在我的应用程序中，如果使用更多的内存，则用户体验会更好，因此，我必须确定是否真的应该释放所有可以使用的内存didReceiveMemoryWarning。根据Split和Jasper Pol的回答，最多使用总设备内存的45％似乎是一个安全的门槛（谢谢大家）。
如果有人想看看我的实际实现：
#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（基于此答案）：
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
}