Dalam aplikasi saya, pengalaman pengguna lebih baik jika lebih banyak memori digunakan, jadi saya harus memutuskan apakah saya benar-benar harus membebaskan semua memori yang saya bisa masuk didReceiveMemoryWarning. Berdasarkan jawaban Split dan Jasper Pol, menggunakan maksimum 45% dari total memori perangkat tampaknya merupakan ambang batas yang aman (terima kasih kawan).
Jika seseorang ingin melihat implementasi aktual saya:
#import "mach/mach.h"
- (void)didReceiveMemoryWarning
{
// Remember to call super
[super didReceiveMemoryWarning];
// If we are using more than 45% of the memory, free even important resources,
// because the app might be killed by the OS if we don't
if ([self __getMemoryUsedPer1] > 0.45)
{
// Free important resources here
}
// Free regular unimportant resources always here
}
- (float)__getMemoryUsedPer1
{
struct mach_task_basic_info info;
mach_msg_type_number_t size = sizeof(info);
kern_return_t kerr = task_info(mach_task_self(), MACH_TASK_BASIC_INFO, (task_info_t)&info, &size);
if (kerr == KERN_SUCCESS)
{
float used_bytes = info.resident_size;
float total_bytes = [NSProcessInfo processInfo].physicalMemory;
//NSLog(@"Used: %f MB out of %f MB (%f%%)", used_bytes / 1024.0f / 1024.0f, total_bytes / 1024.0f / 1024.0f, used_bytes * 100.0f / total_bytes);
return used_bytes / total_bytes;
}
return 1;
}
Swift (berdasarkan jawaban ini ):
func __getMemoryUsedPer1() -> Float
{
let MACH_TASK_BASIC_INFO_COUNT = (sizeof(mach_task_basic_info_data_t) / sizeof(natural_t))
let name = mach_task_self_
let flavor = task_flavor_t(MACH_TASK_BASIC_INFO)
var size = mach_msg_type_number_t(MACH_TASK_BASIC_INFO_COUNT)
var infoPointer = UnsafeMutablePointer<mach_task_basic_info>.alloc(1)
let kerr = task_info(name, flavor, UnsafeMutablePointer(infoPointer), &size)
let info = infoPointer.move()
infoPointer.dealloc(1)
if kerr == KERN_SUCCESS
{
var used_bytes: Float = Float(info.resident_size)
var total_bytes: Float = Float(NSProcessInfo.processInfo().physicalMemory)
println("Used: \(used_bytes / 1024.0 / 1024.0) MB out of \(total_bytes / 1024.0 / 1024.0) MB (\(used_bytes * 100.0 / total_bytes)%%)")
return used_bytes / total_bytes
}
return 1
}