当我使用这段代码时，出现错误“函数‘sysctlbyname’的隐式声明在 C99 中无效”。我以前用过它很好，所以我不知道会发生什么。

我目前只为 iOS7 设备开发并使用应用商店中最新版本的 xcode 5

UIDeviceHardware.m

@interface UIDeviceHardware : NSObject

- (NSString *) platform;
- (NSString *) platformString;

@end

UIDeviceHardware.m

- (NSString *) platform{
    size_t size;
    sysctlbyname("hw.machine", NULL, &size, NULL, 0);
    char *machine = malloc(size);
    sysctlbyname("hw.machine", machine, &size, NULL, 0);
    NSString *platform = [NSString stringWithUTF8String:machine];
    free(machine);
    return platform;
}

- (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 (GSM)";
    if ([platform isEqualToString:@"iPhone3,3"])    return @"iPhone 4 (CDMA)";
    if ([platform isEqualToString:@"iPhone4,1"])    return @"iPhone 4S";
    if ([platform isEqualToString:@"iPhone5,1"])    return @"iPhone 5 (GSM)";
    if ([platform isEqualToString:@"iPhone5,2"])    return @"iPhone 5 (GSM+CDMA)";
    if ([platform isEqualToString:@"iPhone5,3"])    return @"iPhone 5c (GSM)";
    if ([platform isEqualToString:@"iPhone5,4"])    return @"iPhone 5c (GSM+CDMA)";
    if ([platform isEqualToString:@"iPhone6,1"])    return @"iPhone 5s (GSM)";
    if ([platform isEqualToString:@"iPhone6,2"])    return @"iPhone 5s (GSM+CDMA)";
    if ([platform isEqualToString:@"iPhone7,1"])    return @"iPhone 6 GSM";
    if ([platform isEqualToString:@"iPhone7,2"])    return @"iPhone 6 GSM+CDMA";
    if ([platform isEqualToString:@"iPhone8,1"])    return @"iPhone 6+ GSM";
    if ([platform isEqualToString:@"iPhone8,2"])    return @"iPhone 6+ Global";
    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:@"iPod4,1"])      return @"iPod Touch 4G";
    if ([platform isEqualToString:@"iPod5,1"])      return @"iPod Touch 5G";
    if ([platform isEqualToString:@"iPad1,1"])      return @"iPad";
    if ([platform isEqualToString:@"iPad2,1"])      return @"iPad 2 (WiFi)";
    if ([platform isEqualToString:@"iPad2,2"])      return @"iPad 2 (GSM)";
    if ([platform isEqualToString:@"iPad2,3"])      return @"iPad 2 (CDMA)";
    if ([platform isEqualToString:@"iPad2,4"])      return @"iPad 2 (WiFi)";
    if ([platform isEqualToString:@"iPad2,5"])      return @"iPad Mini (WiFi)";
    if ([platform isEqualToString:@"iPad2,6"])      return @"iPad Mini (GSM)";
    if ([platform isEqualToString:@"iPad2,7"])      return @"iPad Mini (GSM+CDMA)";
    if ([platform isEqualToString:@"iPad3,1"])      return @"iPad 3 (WiFi)";
    if ([platform isEqualToString:@"iPad3,2"])      return @"iPad 3 (GSM+CDMA)";
    if ([platform isEqualToString:@"iPad3,3"])      return @"iPad 3 (GSM)";
    if ([platform isEqualToString:@"iPad3,4"])      return @"iPad 4 (WiFi)";
    if ([platform isEqualToString:@"iPad3,5"])      return @"iPad 4 (GSM)";
    if ([platform isEqualToString:@"iPad3,6"])      return @"iPad 4 (GSM+CDMA)";
    if ([platform isEqualToString:@"iPad4,1"])      return @"iPad Air (WiFi)";
    if ([platform isEqualToString:@"iPad4,2"])      return @"iPad Air (GSM)";
    if ([platform isEqualToString:@"iPad4,4"])      return @"iPad Mini Retina (WiFi)";
    if ([platform isEqualToString:@"iPad4,5"])      return @"iPad Mini Retina (GSM)";
    if ([platform isEqualToString:@"i386"])         return @"Simulator";
    if ([platform isEqualToString:@"x86_64"])       return @"Simulator";
    return platform;
}

最佳答案

也许我的回答很简单，但是，你包括了sys/sysctl.h吗？ ？

#import <sys/sysctl.h>

关于ios - sysctlbyname错误函数 'sysctlbyname'的隐式声明在C99中无效，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/20578741/