我有什么方法可以获得iPhone的整体CPU使用率。我见过一些应用程序,如电池医生和iPhone的系统活动监视器,显示整体CPU使用率。
我找到了一个解决方案,(link to the answer)但它只给了我应用的CPU使用率而不是所有整体应用。
答案 0 :(得分:5)
拜托,试试这个,这对你来说可能是一个好的开始。 仅在真实设备上测试过。
processor_info_array_t _cpuInfo, _prevCPUInfo = nil;
mach_msg_type_number_t _numCPUInfo, _numPrevCPUInfo = 0;
unsigned _numCPUs;
NSLock *_cpuUsageLock;
int _mib[2U] = { CTL_HW, HW_NCPU };
size_t _sizeOfNumCPUs = sizeof(_numCPUs);
int _status = sysctl(_mib, 2U, &_numCPUs, &_sizeOfNumCPUs, NULL, 0U);
if(_status)
_numCPUs = 1;
_cpuUsageLock = [[NSLock alloc] init];
natural_t _numCPUsU = 0U;
kern_return_t err = host_processor_info(mach_host_self(), PROCESSOR_CPU_LOAD_INFO, &_numCPUsU, &_cpuInfo, &_numCPUInfo);
if(err == KERN_SUCCESS) {
[_cpuUsageLock lock];
for(unsigned i = 0U; i < _numCPUs; ++i) {
Float32 _inUse, _total;
if(_prevCPUInfo) {
_inUse = (
(_cpuInfo[(CPU_STATE_MAX * i) + CPU_STATE_USER] - _prevCPUInfo[(CPU_STATE_MAX * i) + CPU_STATE_USER])
+ (_cpuInfo[(CPU_STATE_MAX * i) + CPU_STATE_SYSTEM] - _prevCPUInfo[(CPU_STATE_MAX * i) + CPU_STATE_SYSTEM])
+ (_cpuInfo[(CPU_STATE_MAX * i) + CPU_STATE_NICE] - _prevCPUInfo[(CPU_STATE_MAX * i) + CPU_STATE_NICE])
);
_total = _inUse + (_cpuInfo[(CPU_STATE_MAX * i) + CPU_STATE_IDLE] - _prevCPUInfo[(CPU_STATE_MAX * i) + CPU_STATE_IDLE]);
} else {
_inUse = _cpuInfo[(CPU_STATE_MAX * i) + CPU_STATE_USER] + _cpuInfo[(CPU_STATE_MAX * i) + CPU_STATE_SYSTEM] + _cpuInfo[(CPU_STATE_MAX * i) + CPU_STATE_NICE];
_total = _inUse + _cpuInfo[(CPU_STATE_MAX * i) + CPU_STATE_IDLE];
}
NSLog(@"Core : %u, Usage: %.2f%%", i, _inUse / _total * 100.f);
}
[_cpuUsageLock unlock];
if(_prevCPUInfo) {
size_t prevCpuInfoSize = sizeof(integer_t) * _numPrevCPUInfo;
vm_deallocate(mach_task_self(), (vm_address_t)_prevCPUInfo, prevCpuInfoSize);
}
_prevCPUInfo = _cpuInfo;
_numPrevCPUInfo = _numCPUInfo;
_cpuInfo = nil;
_numCPUInfo = 0U;
} else {
NSLog(@"Error!");
}
您还需要以下标题:
#import <sys/sysctl.h>
#import <sys/types.h>
#import <sys/param.h>
#import <sys/mount.h>
#import <mach/mach.h>
#import <mach/processor_info.h>
#import <mach/mach_host.h>