS设备的方法

1.在程序中涉及到全屏的宽高的时候，初始化的时候，代码中不要写320和480

#define Screen_height [[UIScreen mainScreen] bounds].size.height 
#define Screen_width [[UIScreen mainScreen] bounds].size.width  
2.使用宏定义方法判断

#define iPhone5 ([UIScreen instancesRespondToSelector:@selector(currentMode)] ? CGSizeEqualToSize(CGSizeMake(640, 1136), [[UIScreen mainScreen] currentMode].size) : NO) 
if（iPhone5）{ 
/*代码操作*/ 
}else{ 
/*代码操作*/ 
} 
if([ [ UIDevice currentDevice ] userInterfaceIdiom ] == UIUserInterfaceIdiomPhone ){ 
//IPhone设备 
}else{ 
//ipad设备 
}  
网上还有一种方法：

+ (NSString*)deviceString  
{  
// 需要#import "sys/utsname.h"  
struct utsname systemInfo;  
uname(&systemInfo);  
NSString *deviceString = [NSString stringWithCString:systemInfo.machine encoding:NSUTF8StringEncoding];  
if ([deviceString isEqualToString:@"iPhone1,1"]) return @"iPhone 1G";  
if ([deviceString isEqualToString:@"iPhone1,2"]) return @"iPhone 3G";  
if ([deviceString isEqualToString:@"iPhone2,1"]) return @"iPhone 3GS";  
if ([deviceString isEqualToString:@"iPhone3,1"]) return @"iPhone 4";  
if ([deviceString isEqualToString:@"iPhone4,1"]) return @"iPhone 4S";  
if ([deviceString isEqualToString:@"iPhone5,2"]) return @"iPhone 5";  
if ([deviceString isEqualToString:@"iPhone3,2"]) return @"Verizon iPhone 4";  
if ([deviceString isEqualToString:@"iPod1,1"]) return @"iPod Touch 1G";  
if ([deviceString isEqualToString:@"iPod2,1"]) return @"iPod Touch 2G";  
if ([deviceString isEqualToString:@"iPod3,1"]) return @"iPod Touch 3G";  
if ([deviceString isEqualToString:@"iPod4,1"]) return @"iPod Touch 4G";  
if ([deviceString isEqualToString:@"iPad1,1"]) return @"iPad";  
if ([deviceString isEqualToString:@"iPad2,1"]) return @"iPad 2 (WiFi)";  
if ([deviceString isEqualToString:@"iPad2,2"]) return @"iPad 2 (GSM)";  
if ([deviceString isEqualToString:@"iPad2,3"]) return @"iPad 2 (CDMA)";  
if ([deviceString isEqualToString:@"i386"]) return @"Simulator";  
if ([deviceString isEqualToString:@"x86_64"]) return @"Simulator";  
NSLog(@"NOTE: Unknown device type: %@", deviceString);  
return deviceString;  
}